The Algebra Geometry Lexicon
Maximal Ideals
In the context of a non-zero algebra A over a field K, an element a ∈ A is considered algebraic over K if there exists a non-zero polynomial f ∈ K[x] such that f(a) = 0 An algebra A is classified as algebraic over K if every element within A is algebraic This foundational concept is essential for understanding the properties of affine algebras discussed throughout this book.
Lemma 1.1 (Fields and algebraic algebras) LetAbe an algebra over a field
(a) IfA is an integral domain and algebraic overK, thenA is a field. (b) IfAis a field and is contained in an affineK-domain, thenAis algebraic.
To demonstrate that every element \( a \in A \setminus \{0\} \) is invertible in \( A \), we need to establish that \( K[a] \) is a field We can assume \( A = K[a] \) and consider the kernel \( I \subset K[x] \) of the map \( K[x] \to A \) defined by \( f \mapsto f(a) \) This gives us the isomorphism \( A \cong K[x]/I \) Since \( A \) is an integral domain, \( I \) is a prime ideal, and because \( a \) is algebraic over \( K \), \( I \) is non-zero Given that \( K[x] \) is a principal ideal domain, we can express \( I = (f) \) where \( f \in K[x] \) is irreducible, making \( I \) a maximal ideal Consequently, \( A \cong K[x]/I \) is confirmed to be a field.
(b) By way of contradiction, assume that A has an element a1 which is not algebraic By hypothesis, A is contained in an affine K-domain
Let B be defined as K[a₁, , aₙ], potentially including a₁ in the generating set We can rearrange a₂, , aₙ so that {a₁, , aᵣ} constitutes a maximal K-algebraically independent subset of {a₁, , aₙ} The field of fractions Quot(B) forms a finite field extension over the subfield L = K(a₁, , aᵣ) For any element b in Quot(B), its multiplication induces an L-linear endomorphism of Quot(B) By selecting an L-basis for Quot(B), we define a mapping ϕ: Quot(B) → L^m×m that associates each b in Quot(B) with the representation matrix of this endomorphism Additionally, let g be a non-zero polynomial in K[a₁, , aᵣ] that serves as a common denominator for all matrix entries of ϕ(aᵢ), where i ranges from 1 to n.
So ϕ(ai) ∈ K[a1, , ar, g −1 ] m×m for all i Since ϕ preserves addition and multiplication, we obtain ϕ(B)⊆K[a1, , ar, g −1 ] m×m
The ring K[a1, , ar] is isomorphic to a polynomial ring, confirming its factorial nature Consider a factorization of g, where p1, , pk are the irreducible factors of g contained in K[a1] For any irreducible element p in K[a1], its inverse p^(-1) is in A, a field that includes K[a1] Applying the mapping ϕ to p^(-1) results in a diagonal matrix with entries equal to p^(-1), leading to the existence of a non-negative integer s and an f in K[a1, , ar] such that p^(-1) = g^(-s)f, which implies g^s = p f Due to the irreducibility of p, it must be a K-multiple of one of the pi This implies that every element in K[a1] \ K is divisible by at least one pi; however, none of the pi divides the product Qk i=1 pi + 1, resulting in a contradiction Therefore, all elements of A are algebraic.
The following proposition is an important application of Lemma 1.1 A particularly interesting special case of the proposition is that A ⊆ B is a subalgebra andϕis the inclusion.
Proposition 1.2 (Preimages of maximal ideals) Letϕ:A→B be a homo- morphism of algebras over a fieldK, and letm⊂B be a maximal ideal IfB is finitely generated, then the preimageϕ −1 (m)⊆Ais also a maximal ideal.
Proof The map A → B/m, a 7→ ϕ(a) +m has the kernel ϕ −1 (m) =: n.
The subalgebra A/n is isomorphic to B/m, which is algebraic over K, as established in Lemma 1.1(b) Consequently, A/n is also algebraic over K and qualifies as an integral domain Furthermore, according to Lemma 1.1(a), A/n is a field, indicating that n is a maximal ideal.
Example 1.3 We give a simple example which shows that intersecting a max- imal ideal with a subring does not always produce a maximal ideal Let
A=K[x] be a polynomial ring over a field and letB=K(x) be the rational function field Thenm := {0} ⊂B is a maximal ideal, but A∩m={0} is not maximal inA /
Before drawing a “serious” conclusion from Proposition 1.2 in Proposi- tion1.5, we need a lemma.
Lemma 1.4 Let K be a field and P = (ξ 1 , , ξ n ) ∈ K n a point in K n Then the ideal mP := (x1−ξ1, , xn−ξn)⊆K[x1, , xn] in the polynomial ringK[x1, , xn] is maximal.
Every polynomial \( f \in K[x_1, \ldots, x_n] \) is congruent to \( f(\xi_1, \ldots, \xi_n) \) modulo \( m_P \), indicating that \( m_P \) serves as the kernel of the homomorphism \( \phi: K[x_1, \ldots, x_n] \rightarrow K \), defined by \( f \mapsto f(\xi_1, \ldots, \xi_n) \) Consequently, this leads to the conclusion that \( K[x_1, \ldots, x_n]/m_P \) is isomorphic to \( K \), thereby confirming the result.
In conjunction with Lemma 1.4, the subsequent proposition outlines all maximal ideals within a polynomial ring over an algebraically closed field A field K is defined as algebraically closed if every non-constant polynomial in K[x] possesses at least one root in K.
Proposition 1.5 (Maximal ideals in a polynomial ring) Let K be an alge- braically closed field, and let m ⊂ K[x 1 , , x n ] be a maximal ideal in a polynomial ring over K Then there exists a point P = (ξ1, , ξn) ∈ K n such that m= (x1−ξ1, , xn−ξn).
According to Proposition 1.2, the intersection K[xi] ∩ m is identified as a maximal ideal within K[xi] for each i ranging from 1 to n Given that K[xi] is a principal ideal domain, this intersection can be expressed in the form (pi) K[xi], where pi is an irreducible polynomial Since K is algebraically closed, we can represent (pi) K[xi] as (xi - ξi) K[xi], with ξi belonging to K Consequently, there exist elements ξ1, , ξn in K such that xi - ξi is contained in m Utilizing the notation from Lemma 1.4, it follows that mP is a subset of m, leading to the conclusion that m equals mP as established by Lemma 1.4.
We make a definition before giving a refined version of Proposition1.5. Definition 1.6 LetK[x 1 , , x n ] be a polynomial ring over a field.
(a) For a set S⊆K[x 1 , , x n ] of polynomials, the affine varietygiven by
The indexK n is omitted if no misunderstanding can occur.
(b) A subset X ⊆ K n is called an affine (K-)variety if X is the affine variety given by a setS⊆K[x1, , xn] of polynomials.
In the literature, affine varieties are often considered to be irreducible, and the definition of an affine variety is frequently restricted to cases where the field K is algebraically closed.
Theorem 1.7 (Correspondence points - maximal ideals) LetKbe an alge- braically closed field andS ⊆K[x 1 , , x n ] a set of polynomials Let MS be the set of all maximal ideals m⊂K[x 1 , , x n ] withS⊆m Then the map Φ:V(S)→ M S , (ξ 1 , , ξ n )7→(x 1 −ξ 1 , , x n −ξ n ) is a bijection.
In the proof, we define P as a point in the variety V(S), specifically P := (ξ1, , ξn) According to Lemma 1.4, the ideal Φ(P) is maximal Since all functions f in S yield f(P) = 0, it follows that f belongs to Φ(P), confirming that Φ(P) is an element of the maximal spectrum MS Conversely, for any maximal ideal m in MS, Proposition 1.5 states that m can be expressed as (x1 − ξ1, , xn − ξn) for some point (ξ1, , ξn) in K^n The inclusion S ⊆ m indicates that (ξ1, , ξn) belongs to V(S), thereby demonstrating that the mapping Φ is surjective.
To show injectivity, let P = (ξ1, , ξn) and Q = (η1, , ηn) be points in V(S) with Φ(P) = Φ(Q) =: m For each i, we have xi −ξi ∈ m and also x i −η i ∈ m, so ξ i −η i ∈ m This implies ξ i = η i , since otherwise m=K[x 1 , , x n ] ut
Corollary 1.8 (Hilbert’s Nullstellensatz, first version) Let K be an alge- braically closed field and let I$K[x 1 , , x n ]be a proper ideal in a polyno- mial ring Then
To demonstrate the existence of a maximal ideal containing a proper ideal I in the ring K[x1, , xn], we consider the set of all proper ideals J that include I By applying Zorn's lemma, we establish that this set possesses a maximal element, denoted as m Alternatively, we could invoke the Noetherian property of K[x1, , xn] to reach the same conclusion; however, the axiom of choice, which is equivalent to Zorn's lemma, would still be necessary to avoid any shortcuts in the proof Consequently, we find that m is a maximal ideal with I contained within it, leading us to conclude that V(I) is non-empty.
Remark (a) To see that the hypothesis thatKbe algebraically closed can- not be omitted from Corollary 1.8, consider the example K = R and
(b) Hilbert’s Nullstellensatz is really a theorem about systems of polynomial equations Indeed, letf1, , fm∈K[x1, , xn] be polynomials If there exist polynomialsg1, , gm∈K[x1, , xn] such that m
If the sum of the functions \( g_i \) equals one, the system of equations \( f_i(\xi_1, \ldots, \xi_n) = 0 \) for \( i = 1, \ldots, m \) has no solutions The existence of \( g_1, \ldots, g_m \) that satisfy this condition is equivalent to the functions \( (f_1, \ldots, f_m) \) being contained within the polynomial ring \( K[x_1, \ldots, x_n] \) Hilbert's Nullstellensatz states that if this condition does not present an obstacle and \( K \) is an algebraically closed field, the system of equations is indeed solvable Therefore, for algebraically closed fields, the only barrier to solving systems of polynomial equations is the aforementioned condition In Chapter 9, we will explore how to algorithmically determine the existence of this obstacle.
