Theorem 1.5.1(The Fundamental Theorem of Arithmetic). Every integer different from0and units is the product of finitely many irreducible numbers and this decom- position is unique apart from the order of the factors and associates. (Uniqueness means that if
𝑎 = 𝑝1𝑝2. . . 𝑝𝑟= 𝑞1𝑞2. . . 𝑞𝑠
where all𝑝𝑖and𝑞𝑗are irreducible, then𝑟 = 𝑠and the numbers𝑝𝑖and𝑞𝑗can be coupled
into associate pairs.) ♣
Remarks: (1) The units and0had to be excluded because these cannot be decomposed into the product of irreducible numbers: the units can be written only as a product of units, and writing0as a product at least one of the factors must be0(and then this factor is not irreducible).
(2) To interpret the theorem for an irreducible number, it should be considered as a product of a single factor.
(3) A few remarks concerning the uniqueness. Assume that the integer𝑎is the prod- uct𝑎 = 𝑝1𝑝2. . . 𝑝𝑟of irreducible numbers. Then changing the order of the factors we obtain the same product. Also, if𝜀1, . . . ,𝜀𝑟are arbitrary units whose product is1, then𝜀1𝑝1, . . . ,𝜀𝑟𝑝𝑟are irreducible as well and their product is𝑎again. The uniqueness part of the theorem claims that apart from these trivial variants there is no other way to write𝑎as the product of irreducible elements. Taking e.g.12, a few such decompositions are
12 = 2 ⋅ 2 ⋅ 3 = 2 ⋅ (−3) ⋅ (−2) = 3 ⋅ (−2) ⋅ (−2).
(4) When stating the theorem, we should definitely use the notion of irreducible numbers since the theorem declares that (nearly) every integer can be assembled essentially in a unique way from these bricks. For clarity, we shall strictly distin- guish the notions irreducible and prime during the proof. We shall see that their equivalence is crucial for the validity of the Fundamental Theorem.
(5) The Fundamental Theorem is false in many sets of numbers (and integral do- mains). Taking e.g. the even numbers,100has two essentially different decom- positions into the product of irreducible elements:100 = 2 ⋅ 50 = 10 ⋅ 10. We shall see further examples in Chapter 10.
1.5. The Fundamental Theorem of Arithmetic 25
Now we turn to the proof of the Fundamental Theorem. We shall give two proofs for the uniqueness part.
Proof of decomposability. Consider an integer𝑎different from0and units. If𝑎is irreducible, then we are done.
If𝑎is composite, then it has a non-trivial irreducible divisor since its smallest non- trivial positive divisor must be irreducible (see Exercise 1.4.7b). Then𝑎 = 𝑝1𝑎1where 𝑝1is irreducible and𝑎1is not a unit.
If𝑎1is irreducible, then we are done; otherwise there exists an irreducible number 𝑝2satisfying𝑎1= 𝑝2𝑎2where𝑎2is not a unit.
We proceed similarly with𝑎2, etc. Our algorithm must terminate in finitely many steps since the integers|𝑎𝑖|are positive and form a strictly decreasing sequence:
|𝑎| > |𝑎1| > |𝑎2| > . . . , hence some𝑎𝑘must be irreducible:𝑎𝑘 = 𝑝𝑘+1.
Then we get the decomposition𝑎 = 𝑝1𝑝2. . . 𝑝𝑘+1. □ First proof of uniqueness. Our main tool is that every irreducible number is a prime (Theorem 1.4.3).
The proof is by contradiction. Assume that a certain𝑎has (at least) two essentially different decompositions into the product of irreducible elements:
(1.5.1) 𝑎 = 𝑝1𝑝2. . . 𝑝𝑟= 𝑞1𝑞2. . . 𝑞𝑠.
If some𝑝𝑖is an associate of a𝑞𝑗, e.g.𝑝1= 𝜀𝑞1where𝜀is a unit, then cancellation by𝑞1yields
𝑎′= 𝑎 𝑞1
= (𝜀𝑝2)𝑝3. . . 𝑝𝑟= 𝑞2𝑞3. . . 𝑞𝑠,
hence also𝑎′has two essentially different decompositions into the product of irre- ducible elements.
Continuing the process, we get finally an integer where the two decompositions do not share associate factors. Without loss of generality, we may assume that this is the case in (1.5.1), i.e.𝑝𝑖≠ 𝜀𝑞𝑗.
Using (1.5.1), we have𝑝1 ∣ 𝑞1𝑞2. . . 𝑞𝑠. Since𝑝1is irreducible, therefore it is a prime by Theorem 1.4.3; thus𝑝1must divide at least one of the factors𝑞𝑗.
If𝑝1 ∣ 𝑞𝑗, then the irreducibility of𝑞𝑗implies that𝑝1is a unit or it is an associate
of𝑞𝑗, and both are impossible. □
Second proof of uniqueness. This proof uses induction on|𝑎|.
Since associates behave equivalently in every divisibility relation, we may restrict ourselves to the decompositions of positive integers into positive irreducible numbers.
For𝑎 = 2, the uniqueness holds as2is irreducible.
Assuming now that every integer1 < 𝑎 < 𝑛has a unique decomposition into the product of irreducible numbers, we show that then the decomposition of𝑎 = 𝑛is
unique. If not, then𝑛has (at least) two essentially different decompositions into the product of irreducible numbers:
(1.5.2) 𝑛 = 𝑝1𝑝2. . . 𝑝𝑟= 𝑞1𝑞2. . . 𝑞𝑠.
