31 Prime elements

31.1 Definition. Let R be an integral domain, and let a, b ∈ R. We say that a divides b if b = ac for some c ∈ R. We then write: a | b.

31.2 Proposition. If R is an integral domain and a, b ∈ R then a ∼ b iff a | b and b | a.

Proof. Exercise.

31.3 Definition. Let R is an integral domain. An element a ∈ R is a prime element if p 6= 0, p is a non-unit and if a | bc then either a | b or a | c.

31.4 Example.

1) In Z we have {prime elements} = {± prime numbers} = {irreducible elements}

√ √ 2) By the proof of Proposition 30.9 in Z[ −5] the element α = 2 + 5i is irreducible. On the other hand α is not a prime element:

α | (3 · 3) but α - 3

31.5 Proposition. If R is an integral domain and a ∈ R is a prime element then a is irreducible.

Proof. Let a ∈ R be a prime element and let a = bc. We want to show that either b or c must be a unit in R.

124 We have a | (bc), and since a is a prime element it implies that a | b or a | c.

We can assume that a | b. Since also b | a, thus by (31.2) we obtain that a ∼ b, i.e. a = bu for some unit u ∈ R. Therefore we have

bc = a = bu

By (25.6) this gives u = c, and so c is s unit.

31.6 Proposition. If R is a UFD and a ∈ R then a is an irreducible element iff a is a prime element.

Proof.

(⇐) Follows from Proposition 31.5.

(⇒) Assume that a ∈ R is irreducible and that a | (bc). We want to show that either a | b or a | c.

If b = 0 then b = a · 0 so a | 0. If b is a unit then c = b−1bc so a | c.

As a consequence we can assume that b, c are non-zero, non-units.

Since a | (bc) there is d ∈ R such that bc = ad. Assume that d is not a unit. Since R is a UFD we have decompositions:

b = b1 · ... · bm, c = c1 · ... · cn, d = d1 · ... · dp where bi, cj, dk are irreducible. This gives

b1 · ... ·bm · c1 · ... ·cn = a · d1 · ... ·dp

By the uniqueness of decomposition in UFDs this implies that either a ∼ bi for some i or a ∼ cj for some j. In the first case we get a | b, and in the second case a | c.

If d is a unit the argument is similar

125 31.7 Theorem. An integral domain R is a UFD iff 1) every non-zero, non-unit element of R is a product of irreducible elements 2) every irreducible element in R is a prime element.

Proof.

(⇒) Follows from the definition of UFD (30.7) and Propositon 31.6.

(⇐) Assume that R satisfies conditions 1)-2) of the theorem. We only need to show that if b1, . . . , bk, c1, . . . , cl are irreducible elements in R such that

b1 · ... · bk = c1 · ... · cl then k = l, and after reordering factors we have b1 ∼ c1, ... , bk ∼ ck.

We argue by induction with respect to k.

If k = 1 then we have b1 = c1 · ... · cl. Since b1 is irreducible this implies that l = 1, and so b1 = c1.

Next, assume that the uniqueness property holds for some k and that we have

b1 · ... · bk · bk+1 = c1 · ... · cl where bi, cj are irreducible elements. This implies that bk+1 | (c1 · ... · cl). By condition 2) of the theorem bk+1 is a prime element. It follows that bk | cj for some 1 ≤ j ≤ l. We can assume that bk+1 | cl. Then cl = abk+1 for some a ∈ R. Also, since cl, bk+1 are irreducible a must be a unit. This shows that bk+1 ∼ cl. Furthermore, we obtain from here that

b1 · ... · bk · bk+1 = c1 · ... · cl−1 · abk+1 Since R is an integral domain we get

b1 · ... · bk = c1 · ... · cl−1a

Since bk is irreducible and a is a unit the product cl−1a is an irreducible element. Therefore by the inductive assumption k = l − 1, and after reordering of factors we have b1 ∼ c1, . . . , bk−1 ∼ ck−1, bk ∼ cl−1a ∼ cl−1

126 32 PIDs and UFDs

32.1 Theorem. If R is a PID then it is a UFD.

32.2 Lemma. If R is a PID and I1,I2,... are ideals of R such that

I1 ⊆ I2 ⊆ ... then there exists n ≥ 1 such that In = In+1 = ... .

S∞ Proof. Take I = i=1 Ii. Check: I is an ideal of R. Since R is a PID we have I = hai for some a ∈ I. Take n ≥ 1 such that a ∈ In. Then we get

I ⊆ In ⊆ In+1 ⊆ ... ⊆ I

It follows that In = In+1 = ... = I.