Jacobson Rings
The main goal of this section is to prove the second version of Hilbert’s Nullstellensatz (Theorem 1.17) We start by defining the spectrum and the maximal spectrum of a ring.
(a) Thespectrumof R is the set of all prime ideals inR:
Spec(R) :={P ⊂R|P is a prime ideal}. (b) Themaximal spectrumof Ris the set of all maximal ideals in R:
Spec max (R) :={P⊂R|P is a maximal ideal}. (c) We also define the Rabinovich spectrumof R as the set
In this section, we introduce the concept of Spec rab (R), defined as Spec rab (R) := {R ∩ m | m ∈ Spec max (R[x])}, where R[x] denotes the polynomial ring over R This definition is specific to our discussion and is not commonly found in standard literature.
The concept of employing an additional indeterminate to demonstrate the second version of Hilbert’s Nullstellensatz originates from J L Rabinovich and is commonly known as Rabinovich’s trick This inspired my student Martin Kohls to propose naming the set defined in Definition 1.9(c) the Rabinovich spectrum.
Spec max (R)⊆Spec rab (R)⊆Spec(R).
In a ring extension S of R, if P is a prime ideal in S, then the intersection of R and P, denoted as R ∩ P, forms a prime ideal in R The proof for the first inclusion can be found in Exercise 1.3, but the focus here will be solely on the significance of the second inclusion.
1.2 Jacobson Rings 23 this book Exercise1.4gives an example where both inclusions are strict The importance of the Rabinovich spectrum is highlighted by Proposition1.11. Recall that for an idealI⊆Rin a ringR, theradical idealofIis defined as √
I:f ∈R| there exists a positive integerk withf k ∈I
I=I For example, a non-zero ideal (a)⊆Z is radical if and only if a is square-free Recall that every prime ideal is a radical ideal.
Lemma 1.10 LetR be a ring,I⊆Ran ideal, andM ⊆Spec(R)a subset.
If there exist noP ∈ Mwith I⊆P, the intersection is to be interpreted as R.
I, soa k ∈I for somek LetP ∈ MI Thena k ∈P Since
P is a prime ideal, it follows thata∈P ut
Proposition 1.11 (The raison d’ˆetre of the Rabinovich spectrum) Let R be a ring and I⊆Ran ideal Then
If there exist noP ∈Spec rab (R)with I⊆P, the intersection is to be inter- preted asR.
Proof The inclusion “⊆” follows from Lemma 1.10 and the fact that Spec rab (R)⊆Spec(R).
To prove the reverse inclusion, letabe in the intersection of allP ∈ MI. Consider the ideal
The ideal J, generated by the set I and the element ax−1, is a subset of R[x] Assuming that J is not equal to R[x], Zorn’s lemma guarantees the existence of a maximal ideal m in Spec(R[x]) such that J is contained within m Since I is also contained in the intersection of R and J, it follows that R ∩ m is in the set of maximal ideals MI Given that a is an element of m and ax−1 is also in m, this leads to the conclusion that m must equal R[x], resulting in a contradiction Therefore, we deduce that J must equal R[x].
X j=1 gjbj+g(ax−1) (1.3) with g, g1, , gn ∈ R[x] and b1, , bn ∈ I Let R[x, x −1 ] be the ring ofLaurent polynomials and consider the mapϕ:R[x]→R[x, x −1 ], f7→f(x −1 ).Applyingϕto both sides of (1.3) and multiplying with somex k yields x k n
Fork≥max{deg(g1), ,deg(gn),deg(g) + 1}, allhjandhlie inR[x], so we may substitutex=ain the above equation and obtain a k n
I This completes the proof ut
We get the following important consequence.
Corollary 1.12 (Intersecting prime ideals) Let R be a ring and I ⊆R an ideal Then √
If there exist noP ∈Spec(R)withI⊆P, the intersection is to be interpreted asR.
Proof This follows from Lemma1.10and Proposition1.11 ut
Theorem 1.13(Intersecting maximal ideals) Let A be an affine algebra andI⊆Aan ideal Then
If there exist nom∈Spec max (A)withI⊆m, the intersection is to be inter- preted asA.
Proof The inclusion “⊆” again follows from Lemma1.10.
LetP ∈Spec rab (A) Then P =A∩m withm∈Spec max (A[x]) But A[x] is finitely generated as an algebra over a field, so by Proposition1.2it follows that P∈Spec max (A) We conclude that
(In fact, equality holds, but we do not need this.) Now the inclusion “⊇” follows from Proposition1.11 ut
We pause here to make a definition, which is inspired by Theorem1.13.Definition 1.14 A ring R is called a Jacobson ring if for every proper idealI$R we have
Theorem 1.13 establishes that every affine algebra qualifies as a Jacobson ring, with the integers Z serving as a prime example This prompts the inquiry into whether the polynomial ring Z[x] also exhibits Jacobson properties, which it does This scenario exemplifies the broader principle that every finitely generated algebra A over a Jacobson ring retains the same characteristics.
A Jacobson ring is defined as a ring in which every maximal ideal is the intersection of all the maximal ideals of its quotient fields Eisenbud provides a proof of this concept, specifically in Theorem 4.19 Additionally, it is noted that if α is the homomorphism that transforms A into an R-algebra, then for every maximal ideal m in the spectrum of A, the preimage α −1 (m) remains a maximal ideal This observation parallels Proposition 1.2.
A typical example of a non-Jacobson ring is the formal power series ring K[[x]] over a fieldK (see Exercise 1.2) A similar example is the ring of all rational numbers with odd denominator.
We can now prove the second version of Hilbert’s Nullstellensatz To for- mulate it, a bit of notation is useful.
Definition 1.15 Let K be a field andX ⊆K n a set of points The(van- ishing) idealof X is defined as
The indexK[x1, , xn]is omitted if no misunderstanding can occur.
Remark 1.16 It is clear from the definition that the ideal of a set of points always is a radical ideal /
Theorem 1.17(Hilbert’s Nullstellensatz, second version) Let K be an al- gebraically closed field and let I⊆K[x1, , xn] be an ideal in a polynomial ring Then
Proof We start by showing the inclusion “⊇”, which does not requireK to be algebraically closed Letf ∈√
V(I) Then f(ξ1, , ξn) k = 0, so f(ξ1, , ξn) = 0 This shows that f ∈
For the reverse inclusion, assume f ∈ I(V(I)) In view of Theorem1.13, we need to show thatf lies in everym∈ MI, where
So letm ∈ MI By Theorem1.7,m= (x1−ξ1, , xn−ξn)K[x 1 , ,x n ] with (ξ1, , ξn)∈ V(I) This impliesf(ξ1, , ξn) = 0, sof ∈m This completes the proof ut
The following corollary is the heart of what we call the Algebra GeometryLexicon We need an (easy) lemma.
Lemma 1.18 Let K be a field andX ⊆K n an affine variety Then
Proof By assumption,X =V(S) withS⊆K[x 1 , , x n ] SoS⊆ I(X), and applyingV yields
Corollary 1.19 (Ideal-variety correspondence) Let K be an algebraically closed field and n a positive integer Then there is a bijection between the sets
A:={I⊆K[x1, , xn]|I is a radical ideal} and
B:={X⊆K n |X is an affine variety}, given by
Both maps reverse inclusions, i.e., for I, J∈ A we have
I⊆J ⇐⇒ V(J)⊆ V(I), and the corresponding statement holds for the inverse map.
If I is a radical ideal within set A, the Nullstellensatz 1.17 indicates that I(V(I)) equals I Conversely, for any element X in set B, I(X) belongs to A, as noted in Remark 1.16, and by Lemma 1.18, V(I(X)) is equal to X This establishes that the mappings are inverses of one another Additionally, it follows that if I is a subset of J, then V(J) is a subset of V(I) for ideals I and J in A, and if X is a subset of Y, then I(Y) is a subset of I(X).
X, Y ∈ B Now applyI andV to get the converse implications ut
Coordinate Rings
The next part of the algebra geometry lexicon is provided by assigning to an affine varietyX an affine algebra, the coordinate ring K[X], which encodes the properties ofX.
Definition 1.20 Let K be a field and X ⊆K n an affine variety LetI :I(X)⊆K[x1, , xn]be the ideal ofX Then the coordinate ringof X is the quotient ring
The coordinate ring is sometimes also called thering of regular functions on X.
Remark 1.21 (a) Every element of the coordinate ring K[X] of an affine variety is a class f +I with f ∈ K[x1, , xn] Such a class yields a well-defined functionX →K, given by (ξ1, , ξn)7→f(ξ1, , ξn), and different classes yield different functions SoK[X] can be identified with an algebra of functions X → K The functions from K[X] are called regular functions They are precisely those functionsX →Kthat are given by polynomials.
(b) If X =V(J) with J ⊆K[x1, , xn] an ideal, then it is not necessarily true thatK[X] =K[x 1 , , x n ]/J However, ifKis algebraically closed, thenK[X] =K[x 1 , , x n ]/√
The following lemma compares ideals in a quotient ring R/I to ideals in
R It is rather boring and elementary, but very important.
Lemma 1.22 (Ideals in quotient rings) Let R be a ring and let I ⊆ R be an ideal Consider the sets
A:={J ⊆R|J is an ideal andI⊆J} and
The map Φ:A → B, J 7→ {a+I|a∈J}=J/I is an inclusion-preserving bijection with inverse map Ψ:B → A, J 7→ {a∈R|a+I∈ J }.
