Example 2

Now suppose that the field K has characteristic $p\not= 0$. Let $c$ be an element of K. The polynomial $f= X^p-X-c$ is prime to its derivative $f'=-1$, so the algebra $L = K[X]/(f)$ is étale (V, §7, No. 2, p. 37, Proposition 3). We denote the image of X in L by $x$; we have the relation $x^p=x+c$. The sequence $(1, x, . . . , x^{p-1})$ is a basis of L viewed as a vector space over K.

Let P be the additive group of the prime subfield of K; it is a cyclic group of order $p$, generated by the unit element 1 of K. For every $j$ in P, we have $j^p=j$ (V, §1, No. 3, p. 4, formula (4)) and therefore $f(X +j) =f(X)$. Hence there exists an automorphism $\gamma (j)$ of the algebra L characterized by the relation $\gamma (j)(x) =x+j$; moreover, the resulting mapping $\gamma$ is an injective homomorphism from P to Aut$_K(L)$.

Let Ω be an algebraically closed extension of K, and let $\xi$ be a root of the polynomial $f$ in Ω. We have $\xi^p=\xi +c$, hence

by V, §12, No. 1, p. 94, formula (1). For every $j$ in P, there exists a unique algebra homomorphism $\chi_j: L\rightarrow \Omega$ that sends $x$ to $\xi +j$; moreover, every homomorphism from L to Ω is one of the $\chi_j$, and we have the relation $\chi_j=\chi_0\circ \gamma (j)$. The algebra L endowed with $\gamma$ has property (ii) of Theorem 2 of VIII, p. 308 and is therefore a Galois algebra over K.

To describe the structure of L, we must distinguish between two cases:

a) We have $\xi \notin K$. Then the polynomial $f(X)$ is irreducible in K[X] (V, §11, No. 9, p. 93, Example 3). In this case, L is a cyclic extension of K of degree $p$, and $\gamma$ is an isomorphism from P to Gal(L$/K$).

b) We have $\xi \in K$. Then the mapping $\psi :y\mapsto (\chi_j(y))_{j\in P}$ is an isomorphism from the algebra L to the product algebra $K^P$; moreover, $\psi \circ \gamma (k)\circ \psi^{-1}$ is the automorphism $(x_j)_{j\in P}\mapsto (x_{j+k})_{j\in P}$ of $K^P$ for every $k\in P$.