18 1. Elementary Prime Number Theory, I
This shows that every prime divisor of
+ n + A exceeds its square root,
so that
+ n + A is prime.
The proof of the remaining implication requires one more preliminary
Lemma 1.13. If π is an element of Z[η] whose norm is a rational prime p,
then π is prime in Z[η].
Proof. We claim that Z[η]/(π) is isomorphic to Z/pZ. Since Z/pZ is a
field, this implies that π generates a prime ideal of Z[η], which in turn
implies that π is prime. Let ψ : Z Z[η]/(π) be the ring homomorphism
defined by mapping n to n mod π. Since p = ππ 0 (mod π), the kernel of
ψ contains the ideal pZ. Since pZ is a maximal ideal, either ψ is identically
zero or the kernel of ψ is precisely pZ. Since π is not a unit in Z[η], ψ(1) is
nonzero, and so the kernel of ψ is precisely pZ. Hence Z/pZ is isomorphic to
the image of ψ. So the proof will be complete if we show that ψ is surjective.
Write π = r + for integers r and s, and let x + be an arbitrary
element of Z[η]. We can choose integers a and b for which
m := x + π(a + bη) Z.
Indeed, a short computation shows that this containment holds precisely
b(r s) + as = y,
which is a solvable linear Diophantine equation in a and b since gcd(r−s, s) =
gcd(r, s) = 1. Then m x + (mod π), and so ψ(m) = x + mod π.
Since x + was arbitrary, ψ is surjective as claimed.
Proof that (ii) (iii). Suppose that
+ n + A is prime for all
0 n

We are to prove that Z[η] possesses unique factorization. Suppose other-
wise, and let α be a nonzero, nonunit of minimal norm with two distinct
factorizations into irreducibles, say
α = π1 · · · πk = ρ1 · · · ρj.
(Here distinct means that either k = j, or that k = j, but there is no
way to reorder the πi so that each πi is a unit multiple of ρi.) By the
minimality of N (α), it is easy to see that none of the irreducibles in the
first factorization can be a unit multiple of an irreducible in the second
factorization. Consequently, none of the irreducibles appearing in either
factorization can be prime in Z[η].
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