J is a prime ideal ⇐⇒ Φ(J)is a prime ideal and
J is a maximal ideal ⇐⇒ Φ(J)is a maximal ideal hold Moreover, if J = (a1, , an)R with ai ∈ R, then Φ(J) = (a1 +
The maps Φ and Ψ are confirmed to be inclusion-preserving, satisfying the conditions Ψ◦Φ = id A and Φ◦Ψ = id B The isomorphism (1.4) is established as Φ(J) represents the kernel of the epimorphism from R/I to R/J, defined by a + I mapping to a + J Consequently, both equivalences are derived from (1.4), and the final statement is evidently supported.
If X ⊆ K n is an affine variety, then a subvariety is a subset Y ⊆ X which is itself an affine variety in K n We can now prove a correspondence between subvarieties of a variety and radical ideals in the coordinate ring.
Theorem 1.23(Correspondence subvarieties - radical ideals) Let X be an affine variety over an algebraically closed fieldK Then there is an inclusion- reversing bijection between the set of subvarietiesY ⊆Xand the set of radical ideals J ⊆K[X] The bijection is given by mapping a subvariety Y ⊆X to I(Y)/I(X)⊆K[X], and mapping an ideal J ⊆K[X] to
If J ⊆K[X]is the ideal corresponding to a subvariety Y, then
K[Y]∼=K[X]/J, with an isomorphism given byK[X]/J→K[Y],f+J 7→f| Y
Restricting our bijection to subvarieties consisting of one point yields a bijection
Proof All claims are shown by putting Corollary 1.19and Lemma1.22to- gether ut
In Exercise 1.11, we explore the relationship between points and algebraic objects linked to the coordinate ring The subsequent theorem identifies the types of rings that can serve as coordinate rings for affine algebras To properly articulate this theorem, we first require a clear definition.
(a) An element a∈R is callednilpotent if there exists a positive integerk witha k = 0.
(b) The set of all nilpotent elements is called thenilradicalofR, written as nil(R) (So the nilradical is equal to the radical ideal p
{0} of the zero- ideal, which by Corollary1.12 is the intersection of all prime ideals.) (c) Ris calledreducedifnil(R) ={0} (In particular, every integral domain is reduced.)
Theorem 1.25(Coordinate rings and reduced algebras) LetK be a field.
(a) For every affineK-varietyX, the coordinate ringK[X]is a reduced affine K-algebra.
If K is an algebraically closed field and A is a reduced affine K-algebra, then there exists an affine K-variety X such that K[X] is isomorphic to A Specifically, for an ideal I associated with X, K[X] can be expressed as K[x1, , xn]/I, confirming that K[X] is an affine algebra and is reduced due to I being a radical ideal.
(b) Choose generators a 1 , , a n of A Then the epimorphism ϕ: K[x 1 , , x n ]→A, f 7→f(a 1 , , a n ) yields A∼=K[x 1 , , x n ]/I with
I = ker(ϕ) Since A is reduced,I is a radical ideal SetX := V(I) By the Nullstellensatz1.17,I=I(X), so A∼=K[X] ut
The affine variety X described in Theorem 1.25(b) is not uniquely determined, as its definition relies on the chosen generators of the coordinate ring A However, by applying the appropriate concept of isomorphism for varieties, it can be demonstrated that all affine varieties corresponding to the same coordinate ring A are isomorphic This leads to a bijective correspondence between the isomorphism classes of affine K-varieties and those of reduced affine K-algebras.
1.1 (Some counter examples) Give examples which show that none of the hypotheses in Lemma1.1(a) and (b) and in Proposition1.2can be omit- ted (Solution on page227)
1.2 (Formal power series ring) Consider the formal power series ring
(a) Show thatK[[x]] is an integral domain.
(b) Show that all power series f =P∞ i=0aix i witha0 6= 0 are invertible in K[[x]].
(c) Show thatK[[x]] has exactly one maximal idealm, i.e.,K[[x]] is a local ring.
(d) Show thatK[[x]] is not a Jacobson ring.
L:=nX ∞ i=k aix i |k∈Z, ai ∈Ko of formal Laurent series is a field The fieldL of formal Laurent series is often written asK((x)).
(f) IsK[[x]] finitely generated as a K-algebra?
1.3 (Maximal spectrum and Rabinovich spectrum) LetRbe a ring. Show that
*1.4 (Maximal spectrum, Rabinovich spectrum, and spectrum). LetR=K[[y]] be the formal power series ring over a fieldK, and letS=R[z] be a polynomial ring overR Show that
Spec max (S)$Spec rab (S)$Spec(S).
Hint:Consider the ideals (y)S and (z)S.
To determine whether a ring R is a Jacobson ring, it suffices to verify that each prime ideal P in the spectrum of R, Spec(R), can be expressed as an intersection of maximal ideals This criterion streamlines the process of confirming the Jacobson property of the ring For a comprehensive solution, refer to page 228.
1.6 (Zis a Jacobson Rings) Show that the ringZof integers is a Jacob- son ring (Solution on page229)
1.7 (Explicit computations with a variety) Consider the ideal
(b) IsI a prime ideal? IsI a radical ideal?
(c) Does Hilbert’s Nullstellensatz1.17hold forI?
1.8 (Colon ideals) IfI andJ⊆Rare ideals in a ring, the colon idealis defined as
In this exercise we give a geometric interpretation of the colon ideal. (a) SetM:={P ∈Spec(R)|I⊆P andJ 6⊆P} and show that
(b) Let K be a field andX, Y ⊆K n such thatY is an affine variety Show that
1.9 (A generalization of Hilbert’s Nullstellensatz) Let K be a field andKits algebraic closure LetI⊆K[x1, , xn] be an ideal in a polynomial ring Show that
1.10(Order-reversing maps) This exercise puts Corollary 1.19 and its proof in a more general framework Let A 0 and B 0 be two partially ordered sets Let ϕ: A 0 → B 0 and ψ: B 0 → A 0 be maps satisfying the following properties:
SetA:=ψ(B 0 ) andB:=ϕ(B 0 ), and show that the restriction ϕ|A:A → B is a bijection with inverse mapψ|B.
To prove Corollary 1.9, it is essential to establish that all radical ideals in K[x₁, , xₙ] can be represented as vanishing ideals of point sets in Kⁿ, a conclusion derived from Theorem 1.17 This principle also applies to the relationship between subgroups and intermediate fields in Galois theory.
1.11(Points of a variety and homomorphisms) Let K be a field (which need not be algebraically closed) and X a K-variety Construct a bijection betweenX and the set
HomK(K[X], K) :={ϕ:K[X]→K|ϕis an algebra-homomorphism}.
Remark: In the language of affine schemes, an algebra-homomorphism K[X]→K induces a morphism Spec(K)→Spec(K[X]) Such a morphism is called aK-rational point of the affine scheme associated toX.
This chapter explores the theory of Noetherian and Artinian rings, highlighting the relationship between the two properties The Artin property, while formally analogous to the Noether property, is more specialized and implies the Noether property, as demonstrated in Theorem 2.8 Both properties are also examined in the context of modules The focus then shifts to the Noether property, with significant emphasis on Hilbert’s Basis Theorem 2.13 and its implications Notably, the Noether property facilitates elegant yet non-constructive proofs, exemplified by Hilbert’s proof that the rings of invariants of GL n and SL n are finitely generated, which faced criticism from Gordan, who remarked that such an approach is more theology than mathematics.
The Noether and Artin Property for Rings and Modules
Definition 2.1 LetR be a ring and M anR-module.
(a) M is called Noetherian if the submodules of M satisfy the ascend- ing chain condition, i.e., for submodules M1, M2, M3, ⊆ M with
Mi ⊆ Mi+1 for all positive integers i, there exists an integer n such that M i = M n for all i ≥ n In other words, every strictly ascending chain of submodules is finite.
(b) R is called Noetherian if R is Noetherian as a module over itself In other words,Ris Noetherian if the ideals ofRsatisfy the ascending chain condition.
(c) M is calledArtinianif the submodules ofM satisfy thedescending chain condition, i.e., for submodulesM1, M2, M3, ⊆M withMi+1⊆Mi for
1 “This is theology and not mathematics”
33 all positive integers i, there exists an integer n such that M i =M n for alli≥n.
(d) Ris calledArtinian ifR is Artinian as a module over itself, i.e., if the ideals ofR satisfy the descending chain condition.
The ring of integers, denoted as Z, is classified as Noetherian because any ascending chain of ideals corresponds to a sequence of integers where each subsequent integer is a divisor of the preceding one This relationship is supported by the well-ordering principle of natural numbers, which confirms the Noetherian property.
(2) By the same argument, a polynomial ringK[x] over a field is Noetherian. More trivially, every field is Noetherian.
Let X be an infinite set and K a field, or any non-zero ring The collection R, defined as K^X, represents all functions mapping from X to K and forms a ring under point-wise operations Additionally, for any subset Y of X, the set R can be further analyzed.
IY :={f ∈R|f vanishes onY} is an ideal of R Since there are infinite strictly descending chains of subsets ofX, there are also infinite strictly ascending chains of ideals in
(4) The ringsZandK[x] considered above are not Artinian.
(5) Every field and every finite ring or module is Artinian.
(6) The ringK X , as defined in (3), is Artinian if and only ifX is a finite set.
(7) Let R :=K[x] be a polynomial ring over a field Then S := R/(x 2 ) is
Artinian.S is also Artinian as anR-module /
The ring from Example2.2(3) is a rather pathological example of a non- Noetherian ring In particular, it is not an integral domain The following provides a less pathological counter example.
