Lemma 2

Let P and Q be monic polynomials in K[X] and $s$ is a strictly positive integer. If $P^s= Q^s$, then we have P = Q.

Let $\mathscr{I}$ be the set of irreducible monic polynomials in K[X]. Since P and Q are monic, there exist elements $(a_F)$ and $(b_F)$ of $\mathbf{N}^{(\mathscr{I})}$ such that we have $P =\prod_{F\in\mathscr{I}}F^{a_F}$ and $Q =\prod_{F\in\mathscr{I}}F^{b_F}$. It follows from the equality $P^s= Q^s$ and the uniqueness of the decomposition into irreducible factors that we have $sa_F=sb_F$ for every $F\in \mathscr{I}$. Since $s$ is strictly positive, we consequently have $a_F=b_F$ for every $F\in \mathscr{I}$, and therefore P = Q.

B) Let us now prove the existence of P.

By Theorem 1 of VIII, p. 252, there exist a Galois extension L of K of finite degree and an isomorphism of L-algebras $\theta : A_{(L)}\rightarrow \mathbf{M}_n(L)$. By formula (12) of III, §9, No. 1, p. 542, the polynomial Pc($a; X$) is also the characteristic polynomial of the element $1\otimes a$ of the L-algebra $A_{(L)}$, hence of the element $\theta (1\otimes a)$ of the L-algebra $\mathbf{M}_n(L)$.

Set P(X) = det(X$I_n-\theta (1\otimes a)$); it is a monic polynomial in L[X]. By Example 3 of III, §9, No. 3, p. 545, we have

(19) Pc($a; X$) $= P(X)^n$.

Let G be the Galois group of L over K. For $\sigma \in G$, denote by $\overline{\sigma}$ the automorphism of the ring L[X] that coincides with $\sigma$ on L and fixes X. Then K[X] is the set of polynomials Q of L[X] such that we have $\overline{\sigma}(Q) = Q$ for every $\sigma \in G$ (V, §10, No. 1, p. 56, Theorem 1). Since the polynomial Pc($a; X$) $= P(X)^n$ belongs to K[X], we have $\overline{\sigma}(P)^n= P^n$ for every $\sigma \in G$. By Lemma 2, we therefore have $\overline{\sigma}(P) = P$ for every $\sigma \in G$, so P belongs to K[X].