Clearly,𝑟 ≥ 2,𝑠 ≥ 2and further𝑝𝑖 ≠ 𝑞𝑗since if e.g.𝑝1= 𝑞1, then also the number 1 < 𝑛/𝑝1 < 𝑛would have two different decompositions contradicting the induction hypothesis.
Suppose𝑝1< 𝑞1and consider𝑛1= 𝑛 − 𝑝1𝑞2. . . 𝑞𝑠. We show that
(1.5.3) 1 < 𝑛1< 𝑛
and
(1.5.4) 𝑛1has two different decompositions, which is a contradiction.
By (1.5.2), the expression𝑛1 = 𝑛 − 𝑝1𝑞2. . . 𝑞𝑠can be rewritten as (1.5.5) 𝑛1= 𝑝1(𝑝2. . . 𝑝𝑟− 𝑞2. . . 𝑞𝑠) = 𝑞2. . . 𝑞𝑠(𝑞1− 𝑝1).
Clearly𝑛1 < 𝑛and𝑝1 < 𝑞1implies
𝑛1= 𝑞2. . . 𝑞𝑠(𝑞1− 𝑝1) ≥ 𝑞2⋅ 1 = 𝑞2> 1, thus verifying (1.5.3).
Now, write the last factors in both decompositions in (1.5.5) as a product of irre- ducible numbers:
𝑝2. . . 𝑝𝑟− 𝑞2. . . 𝑞𝑠= 𝑢1. . . 𝑢𝑘 and 𝑞1− 𝑝1 = 𝑣1. . . 𝑣𝑚. Then𝑛1has the following representations as a product of irreducible elements:
(1.5.6) 𝑛1 = 𝑝1𝑢1. . . 𝑢𝑘= 𝑞2. . . 𝑞𝑠𝑣1. . . 𝑣𝑚.
(If eventually𝑞1− 𝑝1 = 1, then the factors𝑣𝑖are missing in which case the argu- ment will be even more valid.)
We show that the two decompositions in (1.5.6) are essentially different. The first one contains𝑝1. But𝑝1is missing from the second one, since on the one hand𝑝1≠ 𝑞𝑗, and on the other hand, if𝑝1= 𝑣𝑖for some𝑖, then
𝑝1∣ 𝑣1. . . 𝑣𝑚= 𝑞1− 𝑝1⟹ 𝑝1∣ 𝑞1,
which is impossible. Thus (1.5.4) is proven. □
Remarks: (1) Analyzing the first proof of uniqueness, we find that the division al- gorithm served as its basis, after all. It made possible the Euclidean algorithm, yielding the existence of a special common divisor based on which we showed (via Theorem 1.3.9) that an irreducible number is always a prime, giving the key step to the proof.
It is true also generally that if in some number sets (or integral domains) we can perform the division algorithm, then the Fundamental Theorem of Arithmetic holds there. Our proof of uniqueness remains valid literally also for the general case, whereas the decomposability may require some more refined arguments in
Exercises 1.5 27
certain sets. We shall see such examples in Chapters 7 and 10. In Section 11.3, using ideals, we shall give a unified proof for the general case that division al- gorithm always implies the Fundamental Theorem (both decomposability and uniqueness).
We note that the relation between the division algorithm and the Fundamental Theorem is not symmetric; there exist sets of numbers where the Fundamental Theorem is true but there do not exist division algorithms of any kind. We shall see an example in Chapter 10.
(2) The second proof of uniqueness did not rely on the theorems of Sections 1.3 and 1.4. Thus we can give new proofs for some of those theorems using the Funda- mental Theorem. We emphasize two important results: the existence of a special common divisor (Theorem 1.3.3) and that every irreducible number is a prime (the “harder” part of Theorem 1.4.3). To derive these from the Fundamental The- orem, consult the proof of Theorem 1.6.4 for the first one, and Exercise 1.5.8 for the second one.
Exercises 1.5
1. Verify that the number of irreducible factors in the decomposition of𝑎is at most log2|𝑎|.
2. Consider the set of even numbers.
(a) Which numbers have an essentially unique decomposition into the product of irreducible elements?
(b) Find a number that has exactly1000essentially distinct decompositions.
3. Analyze the reason why our proofs of uniqueness fail for the even numbers.
4. Demonstrate that the Fundamental Theorem is false among the integers divisi- ble by10and there exist elements with decompositions not even having the same number of irreducible factors.
5. Consider the set𝐹of finite decimal fractions.
(a) Determine the units and the irreducible elements.
(b) Prove that the Fundamental Theorem is valid in𝐹.
* (c) Verify that we can perform a division algorithm in𝐹, i.e. we can assign to every 𝑐 ∈ 𝐹a non-negative integer𝑓(𝑐)where𝑓(𝑐) = 0if and only if𝑐 = 0and to every𝑎and𝑏 ∈ 𝐹,𝑏 ≠ 0, there exist𝑞and𝑟 ∈ 𝐹 satisfying𝑎 = 𝑏𝑞 + 𝑟and 𝑓(𝑟) < 𝑓(𝑏).
6. There are many variants of the second proof of uniqueness. Elaborate the argu- ment if we work with𝑛1= 𝑛 − 𝑝1𝑞2.
7. Compute the number of decompositions of a given integer into the product of irre- ducible elements if we count separately those that differ only in the order of factors and/or in associates.
S 8. Derive from the Fundamental Theorem that every irreducible number is a prime.
9. Find all (not necessarily positive and not necessarily distinct) primes (among the integers) satisfying
1
𝑝1− 𝑝2− 𝑝3 = 1 𝑝2 + 1
𝑝3.
S* 10. Determine all positive primes (among the integers) a power of which (with positive integer exponent) is the sum of the cubes of two positive integers.