Example 2.3 LetS:=K[x, y] be the polynomial ring in two indeterminates over a fieldK Consider the subalgebra
It is shown in Exercise2.1thatR is not Noetherian / The following proposition shows that the Noether property and the Artin property behave well with submodules and quotient modules.
Proposition 2.4 (Submodules and quotient modules) Let M be a module over a ringR, and letN ⊆M be a submodule Then the following statements are equivalent.
(b) Both N and the quotient moduleM/N are Noetherian.
In particular, every quotient ring of a Noetherian ring is Noetherian.All statements of this proposition hold with “Noetherian” replaced by “Ar- tinian”.
2.1 The Noether and Artin Property for Rings and Modules 35
Proof First assume that M is Noetherian It follows directly from Defini- tion 2.1 that N is Noetherian, too To show that M/N is Noetherian, let
Let \( U_1, U_2, \ldots \subseteq M/N \) be an ascending chain of submodules Define the canonical epimorphism \( \phi: M \to M/N \) and let \( M_i := \phi^{-1}(U_i) \), which creates an ascending chain of submodules in \( M \) According to the hypothesis, there exists an \( n \) such that \( M_i = M_n \) for all \( i \geq n \) Since \( \phi(M_i) = U_i \), it follows that \( U_i = U_n \) for all \( i \geq n \) Thus, we have demonstrated that (a) implies (b).
Now assume that (b) is satisfied To show (a), let M1, M2, ⊆ M be an ascending chain of submodules We obtain an ascending chain ϕ(M1), ϕ(M2), ⊆M/N of submodules of M/N Moreover, the intersections N∩
Mi⊆N yield an ascending chain of submodules ofN By hypothesis, there exists annsuch that fori≥nwe haveϕ(Mi) =ϕ(Mn) andN∩Mi=N∩Mn.
We claim that also Mi =Mn for alli≥n Indeed, let m∈Mi Then there exists anm 0 ∈Mn withϕ(m) =ϕ(m 0 ), so m−m 0 ∈N∩M i =N∩M n ⊆M n
We concludem=m 0 + (m−m 0 )∈M n So the equivalence of (a) and (b) is proved.
To show the statement on quotient rings, observe that the ideals of a quotient ring R/I are precisely the submodules of R/I viewed as an R- module.
To establish the proof for Artinian modules, one can substitute every instance of "ascending" with "descending" in the argument, while also swapping "Mi" and "Mn" in the demonstration of Mi = Mn.
We need the following definition to push the theory further.
Definition 2.5 (Ideal product) Let R be a ring, I ⊆R and ideal, and M an R-module.
(a) The product of I and M is defined to be the abelian group generated by all productsaãmof elements fromI and elements fromM So
An intriguing special case arises when M equals J, where both are ideals of R In this scenario, the product IJ is referred to as the ideal product It is evident that the formation of the ideal product exhibits both commutative and associative properties.
(c) Forn∈N0,I n denotes the product ofn copies ofI, with I 0 :=R.The following lemma gives a connection between ideal powers and radical ideals.
Lemma 2.6 (Ideal powers and radical ideals) LetRbe a ring andI, J⊆R ideals If I is finitely generated, we have the equivalence
Proof We have I = (a1, , an) Suppose that I ⊆ √
J Then there exists m >0 witha m i ∈J fori= 1, , n Setk:=nã(m−1) + 1 We need to show that the product ofkarbitrary elements fromIlies inJ So letx1, , xk ∈I and write xi n
When multiplying the product \(x_1 \cdots x_k\), each summand includes a subproduct \(a_{m_j}\), indicating that \(x_1 \cdots x_k \in J\) This demonstrates that \(I_k \subseteq J\) The converse is evident and does not necessitate finite generation.
Theorem 2.8 establishes a comparison between the Noether property and the Artin property in ring theory For those focused on the Noether property, further insights can be found in Section 2.2 It's important to note that Theorem 2.8 will not be referenced until Chapter 7.
Lemma 2.7 Let R be a ring andm1, ,mn∈Spec max (R)maximal ideals (which are not assumed to be distinct) such that for the ideal product we have m 1 ã ã ãm n ={0}.
ThenR is Artinian if and only if it is Noetherian Moreover,
Ii:=m1ã ã ãmi, we get a chain
In the context of ideals within a ring R, the chain of inclusions \( I_n \subseteq I_{n-1} \subseteq \ldots \subseteq I_2 \subseteq I_1 \subseteq I_0 \) illustrates that R is Noetherian (or Artinian) if and only if every quotient module \( I_{i-1}/I_i \) is also Noetherian (or Artinian) Since the intersection \( m_i \cap (I_{i-1}/I_i) = \{0\} \), it follows that \( I_{i-1}/I_i \) behaves as a vector space over the field.
In the context of R-modules, a subset of I i−1 /I i qualifies as an R-submodule if and only if it functions as a K i -subspace, where K i is defined as R/m i Consequently, the Noether and Artin properties for I i−1 /I i are equivalent to the condition that the dimension of K i (I i−1 /I i) is finite This establishes the desired equivalence.
To prove the second claim, takeP ∈Spec(R) By hypothesis,m1ã ã ãmn⊆
P From the primality ofP and the definition of the ideal product, we con- clude that there existsiwithmi⊆P, soP =mi ut
2.1 The Noether and Artin Property for Rings and Modules 37
Theorem 2.8 (Artinian and Noetherian rings) Let R be a ring Then the following statements are equivalent.
(b) Ris Noetherian and every prime ideal ofR is maximal.
In accordance with Definition 5.1, condition (b) in Theorem 2.8 can be restated as: "R is Noetherian and has a dimension of 0 or -1," with -1 indicating that R is the zero ring This article focuses on proving the implication "(a) ⇒ (b)" and will defer the proof of the converse until the conclusion of Chapter 3 (refer to page 52).
In an Artinian ring \( R \), it can be proven that there are only finitely many maximal ideals Assuming the opposite leads to a contradiction: if there were infinitely many distinct maximal ideals \( m_1, m_2, m_3, \ldots \), then the ideals \( I_i = \bigcap_{j=1}^i m_j \) would form a descending chain By the Artinian property, this chain must stabilize at some \( n \), implying \( I_{n+1} = I_n \) Consequently, \( \bigcap_{j=1}^n m_j \subseteq m_{n+1} \) suggests that one of the \( m_j \) must equal \( m_{n+1} \), contradicting the assumption of distinctness Therefore, it follows that \( R \) contains only finitely many maximal ideals \( m_1, \ldots, m_k \).
I:=m1ã ã ãmk, we obtain a descending chain of idealsI i ,i∈N0, so there existsn∈N0 with
By way of contradiction, assumeJ 6={0} Then the set
The set M, defined as the collection of non-empty ideals J₀ of R such that J₀ is an ideal and J₀ is not equal to {0}, contains a minimal element J_b If J_b did not exist, M would have an infinite, strictly descending chain of ideals By selecting an element x from J_b where xJ is not equal to {0}, we establish that J_b is generated by x, confirming its minimality.
Given the equation (x)J = (x)J² = (x)J⁶ = {0}, we deduce that (x)J = (x) due to the minimality of Jb Consequently, there exists an element y ∈ J such that xy = x According to the definition of J, y is present in every maximal ideal of R, which implies that y⁻¹ is not in any maximal ideal and is therefore invertible The equation (y⁻¹)x = 0 leads to the conclusion that x must be 0, contradicting the condition that xJ ≠ {0} Thus, we arrive at a significant conclusion.
J ={0} So we can apply Lemma2.7and get thatRis Noetherian and that every prime ideal is maximal ut
Theorem 2.8 raises the question whether it is also true that every Ar- tinian module over a ring is Noetherian This is answered in the negative byExercise2.2.
Noetherian Rings and Modules
The following theorem gives an alternative definition of Noetherian modules. There is no analogue for Artinian modules.
Theorem 2.9 (Alternative definition of Noetherian modules) Let R be a ring and M an R-module The following statements are equivalent.
(b) For every subsetS ⊆M there exist finitely many elements m1, , mk ∈
(S) R = (m 1 , , m k ) R (c) Every submodule of M is finitely generated.
In particular, R is Noetherian if and only if every ideal ofR is finitely gen- erated, and then every generating set of an ideal contains a finite generating subset.
In a Noetherian module M, we consider a subset S that fails to meet condition (b) We recursively define finite subsets S_i, beginning with S_1 being empty Assuming S_i has been established, the non-compliance of S with condition (b) allows us to identify an element m_{i+1} in S that is not included in the union of the previous subsets S_i.
The axiom of choice is essential for accurately defining this concept Our construction shows that (Si)R is related to (Si+1)R for every i, which contradicts (a) Therefore, (a) leads to (b), and it is evident that (b) leads to (c).
Assuming condition (c) holds, let M1, M2, , be an ascending chain of submodules within M, and define N as the union of these submodules, N := ∪i∈N Mi It is straightforward to verify that N is indeed a submodule According to condition (c), we can express N as (m1, , mk)R, where each mj is an element of N and belongs to some Mi_j By letting n represent the maximum index among {i1, , ik}, we find that all mj are contained within Mn for indices i ≥ n.
M i ⊆N = (m 1 , , m k ) R ⊆M n ⊆M i , which implies equality Therefore (a) holds ut
Theorem 2.9implies that every Noetherian module over a ring is finitely generated This raises the question whether the converse is true, too But this is clearly false in general: If R is a non-Noetherian ring, then R is not Noetherian as a module over itself, but it is finitely generated (with 1 the only generator) The following theorem shows that if the converse does not go wrong in this very simple way, then in fact it holds.
