A Separable Extension Is Solvable by Radicals Iff It Is Solvable

Introduction

Polynomial is of great interest in various fields, such as analysis, geometry and algebra. Given a polynomial, we try to extract as many information as possible. For example, given a polynomial, we certainly want to find its roots. However this is not very realistic. Abel-Ruffini theorem states that it is impossible to solve polynomials of degree in general. For example, one can always solve the polynomial for arbitrary , but trying to solve over is not possible. Galois showed that the flux of solvability lies in the structure of the Galois group, depending on whether it is solvable group-theoretically.

In this post, we will explore the theory of solvability in the modern sense, considering extensions of arbitrary characteristic rather than solely number fields over .

Solvable Extensions

Definition 1. Let be a separable and finite field extension, and the smallest Galois extension of containing . We say is solvable if (the Galois group of over ) is solvable.

Throughout we will deal with separable extensions because without this assumption one will be dealing with normal extensions instead of Galois extensions. Although we will arrive at a similar result.

Proposition 1. Let be a separable extension. Then is solvable if and only if there exists a solvable Galois extension such that .

Proof. If is solvable, it suffices to take to be the smallest Galois extension of containing . Conversely, Suppose is a solvable and Galois such that . Let be the smallest Galois extension of containing , i.e. we have . We see is a homomorphism image of and it has to be solvable.

Next we introduce an important concept concerning field extensions.

Definition 2. Let be a certain class of extension fields . We say that is distinguished if it satisfies the following conditions:

1. Let be a tower of fields. The extension is in if and only if is in and is in .
2. If is in and if is any given extension of , and are both contained in some field, then is in too. Here is the compositum of and , i.e. the smallest field that contains both and .
3. If and are in and are subfields of a common field, then is in .

When dealing with several extensions at the same time, it can be a great idea to consider the class of extensions they are in. For example, Galois extension is not distinguished because normal extension does not satisfy 1. That’s why we need to have the fundamental theorem of Galois theory, a.k.a. Galois correspondence, because not all intermediate subfields are Galois. Separable extension is distinguished however. We introduce this concept because:

Proposition 2. Solvable extensions form a distinguished class of extensions. (N.B. these extensions are finite and separable by default.)

Proof. We verify all three conditions mentioned in definition 2. To make our proof easier however, we first verify 2.

Step 1. Let be solvable. Let be a field containing and assume are subfields of some algebraically closed field. We need to show that is solvable. By proposition 1, there is a Galois solvable extension such that . Then is Galois over and is a subgroup of . Therefore is a Galois solvable extension and we have , which implies that is solvable.

Step 2. Consider a tower of extensions . Assume now is solvable. Then there exists a Galois solvable extension containing , which implies that is solvable because . We see is also solvable because and we are back to step 1.

Conversely, assume that is solvable and is solvable. We will find a solvable extension containing . Let be a Galois solvable extension such that , then is solvable by step 1. Let be a Galois solvable extension of containing . If is any embedding of over in a given algebraic closure, then and hence is a solvable extension of . [This sentence deserves some explanation. Notice that is not necessarily Galois, therefore is not necessarily an automorphism of and in general . However, since is Galois, the restriction of on is an automorphism so therefore . The extension is solvable because is isomorphic to and .]

We let be the compositum of all extensions for all embeddings of over . Then is Galois and so is [note: this is the property of normal extension; besides, is finite]. We have which is a product of solvable groups. Therefore is solvable, meaning is a solvable extension. We have a surjective homomorphism (given by ) and therefore has a normal subgroup whose factor group is solvable, meaning is solvable. Since , we are done.

Step 3. If and are solvable and are subfields of a common field, we need to show that is solvable over . By step 1, is solvable. By step 2, is solvable.

Solvable By Radicals

Definition 2. Let be a finite and separable extension. We say is solvable by radicals if there exists a finite extension of containing , and admitting a tower decomposition

such that each step is one of the following types:

1. It is obtained by adjoining a root of unity.
2. It is obtained by adjoining a root of a polynomial with and prime to the characteristic.
3. It is obtained by adjoining a root of an equation with if is the characteristic .

For example, is solvable by radicals. We consider the polynomial . We know its roots are and . However let’s see the question in the sense of field theory. Notice that

Therefore is equivalent to . Then and in are two equations that make perfect sense. Thus we obtain our desired roots. The field gives us the liberty of basic arithmetic, and the radical extension gives us the method to look for a radical root.

