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Considering an integer matrix $M$ over $\mathbb C$. Prove that its minimal polynomial written as a monic has integer coefficients.

I am struggling with this question; as an alternative, I thought that one could consider the question in the rational field, in which case we can still find the RCF of the matrix. It remains to show that at least the minimal polynomial block has integer coefficients.

I thought maybe we can take the conjugate matrix $Q$, which must have rational coefficients, and try to multiply by a factor to turn into an integer matrix and then rewrite RCF as $QMQ^{-1}$, but that requires $Q^{-1}$ to also have integer entries. However that does not seem guaranteed (e.g., if try to wipe out the denominators of the entries of $Q$, we still need to cause the determinant to equal $1$ to have an integer valued inverse or said in another way if we multiply by a factor of $\alpha$, then we have to multiply $Q^{-1}$ with $1/\alpha$, which may not have integer entries).

user1551
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categoryisschemes
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2 Answers2

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Let $\chi\in\mathbb Z[x]$ be the characteristic polynomial of $M$, $m\in\mathbb Q[x]$ be the minimal polynomial and $q\in\mathbb Q[x]$ be the quotient of dividing $\chi$ by $m$. Since $\chi$ and $m$ are monic, so must be $q$. Hence the contents of $m$ and $q$ are reciprocals of positive integers, i.e., we may write $m=\frac{f}{L}$ and $q=\frac{g}{L'}$ for some primitive polynomials $f,g\in\mathbb Z[x]$ with positive integer leading coefficients $L$ and $L'$ respectively. Therefore $fg=LL'\chi$. Since $fg$ is primitive (Gauss’ lemma) and $\chi$ has integer coefficients, $L$ must be equal to $1$. Hence $m$ has integer coefficients.

user1551
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Let $\lambda_1,\dotsc,\lambda_n$ be the eigenvalues of $M$ in $\mathbb{C}$ (with multiplicities). They are the roots of the characteristic polynomial of $M$, which is a monic polynomial over $\mathbb{Z}$, i.e., they are algebraic integers. The minimal polynomial is a product of linear factors of the form $T - \lambda_i$. It follows that its coefficients are algebraic integers as well. On the other hand, since the minimal polynomial can be computed independently from the field (SE/66443), the coefficients are rational. Since $\mathbb{Z}$ is integrally closed, the coefficients are integers.

More generally, if $D$ is an integrally closed domain, and $M \in M_n(D)$, then its minimal polynomial over (any extension of) $Q(D)$ has coefficients in $D$.

Martin Brandenburg
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