Theorem 2.10(Noetherian modules and finite generation) Let R be a Noetherian ring and M an R-module Then the following statements are equivalent.
In particular, every submodule of a finitely generatedR-module is also finitely generated.
To prove the implication that (b) leads to (a), we focus on the case where M = (m1, , mk)R, applying induction on k For the base case of k = 0, no proof is needed, so we proceed with the assumption that k > 0 We then examine the submodule to establish the desired relationship.
By induction, we establish that N is Noetherian The surjective homomorphism ϕ: R → M/N, defined by a7 → amk + N, implies that M/N is isomorphic to R/ker(ϕ) Given the hypothesis and Proposition 2.4, R/ker(ϕ) is Noetherian, which leads to the conclusion that M/N is also Noetherian Furthermore, applying Proposition 2.4 once more confirms that M is Noetherian as well.
The following theorem is arguably the most important result on Noethe- rian rings.
Theorem 2.11(Polynomial rings over Noetherian rings) Let R be a Noe- therian ring Then the polynomial ring R[x]is Noetherian, too.
Proof LetI⊆R[x] be an ideal By Theorem2.9, we need to show thatI is finitely generated For a non-zero integeri, set
Clearly Ji ⊆ R is an ideal Let ai ∈ Ji with f = Pi j=0ajx j ∈ I Then
The ideals \( J_i \) in the ring \( R \) form an ascending chain, expressed as \( I_3xf = \sum_{j=0}^{i} a_j x^{j+1} \) for \( i \in J_{i+1} \) According to the hypothesis, there exists an integer \( n \) such that for all \( i \geq n \), the ideals satisfy \( J_i = J_n \) Additionally, it is given that each ideal \( J_i \) is finitely generated.
Ji=Jn= (an,1, , an,m n ) R for i > n (2.3)
By the definition of Ji, there exist polynomials fi,j ∈I of degree at mosti whosei-th coefficient isai,j Set
We assert that I equals I zero To demonstrate this, let us examine a polynomial f belonging to the ideal I, expressed as Pd i=0b i x i, with a degree of d We will employ induction on d, starting with the scenario where d is less than or equal to n Given that b d is an element of J d, we can apply equation (2.2) to represent b d as a sum of terms, specifically b d = Pm d j=1r j a d,j, where r j belongs to R Consequently, we define fe as the difference between f and m d.
X j=1 rjfd,j lies inIand has degree less thand, so by inductionfe∈I 0 This impliesf ∈I 0 Now assumed > n Then we can use (2.3) and write b d =Pm n j=1r j a n,j with r j ∈R So fe:=f− m n
X j=1 rjx d−n fn,j lies inIand has degree less thand, so by inductionfe∈I 0 Again we conclude f ∈I 0 So indeedI=I 0 is a finitely generated ideal ut
The corresponding statement for formal power series rings is contained in Exercise2.4 By applying Theorem2.11repeatedly and using the second statement of Proposition2.4, we obtain
Corollary 2.12 (Finitely generated algebras) Every finitely generated al- gebra over a Noetherian ring is Noetherian In particular, every affine algebra is Noetherian.
A special case is the celebrated Basis Theorem of Hilbert.
Corollary 2.13 (Hilbert’s Basis Theorem) Let K be a field Then K[x1, , xn] is Noetherian In particular, every ideal in K[x1, , xn] is finitely generated.
The name Basis Theorem comes from the fact that generating sets of ideals are sometimes calledbases One consequence is that every affine variety
X ⊆K n is the solution set of afinite system of polynomial equations:X V(f1, , fm).
The polynomial ring \( S = K[x, y] \) and the subalgebra \( R = K + S\hat{x} \) serve as a notable example of a non-Noetherian ring It can be demonstrated that \( R \) is not Noetherian, which leads to the conclusion that \( R \) is not finitely generated as an algebra This example highlights the significant implications of the properties of non-Noetherian rings in algebraic structures.
• Subrings of Noetherian rings need not be Noetherian.
• Subalgebras of finitely generated algebras need not be finitely generated.
In Exercise 7.4, we will demonstrate that Krull's Principal Ideal Theorem fails for the ring R Additionally, in Exercise 2.6, we will investigate whether Example 2.3 represents the smallest instance of its category (Refer to the solution on page 231.)
2.2 (An Artinian module that is not Noetherian) Let p ∈ N be a prime number and consider theZ-modules
Zp:=n a/p n ∈Q|a, n∈Z o⊂Q and M :=Zp/Z. Show thatM is Artinian but not Noetherian (Solution on page232)
2.3 (Modules over an Artinian ring) Show that a finitely generated moduleM over an Artinian ringRis Artinian (Solution on page232)
2.4 (The Noether property for formal power series rings) LetRbe a Noetherian ring and
R[[x]] :=nX ∞ i=0 a i x i |a i ∈Ro the formal power series ring overR Show thatR[[x]] is Noetherian.(Solution on page232)
In the context of a field K and a sub-algebra A of the polynomial algebra K[x1, , xn], a subset S ⊆ A is termed (A-)separating if it can distinguish between different points in K^n Each polynomial in A corresponds to a function from K^n to K, allowing the subset S to effectively separate points based on the values these polynomials produce.
If there exists f ∈ A with f (P 1 ) 6= f (P 2 ), then there exists f ∈ S with f (P 1 ) 6= f (P 2 ).
(Loosely speaking, this means thatS has the same capabilities of separating points asA.)
If a subset S of A generates A as an algebra, it follows that S is a separating set This demonstrates that the condition of being separating is less stringent than that of generating Furthermore, examples in sections (b) and (c) illustrate that the requirement for S to be separating is significantly weaker than the requirement for it to generate A.
*(b) Show thatA has a finite separating subset.
(c) Exhibit a finite R-separating subset of the algebra R ⊂ K[x, y] from Example2.3.
*2.6 (Subalgebras ofK[x]) LetKbe a field and K[x] a polynomial ring in one indeterminate Is every subalgebra ofK[x] finitely generated? Give a proof or a counter example (Solution on page234)
2.7 (Graded rings) A ringR is called graded if it has a direct sum de- composition
(as an Abelian group) such that for all a ∈ R i and b ∈ R j one has ab ∈
R i+j Then an element from R d is called homogeneous of degree d A standard example isR=K[x 1 , , x n ] withR d the space of all homogeneous polynomials of degreed(including the zero-polynomial) LetRbe graded and set
Rd, which obviously is an ideal.I is sometimes called theirrelevant ideal Show the equivalence of the following statements.
(b) R0 is Noetherian andI is finitely generated.
(c) R 0 is Noetherian andRis finitely generated as anR 0 -algebra.
According to Corollary 2.12, a finitely generated algebra over a Noetherian ring is Noetherian However, it is important to note that not all Noetherian algebras are finitely generated In the context of graded rings, there exists a special case where this converse is true, highlighting a unique relationship between these algebraic structures.
In Exercise 2.1, we observed that the Noether property typically does not transfer to subrings However, this exercise explores a specific scenario where the Noether property is preserved in subrings.
(a) LetS be a Noetherian ring andR⊆S a subring such that there exists a homomorphismϕ:S→RofR-modules withϕ| R = id R Show thatR is Noetherian, too.
(b) Show that for a ringR, the following three statements are equivalent: (i)
Ris Noetherian; (ii)R[x] is Noetherian; (iii)R[[x]] is Noetherian. (Solution on page235)
2.9 (Right or wrong?) Decide if the following statements are true or false. Give reasons for your answers.
(a) Every finitely generated module over an Artinian ring is Artinian. (b) Every Artinian module is finitely generated.
(c) Every ring has a module that is both Noetherian and Artinian.
(d) The set of all ideals of a ring, together with the ideal sum and ideal product, forms a commutative semiring (i.e., we have an additive and a multiplicative commutative monoid, and a distributive law).
The ring R consists of all analytic functions mapping from R to R, defined by power series that converge throughout R It can be demonstrated that this ring is not Noetherian, indicating that it does not satisfy the ascending chain condition on ideals.
Can your argument be used for showing that other classes of functions
R→Rform non-Noetherian rings, too? (Solution on page236)
Affine Varieties
In this section we define the Zariski topology onK n and on its subsets We first need a proposition.
Proposition 3.1 (Unions and intersections of affine varieties).
Let K[x1, , xn]be a polynomial ring over a field K.
(a) LetI, J ⊆K[x1, , xn] be ideals Then
(b) LetM be a non-empty set of subsets ofK[x 1 , , x n ] Then
Proof We first prove (a) It is clear thatV(I)∪ V(J)⊆ V(I∩J) To prove the reverse inclusion, let P ∈ V(I∩J) Assume P /∈ V(I), so there exists
43 f ∈ I with f(P)6= 0 We need to show that P ∈ V(J), so let g ∈J Then f g∈I∩J, sof(P)g(P) = 0 But this impliesg(P) = 0.
Proposition 3.1 states that finite unions and arbitrary intersections of affine varieties in \( K^n \) are also affine varieties Given that both \( K^n \) and the empty set are affine varieties, this leads us to define a topology where affine varieties serve as closed sets This approach will be the focus of our discussion.
Definition 3.2 LetK be a field andna positive integer Then theZariski topology is defined on K n by saying that a subset X ⊆ K n is (Zariski-) closed if and only ifX is an affine variety On a subsetY ⊆K n , we define the Zariski topology to be the subset topology induced by the Zariski topology onK n , i.e., the closed subsets in Y are the intersections of closed subsets in
Remark 3.3 (a) By definition, the closed subsets ofK n have the formV(S) withS ⊆K[x 1 , , x n ] a subset By Lemma 1.18, we may substitute S byI(X), i.e., we may assumeS to be an ideal, and in fact even a radical ideal.
