# Algebraic integers and characteristic polynomial

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?

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 $$f(X) = \det T(X)\in\Bbb Z[X]$$.