Section1.9has pointed out that, already in the 17th century, the tangent to a curve could be computed using the idea of a double point of intersection between the line and the curve. However, let us be careful: having a double point of intersection with a curve does not mean a line is tangent to the curve.
Counterexample 7.4.1 The curve with equation X3+Y2Z−X2Z=0
has a double point of intersection with the line of equationX=0.
Proof Consider the polynomial in the statement and its partial derivative with re- spect toY
P (X, Y, Z)=X3+Y2Z−X2Z, ∂P
∂Y(X, Y, Z)=2Y Z.
Viewing these as polynomials with coefficients inC[X, Z]; their resultant is
R(X, Z)=det
⎛
⎝X3−X2Z 0 Z
0 2Z 0
0 0 2Z
⎞
⎠=4Z2X2(X−Z)=0.
By CorollaryD.1.5,P (X, Y, Z)is a simple equation of the curve.
Putting
A=
⎡
⎣0 0 1
⎤
⎦, B=
⎡
⎣0 1 0
⎤
⎦
in the proof of Lemma7.3.2, we end up with the equation t2s=0
which indeed admitst=0 as a double root. The line throughAandB is that with equationX=0 andt=0 corresponds to the pointA.
An easy adaptation of this proof shows that in fact, every line throughAadmits Aas (at least) a double point of intersection with the curve.
Figure7.4represents the real affine part of the curve and the line involved in Counterexample7.4.1, when choosingZ=0 as the line at infinity. Of course, no- body would describe this line as “tangent to the curve”. Intuitively, there is a double point of intersection because the line cuts the curve at a point where two “branches”
of the curve meet: and the line has a “single” intersection point with each of these two “branches”. Thus being tangent is more subtle than just having a double point of intersection.
Fig. 7.4
Fig. 7.5
Investigating this example further—in a purely intuitive way, for the time being—
we would rather say that the curve admits two tangents at the point considered: one tangent for each branch, as pictured in Fig.7.5. However, if we consider that a tangent to a “branch” has (at least) a double point of intersection with that branch, since this tangent also cuts the other branch, its point of intersection with the curve should thus be (at least) triple. Indeed, consider the following.
Example 7.4.2 The two lines with equations X=Y, X= −Y
admit a triple point of intersection with the curve with equation X3+Y2Z−X2Z=0.
278 7 Algebraic Curves Proof Consider the case of the line with equationX=Y. Choosing on this line
A=
⎡
⎣0 0 1
⎤
⎦, B=
⎡
⎣1 1 1
⎤
⎦
in the proof of Lemma7.3.2, we obtain the equation t3+t2(s+t )−t2(s+t )=0 that is simply
t3=0
which indeed admitst=0, corresponding to the pointA, as triple root. An analo- gous argument holds for the line with equationX= −Y.
An easy adaptation of this proof shows that in fact, only the two lines indicated admit with the curve a triple intersection at the pointA.
The considerations above are the key for defining the tangent(s) to a curve.
Lemma 7.4.3 Consider an algebraic curve C admitting the simple equation P (X, Y, Z)=0 in some system of homogeneous coordinates. Suppose that this curve contains the pointAbelow and consider further the polynomialp(X, Y )
p(X, Y )=P (X, Y,1), A=
⎡
⎣a b 1
⎤
⎦.
The following conditions are equivalent:
1. all the partial derivatives ofP (X, Y, Z), up to the orderk, are zero at the pointA;
2. all the partial derivatives ofp(X, Y ), up to the orderk, are zero at(a, b).
Proof (1⇒2)is obvious, since the partial derivatives ofp(X, Y )are obtained by puttingZ=1 in the partial derivatives ofP (X, Y, Z).
The converse implication follows at once from Euler’s formula (see Theo- remC.1.5). Indeed if
P (a, b,1)=0, ∂P
∂X(a, b,1)= ∂p
∂X(a, b)=0,
∂P
∂Y(a, b,1)= ∂p
∂Y(a, b)=0 then Euler’s formula implies at once
∂P
∂Z(a, b,1)=0.
It remains to repeat the argument, replacingP by one of its partial derivatives.
