31 Prime elements 31.1 Definition. Let R be an integral domain, and let a; b 2 R. We say that a divides b if b = ac for some c 2 R. We then write: a j b. 31.2 Proposition. If R is an integral domain and a; b 2 R then a ∼ b iff a j b and b j a. Proof. Exercise. 31.3 Definition. Let R is an integral domain. An element a 2 R is a prime element if p 6= 0, p is a non-unit and if a j bc then either a j b or a j c. 31.4 Example. 1) In Z we have fprime elementsg = {± prime numbersg = firreducible elementsg p p 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 j (3 · 3) but α - 3 31.5 Proposition. If R is an integral domain and a 2 R is a prime element then a is irreducible. Proof. Let a 2 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 j (bc), and since a is a prime element it implies that a j b or a j c. We can assume that a j b. Since also b j a, thus by (31.2) we obtain that a ∼ b, i.e. a = bu for some unit u 2 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 2 R then a is an irreducible element iff a is a prime element. Proof. (() Follows from Proposition 31.5. ()) Assume that a 2 R is irreducible and that a j (bc). We want to show that either a j b or a j c. If b = 0 then b = a · 0 so a j 0. If b is a unit then c = b−1bc so a j c. As a consequence we can assume that b; c are non-zero, non-units. Since a j (bc) there is d 2 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 j b, and in the second case a j 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 j (c1 · ::: · cl). By condition 2) of the theorem bk+1 is a prime element. It follows that bk j cj for some 1 ≤ j ≤ l. We can assume that bk+1 j cl. Then cl = abk+1 for some a 2 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 = ::: . S1 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 2 I. Take n ≥ 1 such that a 2 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 2 R is irreducible iff hai is a maximal ideal of R. Proof. Exercise. 32.4 Lemma. If R is an integral domain then a 2 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 2 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 2 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 2 R. By induction we obtain that for i = 1; 2;::: there exists non-zero, non-unit elements ai; bi 2 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 2 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 2 R be an irreducible element and let a j (bc). We need to show that either a j b or a j c. Assume that a - b. This implies that b 62 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 2 hai + hbi. Therefore 1 = ar + bs for some r; s 2 R, and so c = a(rc) + (bc)s Since a j a(rc) and a j (bc)s we obtain from here that a j c. 129 33 Application: sums of two squares Recall. The ring of Gaussian integers: Z[i] = fa + bi j a; b 2 Zg 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 2 Z[i] we have N(αβ) = N(α)N(β) 33.2 Proposition. An element α 2 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 2 Z is a prime number and for some α 2 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 2 Z 2) p ≡ 1 (mod 4) 33.6 Lemma. If p 2 Z is a prime number and p ≡ 1 (mod 4) then there is m 2 Z such that m2 ≡ −1 (mod p) 33.7 Proposition. If p 2 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 j (p − 1), and so the group (Z=pZ)∗ has a cyclic subgroup of order 4. Let a 2 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 2 π−1(a). Then π(m2) = a2 = π(−1) which gives m2 ≡ −1 (mod p) Proof of Theorem 33.5.