If the matrix is integral, its characteristic polynomial is monic with integer coefficients, and thus the eigenvalues are algebraic integers.

If \(\alpha \) is an algebraic integer, then its an algebraic number, so I claim that \(m_\alpha \) is monic with integer coefficient. We know \(\alpha \) satisfies some some monic polynomial \(f(x)\) with integer coefficients and therefore \(m_\alpha \) divides \(f(x)\). Now use the monic Gauss lemma 1.2.11 to get that \(m_\alpha \) is in \(\mathbb {Z}[x]\).

Conversely, if \(m_\alpha \) is has integer coefficients, then \(\alpha \) satisfies a monic polynomial with integer coefficients, thus is an algebriac integer.

This follows from Corollary 1.6.8 and Proposition 2.1.8.

Let \(A_\alpha \) be the standard representation of \(\alpha \). Then this is a matrix with rational entries. So if we clear denominators by multiplying with a suitable integer \(n\) we get that \(n A_\alpha =A_{n\alpha }\) is an matrix with integer entries and therefore \(n\alpha \) is an algebraic integer by Lemma 2.1.2.

Let \(\alpha =a+b\sqrt{d}\) with \(a,b \in \mathbb {Q}\). If \(b=0\) then we are just asking which rational numbers are algebraic integers, which we have seen are only the integers. So assume \(b \neq 0\). Then the minimal polynomial of \(\alpha \) over \(\mathbb {Q}\) is

Therefore, \(\alpha \) is an algebraic integer if and only if \(2a\) and \((a^2-db^2)\) are integers.

Now, since \(2a\) is an integer, we have \(4a^2\) is an integer, and thus \(d(2b)^2\) must be an integer. If \(2b\) was not an integer, then a \(p^2\) (for some prime \(p\)) would appear in the denominator of \((2b)^2\). This would force \(d\) to be divisible by \(p^2\) contradicting the square-free assumption. Thus \(2b \in \mathbb {Z}\).

So let \(u=2a,v=2b\). Then we have \(u^2-dv^2 \equiv 0 \pmod4\). Now, if \(v\) is even, then so is \(u\) in which case \(a,b \in \mathbb {Z}\). So assume \(v\) is odd, we need to show this can only happen if \(d \equiv 1 \mod 4\). Now, \(v^2 \equiv 1 \pmod4\), and \(u^2\) is either \(0,1 \pmod4\). But note it can’t be zero since \(d\) cant be \(0 \pmod4\) as its square-free, therefore \(u^2 \equiv 1 \mod 4\) and hence \(d \equiv 1 \mod 4\).

\((1) \implies (2)\) : Since \(\alpha \) is an algebraic integer, it is the root of the monic polynomial \(m_\alpha \) of degree \(n\), say, which has integer coefficients. In this case we claim that \(\{ 1,\alpha ,\cdots ,\alpha ^{n-1}\} \) is generates \(\mathbb {Z}[\alpha ]\). For this we note that if \(f(\alpha ) \in \mathbb {Z}[\alpha ]\), then by doing polynomial long division we have \(f(x)=q(x)m_\alpha (x)+r(x)\) with \(\deg (r){\lt}n\). Now if evaluate at \(\alpha \) we get \(f(\alpha )=r(\alpha )\). But since \(\deg (r){\lt}n\) we see that \(r(\alpha )\) is in the \(\mathbb {Z}\) span of \(\{ 1,\alpha ,\cdots ,\alpha ^{n-1}\} \). Thus giving the claim.

\((2)\implies (3) \implies (4)\) : This is trivial.

It remains to prove \((4) \implies (1)\). Let \(\{ a_1,\dots ,a_r\} \) be a basis of \(A\). Then as each \(\alpha a_i \in A\) we can again write it in terms of this basis, so we get a system of equations

with the \(x_{i,j} \in \mathbb {Z}\). If we write this in matrix form by setting \(X={x_{i,j}}\), \(a=(a_1,\dots ,a_r)\) then we have

(here \(()^T\) denotes transpose). Therefore, \(\alpha \) is an eigenvalue of \(X\), which is a matrix with integer entries. Thus by Lemma 2.1.2 we must have that \(\alpha \) is an algebraic integer.

Since \(\alpha ,\beta \) are algebraic integers, then by Theorem 2.1.15 \((2)\) we have \(\mathbb {Z}[\alpha ]\) and \(\mathbb {Z}[\beta ]\) are finitely generated (as additive groups). Moreover, the ring \(\mathbb {Z}[\alpha ,\beta ]\) (take this to be defined as the smallest ring containing \(\mathbb {Z},\alpha ,\beta \)) is also finitely generated, since if \(\{ a_1,\dots ,a_n\} \) generate \(\mathbb {Z}[\alpha ]\) and \(\{ b_1,\dots ,b_m\} \) generates \(\mathbb {Z}[\beta ]\) then \(\{ a_ib_j\} _{i,j}\) generates \(\mathbb {Z}[\alpha ,\beta ]\). Now, \(\mathbb {Z}[\alpha ,\beta ]\) contains both \(\alpha +\beta \) and \(\alpha \beta \), then 2.1.15 \((3)\) tells us that they must also be algebraic integers.

This follows at once from the above, since if \(\alpha ,\beta \) are algebraic integers in \(K\), then \(\alpha +\beta \in K, \alpha \beta \in K\) and by the above, they are both algebraic integers.

By Corollary 2.1.9 we know \(C_\alpha \) has integer coefficients and therefore by Proposition 1.7.3 we get the result.

Since \(\alpha \) is an algebraic integer and by definition it is contained in \(K\), we have \(\alpha \in \mathcal{O}_K\), from which the result follows by Definition 2.1.13.

We know that \(R \subset \mathcal{O}_K\), we will show the opposite inclusion. Let \(\alpha \in \mathcal{O}_K\). Then since \(K\) is the field of fractions of \(R\) we can find \(\delta ,\gamma \in R\) such that \(\alpha =\frac{\gamma }{\delta }\), with \(\delta ,\gamma \) sharing no common factors other than a unit.

Now, since \(\alpha \) is an algebraic integer, we can find some monic polynomial such that

If we now multiply through by \(\delta ^n\) we get

Since \(R\) is assumed to have unique factorization, we see that if \(\varpi \) is an irreducible factor of \(\delta \), then \(\varpi \) is an irreducible factor of \(\gamma ^n\) and thus is a factor of \(\gamma \). But we assumed that \(\delta ,\gamma \) shared no common factors other than units. So \(\delta \) has no irreducible factors and is therefore a unit. So \(\mathcal{O}_K \subset R\), giving the result.