Let $K/\mathbb{Q}$ be a number field of degree $n$. For $\beta \in K$ define the linear map (viewing $K$ as a $\mathbb{Q}$-vector space) $\theta_{\beta} : K \to K$ via $\theta_{\beta}(\gamma) = \beta\gamma$. Prove that $\beta$ is an algebraic integer if and only if the characteristic polynomial $\det(xI_n – \theta_{\beta})$ has integer coefficients.
So with $f(x) = \det(xI_n – \theta_{\beta})$ I was able to show that $f(\beta) = 0$ using the Cayley-Hamilton Theorem. So if $f$ has integer coefficients, then by definition, $\beta$ must be algebraic. However what about the converse? How to use the minimal polynomial (for which is easy to be shown that it has integer coefficients) to show that $f$ also does have integer coefficients?
Best Answer
Let $f(X) = \det(XI-\beta)$ as in the OP. (But $X$ is a transcendent variable.)
I will write $!K$ for the vector space over $\Bbb Q$ obtained from $K$ by applying the forgetful functor $!$, so $T:=(\beta I-\theta_\beta):!K\to !K$ is the zero morphism, so $\det T=0$. (We do not need Cayley-Hamilton. And moreover...)
This shows that assuming $f\in\Bbb Z[X]$, because also $f(\beta)=0$, the minimal polynomial of $\beta$ is a divisor of $f$, so it is in $\Bbb Z[X]$, gaussian Lemma, so $\beta$ is algebraic.
The converse, and the question.
Let us assume now that $\beta $ is algebraic.
Let $L$ be the subfield of $K$ generated by $\beta$ over $\Bbb Q$. Then $1,\beta,\dots,\beta^{k-1}$ is a basis of $!L$ over $\Bbb Q$, $k$ being the degree of $\beta$ over $\Bbb Q$.
Fix now some $x\in \Bbb Q$, and let us write the matrix of $$T(x):=(xI-\theta_\beta)\ ,\qquad x\in\Bbb Q\ ,$$ seen first as a morphism $!L\to\!L$, (later as $!K\to!K$,) w.r.t. this basis.
This matrix is $xI$ minus a companion matrix, and the companion matrix has only integer entries, since $\beta$ is algebraic.
Now consider a basis of $K:L$, and for each $\gamma$ in this basis the system $\gamma,\gamma\beta,\dots\gamma\beta^{k-1}$. The matrix of $T(x)$, restricted to the subspace generated by this system is again the same matrix, $xI$ minus integral companion matrix.
Pasting these system together, we get a basis of $K:\Bbb Q$, and the matrix of $T(x)$ is a diagonal repetition of the same block. Now $f(x) = \det T(x)$ holds for every $x\in\Bbb Q$, thus $f(X) = \det T(X)\in\Bbb Z[X]$.