The ring of algebraic integers in QQ(i) is ZZ[i]

Prove that the ring of algebraic integers in \mathbb{Q}(i) is \mathbb{Z}[i].


Recall that an element in an algebraic extension of \mathbb{Q} is called an algebraic integer if its minimal polynomial over \mathbb{Q} has only integer coefficients.

First we will show that the minimal polynomial of every element of \mathbb{Z}[i] has only integer coefficients. To that end, let z = a+bi \in \mathbb{Z}[i]. If b = 0, then certainly z = a \in \mathbb{Z} is an algebraic integer, since its minimal polynomial is simply x - a. Suppose b \neq 0; in this case z \notin \mathbb{Q}, and so its minimal polynomial has degree at least 2. Note that z is a root of p(x) = x^2 - 2ax + (a^2 + b^2), and p(x) \in \mathbb{Z}[x]; so z has degree exactly 2, and in fact p(x) is its minimal polynomial. So z is an algebraic integer.

Conversely, suppose z = s+ti \in \mathbb{Q}(i) is an algebraic integer. If t = 0, then s \in \mathbb{Q} is an algebraic integer, and so s \in \mathbb{Z}. Certainly then z \in \mathbb{Z}[i]. Suppose now that t \neq 0. If s = 0, then z = ti is a root of x^2 + t^2, which is irreducible over \mathbb{Q} and hence the minimal polynomial of z. So t^2 \in \mathbb{Z}. If t is not an integer, then neither is t^2; hence t \in \mathbb{Z}, and so z \in \mathbb{Z}[i]. So we assume t \neq 0.

Now z \notin \mathbb{Q}, so the minimal polynomial of z over \mathbb{Q} has degree at least 2. Since z is (evidently) a root of p(x) = x^2 - 2sx + s^2+t^2, this must be the minimal polynomial of z. Since z is an algebraic integer, we have 2s, s^2+t^2 \in \mathbb{Z}.

Let us say s = a/b and t = c/d, where a,b,c,d \in \mathbb{Z} and \mathsf{gcd}(a,b) = \mathsf{c,d} = 1. Then (rewriting) we have \frac{2a}{b} \in \mathbb{Z}, so that b|2a. Let p be a prime factor of b; then p|2a. Since a and b are relatively prime, p|2, and in fact p = 2. So b = 2^k for some k \geq 0. However, if k \geq 2, then we have 2|a, a contradiction since a and b are relatively prime. So b = 1 or b = 2. Suppose b = 2.

We also have that s^2+t^2 \in \mathbb{Z}, so that b^2d^2|(a^2d^2 + c^2b^2). Say 4d^2k = a^2d^2 + 4c^2; rearranging, we have d^2(4k - a^2) = 4c^2. Let p be a prime factor of d. Again because c and d are relatively prime, we have p|4, so that p = 2. So d = 2^k for some k \geq 0. If k \geq 2, then we have 2|c$, a contradiction. So d = 1 or d = 2. Suppose d = 2.

Now (a^2+c^2)/4 \in \mathbb{Z}, so that a^2 + c^2 = 4k for some k \in \mathbb{Z}. In particular, a^2 + c^2 \equiv 0 mod 4. Note that the squares mod 4 are precisely 0 and 1, and the only solutions of this equivalence are (a,c) \in \{(0,0), (2,0), (0,2), (2,2)\}. In any case, a and c are either 0 (a contradiction) or divisible by 2 (a contradiction since they are relatively prime to b and d, respectively.) Now d = 1. Again, we have a^2 + 4c^2 = 4k for some integer k, so that a^2 \equiv 0 mod 4, a contradiction. So b = 1.

Thus z = s+ti \in \mathbb{Z}[i] as desired.

Hence, the ring of algebraic integers in \mathbb{Q}(i) is precisely \mathbb{Z}[i].

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