32.3 Lemma. Let R be a PID. An element a ∈ R is irreducible iff hai is a maximal ideal of R.

Proof. Exercise.

32.4 Lemma. If R is an integral domain then a ∈ R is a prime element iff hai is a non-zero prime ideal of R.

Proof. Exercise.

Proof of Theorem 32.1. Let R be a PID. By Theorem 31.7 it suffices to show that 127 1) every non-zero, non-unit element of R is a product of irreducible elements 2) every irreducible element in R is a prime element.

1) We argue by contradiction. Assume that a0 ∈ R is a non-zero, non-unit element that is not a product of irreducibles. This implies that

a0 = a1b1 for some non-zero, non-unit elements a1, b1 ∈ R.

Next, if both a1 and b1 were products of irreducibles, then a0 would be also a product of irreducibles, contradicting our assumption. We can then assume that a1 is not a product of irreducibles, and so in particular we have

a1 = a2b2 for some non-zero, non-unit elements a2, b2 ∈ R.

By induction we obtain that for i = 1, 2,... there exists non-zero, non-unit elements ai, bi ∈ R such that ai = ai+1bi+1 for all i ≥ 0.

Consider the chain of ideals

ha0i ⊆ ha1i ⊆ ...

By Lemma 32.2 we obtain that hani = han+1i for some n ≥ 0. This means that an = an+1u for some unit u ∈ R (check!). As a consequence we obtain

an+1bn+1 = an = an+1u and so bn+1 = u. This is a contradiction, since bn+1 is not a unit.

2) Let a ∈ R be an irreducible element and let a | (bc). We need to show that either a | b or a | c.

Assume that a - b. This implies that b 6∈ hai and so hai= 6 hai + hbi

128 Since by Lemma 32.3 the ideal hai is a maximal ideal we obtain then that hai + hbi = R, and so in particular 1 ∈ hai + hbi. Therefore

1 = ar + bs for some r, s ∈ R, and so c = a(rc) + (bc)s Since a | a(rc) and a | (bc)s we obtain from here that a | c.

129 33 Application: sums of two squares

Recall. The ring of Gaussian integers:

Z[i] = {a + bi | a, b ∈ Z}

Z[i] is a Euclidean domain with the norm function N(a + bi) = a2 + b2. In particular it follows that Z[i] is a UFD.

33.1 Note. For α, β ∈ Z[i] we have N(αβ) = N(α)N(β)

33.2 Proposition. An element α ∈ Z[i] is a unit iff N(α) = 1.

Proof. Exercise.

33.3 Corollary. The only units in Z[i] are ±1 and ±i.

33.4 Proposition. If p ∈ Z is a prime number and for some α ∈ Z[i] we have N(α) = p then α is an irreducible element in Z[i].

Proof. Assume that α = βγ. Then we have

p = N(α) = N(β)N(γ)

Since p is a prime number we get that either N(β) = 1 or N(γ) = 1, and so either β or γ is a unit in Z[i].

130 33.5 Theorem. Let p be an odd prime. The following conditions are equivalent.

1) p = a2 + b2 for some a, b ∈ Z 2) p ≡ 1 (mod 4)

33.6 Lemma. If p ∈ Z is a prime number and p ≡ 1 (mod 4) then there is m ∈ Z such that m2 ≡ −1 (mod p)

33.7 Proposition. If p ∈ Z is a prime number then the groups of units (Z/pZ)∗ of the ring Z/pZ is a cyclic group of order (p − 1).

Proof of Lemma 33.6. By Proposition 33.7 we have

∗ ∼ (Z/pZ) = Z/(p − 1)Z

Since p ≡ 1 (mod 4), thus 4 | (p − 1), and so the group (Z/pZ)∗ has a cyclic subgroup of order 4. Let a ∈ Z/pZ be a generator of this subgroup. Then a2 is an element of order 2 in (Z/pZ)∗, so a2 = −1.

Take the canonical epimorphism π : Z → Z/pZ and let m ∈ π−1(a). Then π(m2) = a2 = π(−1) which gives m2 ≡ −1 (mod p)

Proof of Theorem 33.5.

1) ⇒ 2) Assume that p = a2 + b2. Since p is odd we can assume that a is even, a = 2m, and b is odd, b = 2n + 1. Then we have

p = (2m)2 + (2n + 1)2 = 4m2 + 4n2 + 4n + 1

131 and so p ≡ 1 (mod 4).

2) ⇒ 1) Since p ≡ 1 (mod 4) by Lemma 33.6 there exists m ∈ Z such that p | (m2 + 1). In Z[i] we have m2 + 1 = (m + i)(m − i)

Therefore in Z[i] we have p | (m + i)(m − i). On the other hand p - (m ± i) since otherwise for some a, b ∈ Z we would have m ± i = p(a + bi) = pa + pbi and so pb = ±1 which is impossible.

