Proof. Exercise. (Hint: if \(f(X) = X^n + a_{n-1} X^{n-1} + \dots + a_0 \in \mathbb{Q}[X]\) is the minimal polynomial of \(\alpha,\) and \(\beta = N\alpha\) for some \(N,\) then what is the minimal polynomial of \(\beta\)?)

Proof. Let \(f \in \mathbb{Q}[X]\) be the minimal polynomial of \(\alpha.\) If \(f \in \mathbb{Z}[X],\) then clearly \(f\) is an algebraic integer.

Conversely, suppose \(f\) does not have integer coefficients, but there is some (larger-degree) monic integral polynomial \(h\) with \(h(\alpha) = 0.\) Then we must have \(h(X) = f(X) g(X)\) for some \(g \in \mathbb{Q}[X].\)

Let \(C\) be the least common multiple of the denominators of the coefficients of \(f,\) so that \(C f \in \mathbb{Z}[X],\) and similarly \(D\) for \(g.\) Then we clearly have \((C f) (D g) = (CD)h.\) Now let \(p\) be a prime dividing \(CD.\) Clearly at least one coefficient of \(C f\) is not divisible by \(p\) (since otherwise \(C / p\) would be the LCM of the denominators). Similarly at least one of the coefficients of \(D g\) is not divisible by \(p.\) So \(Cf \bmod p\) and \(D g \bmod p\) are non-zero in \(\mathbb{F}_p[X].\) But their product \(C D h\) is zero, since \(p \mid CD\) and \(h\) has integral coefficients. This contradicts the fact that \(\mathbb{F}_p[X]\) is an integral domain. So \(CD\) must in fact be 1, i.e. both \(f\) and \(g\) are integral.

Proof. If \(\alpha\) satisfies a polynomial \(f(X) = X^n + a_{n-1} X^{n - 1} + \dots,\) then \(\alpha^{n}\) is in the \(\mathbb{Z}\)-span of \(1, \dots, \alpha^{n-1},\) and by induction one can show that \(\alpha^{n+1},\) \(\alpha^{n + 2}\) etc are also in this span.

Conversely, if this group is finitely generated, then each generator can only mention finitely many powers of \(\alpha,\) so there is some \(N\) such that \(\{1, \dots, \alpha^N\}\) is a generating set. Hence \(\alpha^{N + 1}\) is in the \(\mathbb{Z}\)-span of \(\{1, \dots, \alpha^N\},\) giving a monic integral polynomial that \(\alpha\) satisfies.

Proof. Suppose \(\alpha,\) \(\beta\) satisfy polynomials of degree \(M, N\) respectively. Consider the subgroup of \(\mathbb{C}\) generated by \(\{ \alpha^i \beta^j : 0 \leqslant i < N, 0 \leqslant j < M\}.\) This is finitely generated and contains \(\alpha^r \beta^s\) for all \(r, s \in \mathbb{N},\) so in particular it contains \((\alpha \beta)^j\) and \((\alpha \pm \beta)^k\) for all \(j, k.\) Since a subgroup of a finitely generated abelian group is finitely generated, the result follows.

Proof. First, note that \(\tfrac{1 + \sqrt{d}}{2}\) is a root of \(X^2 - X + \tfrac{1 - d}{4},\) so it is an algebraic integer iff \(d = 1 \bmod 4.\)

Conversely, let \(\alpha = u + v \sqrt{d}\) with \(u, v \in \mathbb{Q},\) and suppose \(\alpha \in \bar{\mathbb{Z}}.\) Then \(\alpha' = u - v \sqrt{d}\) is also in \(\bar{\mathbb{Z}},\) since it satisfies the same polynomial that \(\alpha\) does; and hence \(\alpha + \alpha' = 2u \in \bar{\mathbb{Z}} \cap \mathbb{Q}= \mathbb{Z}.\) Similarly, \(\alpha - \alpha' = 2v \sqrt{d} \in \bar{\mathbb{Z}}\); thus \((2v)^2 d \in \mathbb{Z},\) but since \(d\) is squarefree, this implies that \(2v \in \mathbb{Z}.\)

So, if \(\alpha\) is an algebraic integer but doesn’t lie in \(\mathbb{Z}[\sqrt{d}],\) then we can subtract a \(\mathbb{Z}\)-linear combination of \(1\) and \(\sqrt{d}\) to deduce that one of \(\{ \tfrac{1}{2}, \tfrac{\sqrt{d}}{2}, \tfrac{1 + \sqrt{d}}{2}\}\) is an algebraic integer. Clearly \(\tfrac{1}{2}\) and \(\tfrac{\sqrt{d}}{2}\) are never algebraic integers (since \(4 \nmid d\)); and \(\tfrac{1 + \sqrt{d}}{2}\) is an algebraic integer iff \(d = 1 \bmod 4.\)

Proof. This is a disguised version of Proposition 9.3. Let \(\gamma = 1 / \alpha.\) Then \(\gamma \in \overline{\mathbb{Q}},\) so there is some \(N \in \mathbb{N}_+\) such that \(N\gamma\) is an algebraic integer. Let \(\beta = N\gamma\) for any such \(N.\) Then \(\beta = N / \alpha\) is in \(K,\) and it’s an algebraic integer, so it’s in \(\mathcal{O}_K\); and we have \(\alpha \beta = N.\)