## Galois Theory: The Primitive Element Theorem

The fundamental theorem of Galois theory immediately implies several nifty results about Galois extensions, which with a little work extend to a greater class of extensions, typically separable extensions. Here I will show that every separable extension L/K admits a primitive element, that is an element a of L such that L = K(a).

First, the normal closure of a separable extension is itself separable. To see why, recall that the normal closure of L/K is the splitting field of all the minimal polynomials of the elements of L. L/K is separable, so each a in L is separable over K; but separability is a property of the minimal polynomial of the element, so all other roots of m(a) are separable, and the normal closure is generated by separable elements.

In addition, the normal closure of a finite extension is finite, since if {a1, …, a(n)} is a basis for L over K, then it’s enough to take the splitting field of the finite set of polynomials m(a(i)).

Second, since the intermediate fields of a Galois extension correspond to subgroups of a finite group, there are only finitely many intermediate fields. This property then extends to extensions that are merely separable, since if M/L/K is a tower of extensions then every intermediate field of L/K is intermediate between M and K.

And third, an element not contained in any proper intermediate field is primitive, because K(a) is an intermediate field strictly containing K for a not in K. If K is finite and L has p^n elements, then just take one of the roots of the polynomial x^(p^(n – 1)) – 1, which will generate all the others. Otherwise, note that every proper intermediate field is a proper subspace of L, and no finite vector space over an infinite field can be covered by finitely many proper subspaces.

Proving that last assertion is an induction argument on the dimension of the vector space K^n. If n = 1, then a proper subspace is the trivial subspace consisting only of the origin, so the theorem obviously holds. If n = 2, then a non-trivial proper subspace is a line generated by scalar multiples of some point (a, b). No line can contain two points of the form (1, a) because then it will necessarily contain (0, ab), hence (0, 1), (0, a), (1, 0), and finally the entire space. But K is infinite so there are infinitely many points of that form, proving the theorem for n = 2.

More generally, suppose the theorem is true for all dimensions less than n. It’s enough to find a proper subspace of dimension n – 1 in K^n that isn’t contained in any of the proper subspaces we’re covering the space by; by a dimension argument, this is equivalent to showing the subspace of dimension n – 1 isn’t one of those proper subspaces. So it suffices to show that there are infinitely many subspaces of dimension n – 1.

But if a and b are distinct, then the subspaces spanned by {(1, 0, 0, …, 0), (0, 1, 0, …, 0), (0, 0, 1, …, 0), …, (0, …, 1, 0, 0), (0, …, 0, 1, a)} and {(1, 0, 0, …, 0), (0, 1, 0, …, 0), (0, 0, 1, …, 0), …, (0, …, 1, 0, 0), (0, …, 0, 1, b)} are distinct, by a similar argument to the one that works for K^2. That’s enough to prove the theorem.

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