(b) For a subset X ⊆ K n , the topological closure (also called the Zariski- closure) is
(c) IfY ⊆K n is an affine variety, then by definition the Zariski topology on
Y has the subvarieties ofY as closed sets.
(d) OnR n andC n , the Zariski topology is coarser than the usual Euclidean topology.
(e) Every finite subset of K n is Zariski-closed In other words, K n is aT1 space This also applies to every subsetY ⊆K n
In the affine line \( K_1 \), the closed subsets consist solely of finite subsets and the entire set \( K_1 \) Consequently, the Zariski topology represents the coarsest topology in which singletons, or one-element sets, are considered closed This highlights the significant coarseness of the Zariski topology when compared to the standard topology.
All polynomials \( f \in K[x_1, \ldots, x_n] \) are continuous functions from \( K^n \) to \( K \) when considered within the Zariski topology This topology is defined as the coarsest topology that ensures the continuity of all polynomials, under the assumption that the set \(\{0\}\) is closed in \( K^1 \).
On the other hand, there exist continuous functionsK n →K which are not polynomials, e.g the functionC→C, x7→x(complex conjugation). (h) The Zariski-open subsets ofK n are unions of solution sets of polynomial inequalities.
(i) Recall that a Hausdorff space (also called a T2 space) is a topological space in which for any two distinct points P1 6= P2 there exist disjoint
3.1 Affine Varieties 45 open sets U 1 and U 2 with P i ∈ U i If K is an infinite field, then K n with the Zariski topology is never Hausdorff In fact, it is not hard to see that two non-empty, open subsetsU 1 , U 2 ⊆K n always intersect This is extended in Exercise3.7, where it is shown that no infinite subset ofK n is Hausdorff /
Further examples of continuous maps are morphisms of varieties, which we deal with now.
Definition 3.4 Let K be a field and let X ⊆ K m and Y ⊆ K n be affine varieties A map f: X → Y is called a morphism (of varieties) if there exist polynomialsf1, , fn∈K[x1, , xm]such that f is given by f(P) = (f1(P), , fn(P)) for P ∈X.
We write Mor(X, Y) for the set of all morphisms X → Y Since composi- tions of morphisms are obviously again morphisms, this definition makes the collection of affineK-varieties into a category.
A morphism f:X →Y is called an isomorphismif there exists a mor- phism g: Y → X with f ◦g = id Y and g◦f = id X In particular, every isomorphism is a homeomorphism (i.e., a topological isomorphism).
In particular, the regular functions on X are precisely the morphisms
Letf: X →Y be a morphism given by polynomials f1, , fn Then we have a homomorphism of K-algebras ϕ: K[Y] →K[X] given as follows: If K[X] =K[x1, , xm]/I(X) andK[Y] =K[y1, , yn]/I(Y), then ϕ(yi+I(Y)) :=fi+I(X).
It is essential to ensure that the definitions are precise The homomorphism ϕ is considered to be induced by the function f By associating coordinate rings with affine varieties and linking induced homomorphisms to morphisms, we establish a contravariant functor that connects the category of affine K-varieties to that of affine K-algebras.
In the context of algebraic structures, we observe a reverse process where an algebra homomorphism ϕ: K[Y] → K[X] leads to polynomials f₁, , fₙ in K[x₁, , xₘ] Here, the relationship ϕ(yi + I(Y)) = fi + I(X) establishes that these polynomials define a morphism f: X → Y, which is independent of the specific choice of fi Furthermore, it can be verified that the correspondence between homomorphisms from K[Y] to K[X] and morphisms from X to Y creates a pair of inverse bijections, represented as Mor(X, Y) ↔ HomK(K[Y], K[X]).
A bijective morphism from X to Y does not guarantee that it is an isomorphism For instance, consider the set X, which is the union of the hyperbola defined by the equation ξ1ξ2=1 and the point (0,1) in K² The first projection function f: X → K¹ serves as a bijective morphism, yet it is not an isomorphism, as illustrated in Figure 3.1.
Figure 3.1 A bijective morphism which is not an isomorphism
Spectra
Theorem 1.23 establishes a one-to-one correspondence between the maximal ideals of K[X] and the points in X, indicating that prime ideals can be viewed as "generalized points." Building on this concept, we will define a topology on Spec(R) for any ring R.
Definition 3.5 LetR be a ring For a subset S⊆R we write
V Spec(R) (S) :={P ∈Spec(R)|S⊆P}. For a subsetX ⊆Spec(R) we write
The Zariski topology on Spec(R) is established by defining closed sets as those of the form VSpec(R)(S), where S is a subset of R According to Proposition 3.6, particularly points (a) and (b), the empty set corresponds to VSpec(R)({1}), while Spec(R) is represented as VSpec(R)(∅) This definition effectively creates a valid topology on Spec(R).
A subset of Spec(R) is equipped with the subspace topology induced from the Zariski topology onSpec(R).
The following proposition contains all the important general facts about the mapsV Spec(R) andI R defined above In particular, part (e) is an analogy to the ideal-variety correspondence in Corollary1.19.
Proposition 3.6 (Properties ofV Spec(R) andIR) Let R be a ring.
.(b) LetM be a non-empty set of subsets ofR Then
(c) LetX ⊆Spec(R) be a subset ThenIR(X)is a radical ideal ofR. (d) LetI⊆R be an ideal Then
(e) We have a pair of inverse bijections between the set of radical ideals ofR and the set of closed subsets ofSpec(R), given byV Spec(R) andIR Both bijections are inclusion-reversing.
Proof (a) If P ∈ V Spec(R) (S), then S ⊆ P, so also (S)R ⊆ P and (S)R∩ (T)R ⊆P The same follows if P ∈ V Spec(R) (T), so in both cases P ∈
(S)R∩(T)R and assume S 6⊆ P So there exists f ∈ S \P Let g ∈ T Then f g ∈ (S)R ∩(T)R, so f g ∈ P Since P is a prime ideal, g ∈ P follows, so
(c) This follows since prime ideals are always radical ideals, and intersections of radical ideals are again radical ideals.
(e) In the light of (c) and (d), we only need to showVSpec(R)(IR(X)) =X forX ⊆Spec(R) a closed subset We haveX =VSpec(R)(S) withS⊆R, soS⊆ IR(X) Since the mapVSpec(R)is inclusion-reversing, we obtain
VSpec(R)(IR(X))⊆ VSpec(R)(S) =X⊆ VSpec(R)(IR(X)).
This completes the proof ut
In Theorem 1.23, we establish a bijection between points of an affine variety and the maximal ideals of its coordinate ring, which is further demonstrated as a homeomorphism in Exercise 3.3 This highlights the notion that prime ideals can be interpreted as generalized points Additionally, it is noteworthy that a ring R qualifies as a Jacobson ring if and only if every closed subset satisfies certain conditions.
Y ⊆ Spec(R) we have that Spec max (R)∩Y is dense in Y (Recall that a subset in a topological space is calleddenseif its closure is the whole space.)
In fact, this is nothing but a translation of the Jacobson property.
In the context of ring theory, each ring R is associated with a topological space known as Spec(R) This relationship can be transformed into a contravariant functor For two rings R and S and a homomorphism ϕ: R→S, the preimage of any prime ideal P in Spec(S) under the mapping ϕ is a prime ideal in R, establishing a function ϕ ∗ : Spec(S)→Spec(R) that maps P to its preimage ϕ −1 (P).
We will often say thatϕ ∗ is induced from ϕ For I ⊆R a subset, we have (ϕ ∗ ) −1 VSpec(R)(I)
= VSpec(S)(ϕ(I)), so ϕ ∗ is continuous Maps between spectra of rings that are induced from ring-homomorphisms are calledmor- phisms.
The transition from ϕ to ϕ ∗ aligns with and expands upon the method of deriving a morphism X → Y of affine varieties from a homomorphism K[Y] → K[X] Notably, this process lacks a reverse path from ϕ ∗ back to ϕ Moreover, distinct ring homomorphisms can produce identical induced maps, even when the rings in question are reduced, as illustrated by the example of complex conjugation.
In algebraic geometry, a morphism between two spectra, or more generally between two ringed spaces, involves a continuous map between the spectra as topological spaces, accompanied by a morphism of sheaves This comprehensive definition of a morphism enables a seamless transition between ring-homomorphisms and morphisms of spectra, addressing previous oversights in assigning ϕ ∗ to ϕ.
Noetherian and Irreducible Spaces
Motivated by the correspondence between ideals and Zariski-closed subsets, we can transport the definition of the Noether property to topological spaces in general.
Definition 3.7 LetX be a topological space.
A topological space X is termed Noetherian if its closed subsets adhere to the descending chain condition, meaning that for any sequence of closed subsets Y1, Y2, Y3, and so forth, where each subset is contained within the previous one, there exists an integer n such that all subsequent subsets are equal to Yn Alternatively, this condition can also be expressed in terms of open subsets, which must satisfy the ascending chain condition.
A set X is considered irreducible if it cannot be expressed as the union of two proper closed subsets and is not empty Additionally, an equivalent condition for irreducibility is that any two non-empty open subsets of X must intersect in a non-empty set, with the stipulation that X itself is not empty.
Example 3.8 (1) R and C with the usual Euclidean topology are neither Noetherian nor irreducible.
(2) Every finite space is Noetherian.
If K is an infinite field, then the Zariski topology on X = K^1 is irreducible, as its only closed subsets are X itself and its finite subsets More broadly, we will demonstrate that K^n is also irreducible, which follows from this property.