Proposition 7.4.4 Consider an algebraic curveC admitting the simple equation P (X, Y, Z)=0 in some system of homogeneous coordinates. Suppose that:
• all the partial derivatives ofP (X, Y, Z), up to the orderk−1, at a given point A, are zero;
• at least one partial derivative ofP (X, Y, Z), of orderk, at the pointA, is non- zero.
Under these conditions, there are exactlyklines, when counted with their multiplic- ities, having with the curve an intersection of multiplicityk+1 at A(see Defini- tion7.3.3).
Proof At least one of the coordinates of the pointAis non-zero; let us assume that it is the third one, which we can then choose to be equal to 1. ViewingZ=0 as the line at infinity, an arbitrary line throughAis entirely determined by its pointBat infinity:
A=
⎡
⎣a b 1
⎤
⎦, B=
⎡
⎣u v 0
⎤
⎦.
Going back to the proof of Lemma7.3.2, we have to consider the polynomial P (sa+t u, sb+t v, s) and study the multiplicity of its root (s, t )=(1,0), cor- responding to the point A. The considerations of Sect. C.2 indicate that we can equivalently puts=1 and study the multiplicity oft=0 as root of the polynomial p(a+t u, b+t v)=P (a+t u, b+t v,1). Moreover, by Lemma7.4.3, the nullity of all the partial derivatives ofP (X, Y, Z)atAis equivalent to the nullity of all the partial derivatives ofp(X, Y )at(a, b).
Let us consider the Taylor expansion of the polynomialp(a+t u, b+t v)around t=0 (see TheoremB.5.3):
p(a+t u, b+t v)=p(a, b) +t
u∂p
∂X(a, b)+v∂p
∂Y(a, b)
+t2 2
u2∂2p
∂X2(a, b)+2uv ∂2p
∂X∂Y(a, b)+v2∂2p
∂Y2(a, b)
+ ã ã ã
We know thatp(a, b)=0 sinceAis on the curve. Thus in any case the right hand side contains at least one factort.
• The right hand side contains at least two factorstif and only if u∂p
∂X(a, b)+v∂p
∂Y(a, b)=0.
If the partial derivatives of p(X, Y )of order 1 are not all zero at (a, b), this is a homogeneous equation of degree 1 inuandv. Up to a constant multiple,
280 7 Algebraic Curves this equation admits a unique solution(u, v), that is, there exists a unique line admitting with the curve an intersection of multiplicity at least 2 atA.
• If now the partial derivatives ofp(X, Y )of order 1 are all zero at(a, b), the right hand side contains at least two factorst. It contains at least three factorstif and only if
u2∂2p
∂X2(a, b)+2uv ∂2p
∂X∂Y(a, b)+v2∂2p
∂Y2(a, b)=0.
If the partial derivatives of order 2 are not all zero at(a, b), this is a homoge- neous equation of degree 2 inuandv. By PropositionC.1.4, this equation admits exactly two roots counted with their multiplicities. That is, there are exactly two (equal or distinct) lines admitting with the curve an intersection of multiplicity at least 3 atA.
And so on.
We thus adopt the following definition:
Definition 7.4.5 Consider a pointAof an algebraic curveCadmitting the simple equation P (X, Y, Z)=0 in some system of homogeneous coordinates. Suppose that:
• all the partial derivatives ofP (X, Y, Z), up to the orderk−1, at the pointA, are zero;
• at least one partial derivative of orderkofP (X, Y, Z)is non-zero at the pointA.
Then:
• The pointAis called a point of multiplicitykof the curve.
• Thek(equal or distinct) lines throughAhaving with the curve an intersection of multiplicityk+1 are called the tangents to the curve atA.
Of course we shall as usual use the terminology:
• simple point for point of multiplicity 1;
• double point for point of multiplicity 2;
• triple point for point of multiplicity 3;
and so on.
Our next concern is to exhibit the equation of the tangent at a simple point.
