Most statements of the Fundamental Theorem of Galois Theory I've seen online state it for finite, Galois extensions. A Galois extension is defined as one that is both separable and normal. So most authors would begin the theorem with something like this:

Let $F$ be a field and $E$ a finite, normal, separable extension of $F$.

My text states:

Let $F$ be either a finite field or a field of characteristic $0$. Suppose $E$ is a finite, normal extension of $F$.

Are these equivalent? I'm having a hard time seeing this.

If $F$ has characteristic $0$: consider any $a \in E.$ Since $E$ is normal it is also algebraic, so that there is some (irreducible) minimal polynomial $f_a(x)$ over $F$ such that $f_a(a) = 0.$ Since $F$ has characteristic $0$, we know that an irreducible polynomial is separable. Thus, for every $a \in E$ we know there is a separable $f_a(x)$ such that $f_a(a) = 0$, and so $E$ must be a separable extension.

If $F$ is a finite field: note the characteristic of $F$ is some prime $p$. The same general argument applies but for finite fields, it is not true that any irreducible polynomial $f(x)$ is separable. Instead we have the additional condition that $f(x)$ is not of the form $g(x^p).$

If we can show that the minimal polynomial $f(x)$ is not of the form $g(x^p)$, then this would prove that $E$ is a separable extension. I'm also having trouble showing the other direction.


2 Answers 2


Finite fields and fields of characteristic zero are examples of perfect fields, which have the property that every finite extension is separable. For fields of characteristic zero this is fairly clear, while for finite fields the important point is that the Frobenius endomorphism $x\mapsto x^p$ is surjective.

So the statement in your book is less general, and was likely chosen to avoid dealing with separability.

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    $\begingroup$ To see that it is strictly weaker, simply let $F$ be infinite of non-zero characteristic, and $E=F$. $\endgroup$ Dec 9, 2016 at 17:22
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    $\begingroup$ To elaborate a bit, surjectivity of Frobenius means that if $f(x)=g(x^p)$, you can take the $p$th root of each of the coefficients of $g$ to get a polynomial $h$ which will then satisfy $f(x)=h(x)^p$. So such an $f$ cannot be irreducible. $\endgroup$ Dec 10, 2016 at 2:55

In a finite field of order $q$, the product of all monic irreducible polynomials of degree dividing $n$ is $x^{q^n}-x$. Therefore, if you prove that $x^{q^n} - x$ is separable, then so is any minimal polynomial.

You can show that a polynomial $f(x)$ has a repeated root at $x = a$ if and only if both $f(a)=0$ and $f'(a) = 0$ (here we define the derivative of a polynomial formally to be what we know it should be from calculus).

Since the derivative of $x^{q^n} - x$ is $-1$, $x^{q^n} -x $ has no repeated roots. I.e. it is separable.


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