The topological spaces that we normally deal with in analysis are almost never Noetherian or irreducible For example, a Hausdorff space can only be
3.3 Noetherian and Irreducible Spaces 49 irreducible if it is a singleton (this is obvious), and it can only be Noetherian if it is finite (see Exercise 3.7) However, the following two theorems show that the situation is much better when we consider spaces with the Zariski topology.
Theorem 3.9 (Noether property of the Zariski topology).
(a) Let K be a field andX ⊆K n a set of points, equipped with the Zariski topology ThenX is Noetherian.
(b) Let R be a Noetherian ring and X ⊆ Spec(R) be a set of prime ideals, equipped with the Zariski topology ThenX is Noetherian.
In a Noetherian topological space X, any subset Y equipped with the subset topology is also Noetherian For part (a), we can assume X = K^n, where a descending chain of closed subsets Y1, Y2, Y3, in K^n leads to an ascending chain of ideals I_i := I_K[x_1, ,x_n](Y_i) According to Hilbert's Basis Theorem, there exists an integer n such that I_i = I_n for all i ≥ n This result is further supported by Lemma 1.18.
Part (b) follows directly from Proposition3.6(e) ut
In particular, Spec(R) is a Noetherian space if R is a Noetherian ring. Exercise3.5deals with the question whether the converse also holds. Theorem 3.10(Irreducible subsets ofK n and Spec(R)).
(a) Let K be a field andX ⊆K n a set of points, equipped with the Zariski topology ThenX is irreducible if and only ifI K[x 1 , ,x n ] (X)is a prime ideal.
(b) Let R be a ring and X ⊆ Spec(R) be a set of prime ideals, equipped with the Zariski topology Then X is irreducible if and only if IR(X) is a prime ideal.
Proof (a) First assume thatX is irreducible Then I :=IK[x 1 , ,x n ](X)$ K[x 1 , , x n ], sinceX 6=∅ To show that I is a prime ideal, letf 1 , f 2 ∈ K[x 1 , , x n ] withf 1 f 2 ∈I Then
X = (X∩ V K n(f 1 ))∪(X∩ V K n(f 2 )), so by the irreducibility of X there exists i ∈ {1,2} with X ⊆ VK n (fi). This impliesfi∈I So indeedI is a prime ideal.
Conversely, assume that I is a prime ideal Then X 6= ∅, since
I K[x 1 , ,x n ] (∅) =K[x1, , xn] To show that X is irreducible, let X X1∪X2withXiclosed inX, soXi=X∩VK n (Ii) withIi⊆K[x1, , xn] ideals Then
X ⊆ VK n (I1)∪ VK n (I2) =VK n (I1∩I2),where we used Proposition3.1(a) for the equality This implies
I 1 ∩I 2 ⊆ IK[x 1 , ,x n ](VK n (I 1 ∩I 2 ))⊆ IK[x 1 , ,x n ](X) =I. SinceIis a prime ideal, there exists iwithI i ⊆I, so
This impliesXi=X ThereforeX is irreducible.
(b) The proof of this part is obtained from the proof of part (a) by changing K[x1, , xn] toR,K n to Spec(R), and “Proposition3.1(a)” to “Propo- sition3.6(a)” ut
The following theorem allows us to view irreducible spaces as the “atoms” of a Noetherian space.
Theorem 3.11(Decomposition into irreducibles) Let X be a Noetherian topological space.
(a) There exist a non-negative integer n and closed, irreducible subsets
X=Z1∪ ã ã ã ∪Zn and Zi6⊆Zj for i6=j (3.1)
(b) If Z 1 , , Z n ⊆ X are closed, irreducible subsets satisfying (3.1), then every irreducible subset Z ⊆X is contained in some Z i (Observe that we do not assumeZ to be closed.)
(c) If Z 1 , , Z n ⊆ X are closed, irreducible subsets satisfying (3.1), then they are precisely the maximal irreducible subsets ofX In particular, the
Zi are uniquely determined up to order.
Every non-empty set of closed subsets in a topological space X possesses a minimal element, preventing the existence of an infinite strictly descending chain If there exists a non-empty closed subset Y within X that cannot be expressed as a finite union of closed, irreducible subsets, we can consider Y to be minimal with this property Since Y is not irreducible, it can be represented as the union of two closed subsets, Y1 and Y2.
Y 1 , Y 2 $Y closed subsets By the minimality ofY, theY i are finite unions of closed irreducible subsets, so the same is true forY This is a contradiction. Hence in particular
X =Y 1 ∪ ã ã ã ∪Y m withY i ⊆X closed and irreducible (wherem= 0 ifX =∅) We may assume theYi to be pairwise distinct By deleting those Yi for which there exists a j6=iwithYi⊆Yj, we obtain a decomposition as in (3.1).
Now assume that (3.1) is satisfied, and letZ⊆Xbe an irreducible subset. Then
The equation Z = (Z ∩ Z1) ∪ ∪ (Z ∩ Zn) indicates that Z is a subset of Zi for some i, demonstrating that Z is contained within the irreducible subsets of X Furthermore, if Z is maximal among these irreducible subsets, it follows that Z must equal Zi, confirming that all maximal irreducible subsets of X are found within the sets Zi This completes the proof of the statement.
3.3 Noetherian and Irreducible Spaces 51 we need to show that every Z i is maximal among the irreducible subsets. Indeed, if Z i ⊆ Z with Z ⊆X irreducible, then by (b) there exists j with
Z ⊆Z j , soZ i ⊆Z ⊆Z j , which by (3.1) impliesi=j and Z=Z i ut
It is important to note that the assumption of X being Noetherian is only necessary for part (a) of the theorem Therefore, even if X is not Noetherian, the existence of a decomposition as described in part (a) ensures that parts (b) and (c) still hold true.
Theorem3.11has statements on existence and uniqueness, which justifies the following definition.
Definition 3.12 LetX be a Noetherian topological space Then theZi from Theorem3.11 are called the irreducible componentsof X.
Example 3.13 Let K be an algebraically closed field andg ∈K[x 1 , , x n ] a non-zero polynomial Letp 1 , , p r be the distinct prime factors ofg Then
The (pi) ⊆ K[x1, , xn] are prime ideals, so by the Nullstellensatz 1.17,
I K[x 1 , ,x n ] (VK n (pi)) = (pi) By Theorem 3.10(a), theVK n (pi) are the irre- ducible components of the hypersurfaceVK n (g) /
In the context of prime ideals within the spectrum of a ring R, a prime ideal P is deemed minimal if no other prime ideal Q can be contained within P without being equal to P itself In the case of an integral domain, there exists a unique minimal prime ideal, which is {0} According to Proposition 3.6(e) and Theorem 3.10(b), minimal prime ideals are associated with the maximal closed irreducible subsets of Spec(R), representing the irreducible components of Spec(R) when it is Noetherian.
Corollary 3.14 (Minimal prime ideals) Let R be a Noetherian ring.
(a) There exist only finitely many minimal prime idealsP1, , Pn ofR. (b) Every prime ideal ofR contains at least one of the P i
(d) LetI⊆Rbe an ideal Then the setV Spec(R) (I)has finitely many minimal elementsQ1, , Qr, and we have
Proof By Proposition3.6(e) and by Theorem3.10(b), the (maximal) closed, irreducible subsets ofX := Spec(R) correspond to (minimal) prime ideals of
R So for (a) and (b), we need to show thatX has only finitely many max- imal closed, irreducible subsets, and that every closed, irreducible subset is contained in a maximal one By Theorem3.9(b),Xis Noetherian, so by The- orem 3.11(c), X has finitely many maximal irreducible subsets Z 1 , , Z n , which are all closed By Theorem3.11(b), every closed, irreducible subset is contained in aZ i
Part (c) follows from (b) and Corollary1.12, and part (d) follows from ap- plying (a) and (c) toR/Iand using the correspondence given by Lemma1.22. u t
Let us remark here that part (b) of Corollary 3.14 generalizes to non- Noetherian rings: It us always true that a prime ideal contains a minimal prime ideal (see Exercise3.6).
Part (d) of Corollary3.14is sometimes expressed by saying that there are only finitely many prime ideals which are minimal overI.
Corollary 3.14 will be utilized extensively in this book, with its first application being the completion of the proof for Theorem 2.8 found on page 36 The implication "(a) ⇒ (b)" of this theorem was previously established in Chapter 2.
To prove the implication "(b) ⇒ (a)" in Theorem 2.8, we start with the assumption that R is Noetherian and that Spec(R) equals Spec max(R) Our goal is to demonstrate that R is Artinian According to Corollary 3.14(c), there are finitely many prime ideals in R, and their intersection corresponds to the nilradical of R, denoted as nil(R) Thus, we conclude that nil(R) is contained within the intersection of these prime ideals.
{0}, so by Lemma2.6there existsk with I k = {0} Therefore we can apply Lemma2.7 and conclude that R is
3.1 (Properties of maps) Let X ={(ξ 1 , ξ 2 )∈ C 2 | ξ 1 ξ 2 = 1} Which of the following maps ϕi: X → X are morphisms, isomorphisms, or Zariski- continuous?
3.2 (Separating sets by polynomials) LetKbe an algebraically closed field and letX, Y ⊆K n be two subsets Show that the following statements are equivalent.
(a) There exists a polynomial f ∈ K[x 1 , , x n ] such thatf(x) = 0 for all x∈X, andf(y) = 1 for ally∈Y.
(b) The Zariski-closures ofX andY do not meet:X∩Y =∅.
