Lemma 4

Let M be a torsion A-module generated by a finite number $x_1, ..., x_h$ of elements; then there exists an element x of M whose annihilator (II, p. 219) is equal to the annihilator of M.

In both cases $A$ is an integral domain and every ideal of $A$ is principal. When $A = \mathbf{Z}$ (resp. $A = K[X]$), we denote by $\mathscr{P}$ the set of prime numbers (resp. the set of irreducible monic polynomials in $K[X]$). For every element $a \neq 0$ of $A$ there exists then an invertible element $u$ of $A$ and a family $(v_p(a))_{p \in \mathscr{P}}$, with finite support, of positive integers such that $a = u \prod_{p \in \mathscr{P}} p^{v_p(a)}$ and $u$ and the integers $v_p(a)$ are uniquely determined (I, p. 51 and IV, p. 13, Prop. 13).

Let $a_i$ be the annihilator of $x_i$, (for $1 \leq i \leq h$) and $a$ the annihilator of M; let $a_1, ..., a_h, a$ be non-zero elements of $A$ such that $a_i = Aa_i$, and $a = Aa$; since $a = a_1 \cap ... \cap a_h$, it follows from what has been said that

Let us write $a$ in the form $up_1^{n(1)} ... p_r^{n(r)}$, with $p_1, ..., p_r$ distinct in $\mathscr{P}$, $n(1) > 0, ..., n(r) > 0$ and $u$ an invertible element of $A$. Let $j = 1, ..., r$; by (10) there exists an integer $c(j)$ such that $1 \leq c(j) \leq h$ and $v_{p_j}(a_{c(j)}) = n(j)$; there exists $b_j$ in $A$ with $a_{c(j)} = p_j^{n(j)} b_j$ and the element $y_j = b_j x_{c(j)}$ has as annihilator the ideal $Ap_j^{n(j)}$.

Let us show that the annihilator b of $y = y_1 + ... + y_r$ is equal to the annihilator a of M. In any case we have $a \subset b$, so b is of the form $Ap_1^{m(1)} ... p_r^{m(r)}$ with $0 \leq m(j) \leq n(j)$ for $1 \leq j \leq r$. If we had $a \neq b$, there would exist an integer jsuch that $1 \leq j \leq r$ and $m(j) < n(j)$, and hence $d_j = a/p_j$ would annihilate $y$. Now we have $d_j y_k = 0$ for $k \neq j$, whence we obtain $d_j y_j = 0$; but the annihilator of $y_j$ is $Ap_j^{m(j)}$ and $d_j$ is not a multiple of $p_j^{m(j)}$. So the hypothesis $a \neq b$ is absurd.

$^1$ The reader can easily convince himself that the normal basis theorem is not used anywhere before the proof of that proposition.

We shall apply Lemma 4 to the case where $A$ is the polynomial ring $K[X]$ and $M$ the abelian group $N$ with the external law defined by $a \cdot x = \sum_{k=0}^m c_k \gamma^k(x)$ for $a = \sum_{k=0}^\infty c_k X^k$ in $K[X]$ and $x \in N$. Let $a$ be the annihilator of $M$, then $\gamma^n = 1$, hence the polynomial $X^n - 1$ lies in $a$. Let $F \in a$; by IV, p. 11, Cor. there exist elements $c_0, c_1, \ldots, c_{n-1}$ of $K$ and $G \in K[X]$ such that

Thus we have $c_0 + c_1 \gamma + \ldots + c_{n-1} \gamma^{n-1} = 0$ in $\operatorname{Hom}_K(N, N)$ and since the automorphisms $1, \gamma, \gamma^2, \ldots, \gamma^{n-1}$ of $N$ are distinct, Dedekind's theorem (V, p. 27, Cor. 2) implies that $c_0 = c_1 = \ldots = c_{n-1} = 0$. Finally, we have

By Lemma 4 there exists an element $x$ of $N$ whose annihilator in $K[X]$ is equal to $(X^n - 1) K[X]$. Since the monomials $1, X, \ldots, X^{n-1}$ form a basis of a vector subspace of $K[X]$ supplementary to $(X^n - 1) K[X]$ (IV, p. 11, Cor.), the elements $x, \gamma(x), \ldots, \gamma^{n-1}(x)$ of $N$ are linearly independent over $K$. Since $[N : K] = n$ (V, p. 66, Th. 3), the sequence $(x, \gamma(x), \ldots, \gamma^{n-1}(x))$ is therefore a (normal) basis of $N$ over $K$.