Proposition 7.4.6 Consider an algebraic curveC admitting the simple equation P (X, Y, Z)=0 in some system of homogeneous coordinates. Given a simple point Awith coordinates
A=
⎡
⎣a b c
⎤
⎦
on this curve, the equation of the tangent toCatAadmits the equation X∂P
∂X(a, b, c)+Y∂P
∂Y(a, b, c)+Z∂P
∂Z(a, b, c)=0.
Proof By Euler’s formula (see TheoremC.1.5), we have a∂P
∂X(a, b, c)+b∂P
∂Y(a, b, c)+c∂P
∂Z(a, b, c)=nP (a, b, c)=0 thusAis indeed a point on the line in the statement.
Now at least one of the coordinates ofAis non-zero. Thus there is no loss of generality in assuming thatc=1, so that we are back in the situation of the proof of Proposition7.4.4, the notation of which we shall use freely. The two quantities (u, v)corresponding to the tangent thus satisfy the equation
u∂P
∂X(a, b,1)+v∂P
∂Y(a, b,1)=0.
Multiplying this last equation byt and adding it to the other equation above (with c=1), we obtain
(a+ut )∂P
∂X(a, b,1)+(b+vt )∂P
∂Y(a, b,1)+∂P
∂Z(a, b,1)=0.
However, again as the proof of Proposition7.4.4indicates, the points with coordi- nates
⎡
⎣a+ut b+vt
1
⎤
⎦
are precisely the affine points of the tangent. Thus the homogeneous equation of the tangent is indeed that indicated in the statement (see Sect.C.2).
As expected:
Corollary 7.4.7 All the points of a line are simple and the tangent to the line at each point is the line itself.
Proof Given a line with equation
aX+bY+cZ=0
at least one of the coefficients—that is, at least one first partial derivative—is non- zero. So all points are simple. The tangent at any point is thus, by Proposition7.4.6
Xa+Y b+Zc=0
that is, the line itself.
282 7 Algebraic Curves More generally:
Proposition 7.4.8 Consider a simple pointAof a curveC. IfChas a linear com- ponent containingA, this is also the tangent atAto the curveC.
Proof The equation ofChas the form
(aX+bY+cZ)P (X, Y, Z)=0
where the first factor corresponds to the linear component throughA. Write
A=
⎡
⎣u v w
⎤
⎦.
The first partial derivative atAis
auP (u, v, w)+(au+bv+cw)∂P
∂X(u, v, w)=auP (u, v, w)
becauseau+bv+cw=0 and analogously for the other two partial derivatives.
Since the point Ais simple, at least one of these partial derivatives is non-zero, proving thatP (u, v, w)=0. Thus by Proposition7.4.6, the equation of the unique tangent atAreduces to
a P (u, v, w) X+b P (u, v, w) Y+c P (u, v, w) z=0.
Simplifying byP (u, v, w)we recapture
aX+bY+cZ=0
as the equation of the tangent.
Of course, when a line is a component of a curve, its points on this curve are no longer necessarily simple.
Example 7.4.9 Consider the curveCcomprisingnlines passing through the same pointA. ThenAis a point of multiplicitynonC.
Proof The equation ofChas the form
$n i=1
(aiX+biY+ciZ)=0.
Each term of each partial derivative up to the leveln−1 still contains at least one of the factors appearing in the equation ofC, thus vanishes atA. SoAis at least of multiplicityn.
Since each factor contains at least one non-zero coefficient, the corresponding partial derivative at the levelnis the product of all these coefficients, thus is non-
zero. ThusAhas multiplicity exactlyn.
In the spirit of the comments at the end of Sect.7.2, we shall sometimes need to consider the following more technical notion, which is not intrinsic, but depends heavily on the choice of a particular—possibly non-simple—equation describing the curve:
Definition 7.4.10 Consider a pointAof an algebraic curveC admitting the arbi- trary equationP (X, Y, Z)=0 in some system of homogeneous coordinates. Sup- pose that:
• all the partial derivatives ofP (X, Y, Z), up to the orderk−1, at the pointA, are zero;
• at least one partial derivative of P (X, Y, Z), at the order k, at the point A, is non-zero.
Then the pointA is called a point of multiplicity k with respect to the equation P (X, Y, Z)=0 of the curve.