Example 1

Let $n$ be a strictly positive integer, prime to the characteristic exponent of K. Suppose that the group $\mu_n$ of $n$-th roots of unity in K has order $n$. Then for every divisor $d$ of $n$, the group $\mu_d$ of $d$-th roots of unity has order $d$. Let $a$ be a nonzero element of K, let L be the algebra $K[X]/(X^n-a)$, and let $x$ be the class of X in L. The sequence $(1, x, . . . , x^{n-1})$ is a basis of the vector space L over the field K, and we have $x^n=a$. Moreover, the polynomial $X^n-a$ is prime to its derivative $nX^{n-1}$, so the algebra L is étale (V, §7, No. 2, p. 37, Proposition 3). For every $\zeta$ in $\mu_n$, the automorphism $P(X)\mapsto P(\zeta X)$ of the ring K[X] defines, by passing to the quotients, an endomorphism $\lambda (\zeta )$ of L because we have $(\zeta X)^n-a= X^n-a$; it is an automorphism. We have

for $0\leqslant i < n$. The mapping $\lambda :\zeta \mapsto \lambda (\zeta )$ is an injective homomorphism from $\mu_n$ to Aut$_K$(L), and the ring of invariants of the group $\lambda (\mu_n)$ in L is equal to $K\cdot 1$. Since the cardinal of $\mu_n$ is equal to $n= [L : K]$, the algebra L endowed with the action of $\lambda$ is a Galois algebra (VIII, p. 308, Theorem 2, (v)).

Let $r$ be the least strictly positive integer such that $a^r$ belongs to $K^{*n}$; it divides $n$, and there exists an element $b$ of $K^*$ such that $a=b^{n/r}$. Then (V, §11, No. 8, p. 91, Remark) the polynomial $X^r-b$ is irreducible, and we have $X^n-a=\prod_{\zeta\in\mu_{n/r}}(X^r-\zeta b)$. Let E be the field $K[Y]/(Y^r-b)$, and let $y$ be the class of Y in E. There exists an isomorphism $\theta$ from $\mu_{n/r}$ to Gal(E$/K$), characterized by the relation $\theta (\xi )(y) =\xi y$ (V, §11, No. 8, p. 91, Example 3). We then verify that the Galois algebra L is isomorphic to the $(K, \mu_n$)-algebra Coind$^{\mu_n}_{\mu_{n/r}}(E)$.