As a consequence we obtain that p is not a prime element in Z[i]. Since Z[i] is a UFD, prime elements in Z[i] coincide with irreducible elements and so p is not irreducible. Therefore p = αβ for some non-units α, β ∈ Z[i]. We have p2 = N(p) = N(α)N(β) By Lemma 33.2 we have N(α) 6= 1 6= N(β) so N(α) = N(β) = p. Therefore, if α = a + bi then p = a2 + b2.

33.8 Note. 1) One can use factorization in Z[i] to show that, in general, a positive integer n is a sum of two squares iff n is of the form

k mi mr 2n1 2ns n = 2 p1 . . .pr q1 . . .qs

where k, mi, nj ≥ 0, and pi, qj are prime numbers such that pi ≡ 1 (mod 4) and qj ≡ 3 (mod 4).

2) The ring Z[i] can be also used to describe all Pythagorean triples, e.i. triples of integers (x, y, z) that satisfy the equation x2 + y2 = z2.

132 34 Application: Fermat’s Last Theorem

34.1 Fermat’s Last Theorem.

For n ≥ 3 there are no integers x, y, z > 0 such that xn + yn = zn.

Kummer’s idea of the proof.

1) It is enough to show that xn + yn = zn has no integral solutions for n = 4 and for n = p where p is an odd prime. Indeed, if n ≥ 3 is any integer then n = mk where either k = 4 or k is an odd prime, and if x = a, y = b, z = c would be a solution of xn + yn = zn then x = am, y = bm, z = cm would be a solution of xk + yk = zk

2) The case m = 4 was proved by Fermat.

3) Take an odd prime p. Let ζ = e2πi/p be a pth primitive root of 1, and let Z[ζ] be the smallest subring of C that contains ζ.

For any x, y ∈ Z we have a factorization in Z[ζ]: p−1 Y xp + yp = (x + ζiy) i=0 Therefore, if xp + yp = zp then in Z[ζ] we have p−1 Y zp = (x + ζiy) i=0

Assuming that Z[ζ] is a UFD we can compare these factorizations and try to get a contradiction.

Problem. For some values of p (e.g. p = 23) the ring Z[ζ] is not a UFD.

133 35 Greatest common divisor

35.1 Definition. Let R be a ring and let a1, . . . an ∈ R. We say that b ∈ R is a greatest common divisor of a1, . . . , an if

1) b | ai for i = 1, . . . , n

2) if c | ai for i = 1, . . . , n then c | b.

In such case we write b ∼ gcd(a1, . . . , an).

35.2 Note.

0 0 1) If b is a greatest common divisor of a1, . . . , an and b ∼ b then b is also a greatest common divisor of a1, . . . , an. √ 2) In general gcd(√a1, . . . , an) need not exists. E.g. gcd(9, 3(2 + 5i)) does not exist in Z[ −5].

35.3 Theorem. If R is a UFD then gcd(a1, . . . , an) exists for any a1, . . . , an.

35.4 Lemma. If R is a ring such that gcd(a1, a2) exists for any a1, a2 ∈ R then gcd(a1, . . . , an) exists for any a1, . . . , an ∈ R.

Proof. Check:

gcd(a1, . . . , an) ∼ gcd(gcd(a1, . . . , an−1), an)

Proof of Theorem 35.3. Let a, b ∈ R. By Lemma 35.4 it is enough to show that gcd(a, b) exists.

134 If a = 0 then gcd(a, b) ∼ b (check!).

If a is a unit then gcd(a, b) ∼ a (check!).

Therefore we can assume that a, b are non-zero, non-unit elements of R. Since R is a UFD in such case we have

k1 k2 km l1 l2 lm a = uc1 c2 . . .cm , b = vc1 c2 . . .cm where u, v are units, c1, . . . , cm are distinct irreducible elements, and ki, li ≥ 0. Check: we have n1 n2 nm gcd(a, b) ∼ c1 c2 . . .cm where ni = min{ki, li} for i = 1, . . . m.

35.5 Definition. Let R be a ring. Elements a1, . . . , an ∈ R are relatively prime if gcd(a1, . . . , an) ∼ 1.

35.6 Note.

1) The elements a1, . . . , an are relatively prime iff every common divisor of a1, . . . , an is a unit (check!). 2) If R is a UFD, a, b, c ∈ R, gcd(a, b) ∼ 1 and a | bc then a | c (check!).

135