In this section, we demonstrate that the bijection Φ: X → Spec max (K[X]), as established in Theorem 1.23, is a homeomorphism when K is an algebraically closed field and X is an affine variety It is important to note that Spec max (K[X]) adopts the subset topology derived from the Zariski topology on Spec (K[X]) This establishes a crucial connection between algebraic geometry and topology, highlighting the topological properties of affine varieties.
In the context of ring theory, consider a ring R and an ideal I within it According to Lemma 1.22, there exists a bijection between VSpec(R)(I) and Spec(R/I) This relationship is further established as a homeomorphism, with VSpec(R)(I) adopting the subset topology that is induced from the Zariski topology on Spec(R) For a detailed solution, refer to page 238.
*3.5 (A converse of Theorem 3.9?) Let R be a ring If Spec(R) is a Noetherian space, does this imply thatRis a Noetherian ring? Give a proof or a counter example (Solution on page238)
The following exercise is due to Martin Kohls.
3.6 (Minimal prime ideals) LetRbe a (not necessarily Noetherian) ring andQ∈Spec(R) a prime ideal Show that there exists a minimal prime ideal
P ∈ Spec(R) with P ⊆ Q In particular, if R 6= {0}, there exist minimal prime ideals in R.
Hint:Use Zorn’s lemma with an unusual ordering (Solution on page238)
3.7 (Hausdorff spaces) Let X be a Noetherian topological space Show that the following two statements are equivalent.
(b) X is finite and has the discrete topology.
In particular, no infinite subsetY ⊆K n with the Zariski topology is Haus- dorff (Solution on page239)
3.8 (Quasi-compact spaces) Recall that a topological spaceX is called quasi-compact if for every setMof open subsets withX =S
U∈MU, there exist U1, , Un∈ MwithX =Sn i=1Ui. (a) Show that a topological spaceX is Noetherian if and only if every subset ofX is quasi-compact.
(b) Let Rbe a ring andX = Spec(R) ThenX is quasi-compact (even if it is not Noetherian).
3.9 (Products of irreducible varieties) Let X ⊆ K m and Y ⊆K n be two irreducible affine varieties over a field Show that the product variety
Hint: This may be done as follows: For X×Y = Z 1 ∪Z 2 with Z i closed, consider the sets X i :={x∈X| {x} ×Y ⊆Z i } Show that X = X 1 ∪X 2 and that theXi are closed (Solution on page240)
In an algebraically closed field K, the set D of all diagonalizable n×n matrices forms a dense subset within the space of n×n matrices, K n×n To demonstrate this, we need to show that the Zariski-closure of D is equal to K n×n, which can be identified with the affine space K n² Additionally, it is important to consider whether D is an open subset in K n×n For further details, refer to the solution on page 240.
True Geometry: Affine Varieties
In this section,Kis assumed to be an algebraically closed field We have the following correspondences between algebraic and geometric objects:
(1) Hilbert’s Nullstellensatz gives rise to a bijective correspondence affine varieties inK n ←→ radical ideals inK[x1, , xn] (4.1)
The assignment of a set \( X \subseteq K^n \) of points to its vanishing ideal \( I(X) \) creates a mapping from the power set of \( K^n \) to the set of radical ideals in \( K[x_1, \ldots, x_n] \) Similarly, assigning a set \( S \subseteq K[x_1, \ldots, x_n] \) of polynomials to the affine variety \( V(S) \) establishes a mapping from the power set of \( K[x_1, \ldots, x_n] \) to the set of affine varieties in \( K^n \) These mappings, when restricted, form a correspondence that is inclusion-reversing An affine variety is considered irreducible if and only if its vanishing ideal is a prime ideal, leading to a sub-correspondence between irreducible affine varieties in \( K^n \) and prime ideals in \( K[x_1, \ldots, x_n] \).
(2) Every affine K-variety X has a coordinate ring K[X], whose elements give rise to regular functions X → K This leads us to identify K[X]
55 with the ring of regular functions onX Assigning to an affineK-variety
X its coordinate ring yields a map affineK-varieties −→ reduced affineK-algebras.
Conversely, every reduced affineK-algebra is isomorphic to the coordi- nate ring of an affineK-variety, which is unique up to isomorphism For
X an affineK-variety, we have the equivalence
X is irreducible ⇐⇒ K[X] is an affine domain.
(3) Let X be an affineK-variety with coordinate ringK[X] Then we have an inclusion-reversing, bijective correspondence
Zariski-closed subsets ofX ←→ radical ideals inK[X] (4.2)
A closed subset of X is deemed irreducible if its corresponding ideal in K[X] is a prime ideal This establishes a relationship where chains of closed, irreducible subsets of X align with chains of prime ideals in K[X], albeit with reversed inclusions Additionally, there exists a sub-correspondence between the irreducible components of X and the minimal prime ideals in K[X].
(4) Given two affineK-varietiesXandY, we have a bijective correspondence morphismsX →Y of varieties ←→ homomorphisms K[Y]→K[X] of K-algebras.
This correspondence translates isomorphisms effectively, although it is less reliable concerning injectiveness Additionally, the composition of two morphisms of varieties aligns with the composition of homomorphisms of the coordinate rings, albeit in reverse order.
We should mention that some parts of the lexicon stay intact if we drop the hypothesis thatK be algebraically closed.
Abstract Geometry: Spectra
There is no variety associated to a general ringR However, we always have the spectrum Spec(R), which is an abstract substitute for an affine variety.
By (4.3), affine varieties over algebraically closed fields are embedded into the
4.2 Abstract Geometry: Spectra 57 spectrum of the coordinate ring, so, taking a somewhat generous view, we can regard the spectrum as a generalization of an affine variety In particu- lar, statements about spectra of rings almost always imply statements about affine varieties as special cases As in the previous section, we will summarize some algebra-geometry correspondences We will also see that they are gen- eralizations of the correspondences from Section4.1 In the sequel,Rstands for a ring.
(1) We have an inclusion-reversing bijective correspondence
Zariski-closed subsets of Spec(R) ←→ radical ideals inR (4.4)
Assigning the intersection of all prime ideals in a subset \( X \subseteq \text{Spec}(R) \) creates a mapping from the power set of \( \text{Spec}(R) \) to the set of radical ideals in \( R \) Similarly, for a set \( S \subseteq R \), the collection of all prime ideals containing \( S \) establishes a map from the power set of \( R \) to the Zariski-closed subsets of \( \text{Spec}(R) \) By restricting both mappings, we obtain a significant correspondence.
A closed subset of Spec(R) is considered irreducible if and only if its corresponding ideal in R is a prime ideal Consequently, chains of closed, irreducible subsets of Spec(R) are linked to chains of prime ideals in R, with all inclusions reversed.
In the special case that R = K[X] is the coordinate ring of an affine variety over an algebraically closed field, we can compose (4.4) with the correspondence (4.2), and get a correspondence
Zariski-closed subsets of Spec (K[X]) ←→
Zariski-closed subsets of X are formed by intersecting a closed subset of Spec(K[X]) with Spec max(K[X]) This process allows for the application of the correspondence outlined in (4.3) to the points Consequently, (4.4) serves as a generalization of (4.2).
A ring homomorphism ϕ: R → S creates a morphism ϕ ∗: Spec(S) → Spec(R), mapping a prime ideal Q in S to its preimage ϕ −1(Q) in R Specifically, if R is a subset of S and ϕ is the inclusion map, then ϕ ∗(Q) equals the intersection of R with Q It's important to note that the relationship between ring homomorphisms and morphisms of spectra is not bijective Additionally, for another ring homomorphism ψ: S → T, the composition of morphisms satisfies (ψ◦ϕ) ∗ = ϕ ∗ ◦ ψ ∗.
In general, ϕ ∗ does not restrict to a map Spec max (S) → Spec max (R); but ifϕis a homomorphism of affineK-algebras, it does If in addition
R=K[Y] andS =K[X] are coordinate rings of affine varieties over an algebraically closed field, then (4.3) translates this restriction ofϕ ∗ into a mapX →Y, which is exactly the morphism from (4) in Section 4.1 corresponding toϕ.
4.1 (Dominant and injective morphisms) LetX andY be affine vari- eties over a field K, and letf:X →Y be a morphism with induced homo- morphismϕ:K[Y]→K[X] We say that f isdominantif the imagef(X) is dense inY, i.e., f(X) =Y.
(a) Show thatf is dominant if and only ifϕis injective.
(b) Show that ifϕis surjective, thenf is injective.
(c) Give examples where f is dominant but not surjective, and where the converse of part (b) does not hold.
4.2 (When isϕ ∗ dominant?) Letϕ:R→Sbe a homomorphism of rings. Show that the following two statements are equivalent.
(a) The mapϕ ∗ : Spec(S)→Spec(R) is dominant.
(b) The kernel ker(ϕ) is contained in the nilradical nil(R) ofR.
In the study of coproducts of spectra and affine varieties, consider rings R1, , Rn, where the direct sum R is defined as the Cartesian product of these rings with component-wise addition and multiplication The projections πi from R to Ri create morphisms fi from Spec(Ri) to Spec(R), illustrating the relationship between the spectra of the individual rings and their coproduct.
In the context of ring theory, if IfS is a ring with morphisms gi: Spec(Ri) → Spec(S), there exists a unique morphism g: Spec(R) → Spec(S) such that g ◦ fi = gi for all i This scenario illustrates that Spec(R), along with the morphisms fi, serves as a coproduct in the category of spectra of rings.
(b) Spec(R) is the disjoint union of the images of thefi For eachi, the image offi is closed in Spec(R).
If X1, , Xn are affine varieties over an algebraically closed field K, then a coproduct X exists within the category of affine K-varieties Similar to the previous case, the image of each Xi in X is isomorphic to Xi The universal property of this coproduct is illustrated in the accompanying diagram.