It is immediate that the class of extensions solvable by radicals is a distinguished class.

In general, we are adding “-th root of something”. However, when the characteristic of the field is not zero, there are some complications. For example, talking about the -th root of an element in a field of characteristic will not work. Therefore we need to take good care of that. The second and third types are nods to Kummer theory and Artin-Schreier theory respectively, which are deduced from Hilbert’s theorem 90’s additive and multiplicative form. We interrupt the post by introducing the respective theorems.

Let be a cyclic extension of degree , that is, is Galois and is cyclic. Suppose is generated by . Then we have the celebrated “Theorem 90”:

Theorem 1 (Hilbert’s theorem 90, multiplicative form). Notation being above, let . The norm if and only if there exists an element in such that .

To prove this, we need Artin’s theorem of independent characters. With this, we see the second type of extension in definition 2 is cyclic.

Theorem 2. Let be a field, an integer prime to the characteristic of , and assume that there is a primitive -th root of unity in .

1. Let be a cyclic extension of degree . Then there exists such that and satisfies an equation for some .
2. Conversely, let . Let be a root of . Then is cyclic over of degree , and is an element of .

All in all, theorem 2 states that a -th root of yields a cyclic extension. However we don’t drop the assumption that is prime to the characteristic of . When this is not the case, we will use Artin-Schreier theorem.

Theorem 3 (Hilbert’s theorem 90, additive form). Let be a cyclic extension of degree . Let be the generator of . Let . The trace if and only if there exists an element such that .

This theorem requires another application of the independence of characters.

Theorem 4 (Artin-Schreier). Let be a field of characteristic .

1. Let be a cyclic extension of of degree . Then there exists such that and satisfies an equation with some .
2. Conversely, given , the polynomial either has one root in , in which case all its roots are in , or it is irreducible. In the latter case, if is a root then is cyclic of degree over .

In other words, instead of looking at the -th root of unity in a field of characteristic , we look at the root of , which still yields a cyclic extension.

Now we are ready for the core theorem of this post.

Theorem 5. Let be a finite separable extension of . Then is solvable by radicals if and only if is solvable.

Proof. First of all we assume that is solvable. Then there exists a finite Galois solvable extension of containing and we call it . Let be the product of all primes such that but . Let where is a primitive -th root of unity. Then is abelian and is solvable by radical by definition.

Since solvable extensions form a distinguished class, we see is solvable. There is a tower of subfields between and such that each step is cyclic of prime order, because every solvable group admits a tower of cyclic groups, and we can use Galois correspondence. By theorem 2 and 4, we see is solvable by radical because extensions of prime order have been determined by these two theorems. It follows that is solvable by radicals: is solvable by radicals, is solvable by radicals is solvable by radicals is solvable by radicals because .

The elaboration of the “if” part is as follows. In order to prove is solvable by radicals, we show that there is a much bigger field containing such that is solvable by radical. First of all there exists a finite Galois solvable extension containing . Next we define a cyclotomic extension with the following intentions

1. should be solvable by radicals.
2. contains enough primitive roots of unity, so that we can use theorem 2 freely.

To reach these two goals, we decide to put where is a -th root of unity and is the radical of divided by the characteristic of when necessary. This field certainly ensures that is solvable by radical. For the second goal, we need to take a look of the subfield between and . Let be a tower of field extensions such that every step is of prime degree [this is possible due to the solvability of ]. These prime numbers can only be factors of Then in the lifted field extension we do not introduce new prime numbers. Why do we consider prime factors of ? Let’s say is a prime number. If then we can use theorem 4. Otherwise we still have so we use theorem 2. However this theorem requires a primitive -th root to be in . Our choice of and guaranteed this to happen because and therefore a primitive -th root of unity exists in . We can make bigger but there is no necessity. The “only if” part does nearly the same thing, with an alternation of logic chain.

Conversely, assume that is solvable by radicals. For any embedding of in over , the extension is also solvable by radicals. Hence the smallest Galois extension of containing , which is a composite of and its conjugates is solvable by radicals. Let be the product of all primes unequal to the characteristic dividing the degree and again let where is a primitive -th root of unity. It will suffice to prove that is solvable over , because it follows that is solvable by and hence is solvable because it is a homomorphic image of . But can be decomposed into a tower of extensions such that each step is prime degree and of the type described in theorem 2 and theorem 4. The corresponding root of unity is in the field . Hence is solvable, proving the theorem.