Spring 2016 - Problem 6

Galois theory

Let KK be a field of characteristic p>0p > 0. For an element aKa \in K, show that the polynomial P(X)=XpX+aP\p{X} = X^p - X + a is irreducible over KK if and only if it has no root in KK. Show also that, if PP is irreducible, then any root of it generates a cyclic extension of KK of degree pp.

Solution.

Notice that over the splitting field of PP, if α\alpha is a root of PP, then for iZ/pZi \in \Z/p\Z, we have

P(α+i)=(α+i)p(α+i)+a=αp+ipαi+a=αpα+a=P(α)=0.\begin{aligned} P\p{\alpha + i} &= \p{\alpha + i}^p - \p{\alpha + i} + a \\ &= \alpha^p + i^p - \alpha - i + a \\ &= \alpha^p - \alpha + a \\ &= P\p{\alpha} \\ &= 0. \end{aligned}

For the third equality, we used the fact that (Z/pZ)×Z/(p1)Z\p{\Z/p\Z}^\times \simeq \Z/\p{p-1}\Z, i.e., ip=ii^p = i. Thus, if PP has a root in KK, then it automatically has all its roots α,α+1,,α+(p1)\alpha, \alpha + 1, \ldots, \alpha + \p{p-1} in KK as well, so PP is irreducible if and only if PP has no roots in KK.

For the second claim, note that the calculation above shows that the splitting field of PP is K(α)K\p{\alpha} where α\alpha is any root of PP. Moreover, it also shows PP is separable, so K(α)/KK\p{\alpha}/K is Galois. Hence, Gal(K(α)/K)=[K(α):K]=p\abs{\Gal\p{K\p{\alpha}/K}} = \br{K\p{\alpha} : K} = p, so Gal(K(α)/K)Cp\Gal\p{K\p{\alpha}/K} \simeq C_p, which was what we wanted to show.