Galois theory-some concepts

The Galois theory on finite algebraic extensions of perfect fields are fairly well understood(at least for me). Yet, when it comes to imperfect fields, things begin to get complicated. I write this post mainly to clarify these closely related, yet not so well distinguished concepts(at least for myself).

First of all, what is a perfect field? Before talking about that, perhaps it is proper to ask what it means by a polynomial being separable. Suppose that k is a field, and P(X)\in k[X] is an irreducible polynomial. So, we say that P(X) is separable if it has no multiple roots in an(and hence in any) algebraic closure \bar{k} of k. There is an equivalent way to decide if P(X) is separable or not, that is, take the formal derivative P'(X) of P(X) and see if P(X) and P'(X) are coprime. If they are coprime, then P(X) is separable, if not, then P(X) is not separable. In this way, we see the separability of a polynomial does not depend on the field containing the coefficients.

P(X) being separable means that all its roots are simple. This is one way extreme. The opposite extreme way is that P(X) has only one root, that is the multiplicity of its root is equal to the degree of P(X). One classic example is this: take k=\mathbb{F}_p(t) the rational function field of one variable over \mathbb{F}_p, and take P(X)=X^p-t. We say that this kind of polynomials are purely inseparable. Using (in)separable polynomials, we can define perfect fields.

We say that k is a perfect field if all the irreducible polynomials in k[X] is separable(the word ‘perfect’ in some sense reflects our nice dreams in doing mathematics, we simply want all the irreducible polynomials have simple roots. Multiple roots are always hard to deal with). k is non-perfect if there is some irreducible polynomial in k[X] inseparable. The fields \mathbb{Q},\mathbb{F}_p are examples of perfect fields. And \mathbb{F}_p(t) is an example of imperfect field.

Note that for any algebraically closed field k=\bar{k}, k is perfect, simply for the raison that the only irreducible polynomials are of degree one, so they are always separable.

Now a problem arises: given a field k, if it is not perfect, yet we know that \bar{k} is always perfect, so is there some minimal field containing k which is perfect? Note that, if two fields k_1,k_2 all contain k as a subfield, if they are both perfect, then their intersection(which is still a field, of course) k_1\bigcap k_2 is still perfect: an irreducible polynomial P(X)\in k_1\bigcap k_2[X] is coprime to its derivative if and only if it is separable. Yet the first part does not depend on the base field. So, in this way, we can take the intersection of all the perfect fields containing k, that is k^p=\bigcap_{k\subset k'}k' where k' is a perfect field. So, this field k^p is a perfect field, and it is the smallest field containing k, we call k^p (the) perfect closure of k.

Now return to the example k=\mathbb{F}_p(t). One way to get the perfect closure of this kind of field is to take all the p^n-th root of the inseparable element t(indeed, all the problems rise due to t), that is k^p=\mathbb{F}_p(t,t^{1/p},t^{1/p^2},...). Sometimes we denote this also as k^p=k^{p^{-\infty}}.

Another important concept is the separable closure. For an algebraic extension k\subset K, we say this extension is separable if all the minimal polynomials associated to the elements of K are separable(over k). If we define a\in K to be separable(over k) if the minimal polynomial(with respect to the extension k\subset K) associated to it is separable(over k), then we can say that k\subset K is a separable extension if all elements in K are separable.

Note that, we can say that k is perfect or not, yet we must say that the extension k\subset K is separable or not.

We can show that the subset of \bar{k} consisting of separable elements over k is again a field, and we call this field the separable closure of k, we write it as k^s.

An important remark is that, for two successive extensions, k\subset L\subset K, an element a\in K can be inseparable in k, yet it can be separable in L(we just have to take the example k\subset L=K). Meanwhile, if a is separable in k, then it is also separable in L. The counterpart is worth written out: if a is inseparable in L, then it is also inseparable in k; if a is separable in L, it is possible to be inseparable in k.

So, for any separable extension k\subset K, we have that k^s=K^s.

The perfect closure and separable closure can be very different. For example, for any field k of characteristic 0, we have that k^p=k since k is always perfect. Meanwhile k^s=\bar{k}. Another example, k=\mathbb{F}_p(t), we knew that k^p=k^{p^{-\infty}}. Yet for any element a\in k^p, we see that the p-th root a^{1/p} of a exists. So, X^p-a=X^p-(a^{1/p})^p=(X-a^{1/p})^p is not separable. So, here we know that the intersection of k^s and k^p is trivial.

We can examine the successive extension k\subset k^s\subset \bar{k}. Clearly k\subset k^s is a separable extension. We can also show that k^s\subset \bar{k} is purely inseparable. The latter means that all the polynomials associated to an element a\in\bar{k}-k^s is purely inseparable. In fact, we can assume that char(k)=p, and the minimal polynomial F(X) of a is not separable, which means that F'(X)=0. So, we have that F(X)=G(X^p) with G(X)\in k[X]. So, since deg_X(G)<deg_X(F), we have that deg(a)>deg(a^p). Thus we must have that a^{p^n}=b\in k^s. This means that X^{p^n}-b takes a as a root, yet this polynomial is purely inseparable, thus a is purely inseparable, which as a result, shows that k^s\subset\bar{k} is purely inseparable.

Another problem remains: what about k\subset k^p\subset \bar{k}? Is k^p\subset \bar{k} separable? In general, we should assume that the extension is normal, some discussions on this kind of problem can be found here, here . These discussions are rather interesting.


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