Solvability by Radicals

Here comes the best known and most important application of Galois theory: to judge whether an equation is solvable by radicals.

One of the most common misconception by folks is that Galois proved equation of degree greater than is not solvable. Since the proof had been given by Abel and Ruffini for near twenty years ago before Galois's work. In fact, Galois proved the sufficient and essential condition, that an equation related to a polynomial is solvable by radicals iff the Galois group of the polynomial is solvable group.

The true insight and meaning of his work is, it has fostered the mind of mathematicians that one can also study the structure of a mathematical objects to solve problems.

And in this text, we will demonstrate what does it mean by studying structure to solve problem. First we will establish the theory about structure of Galois group of polynomial whose roots are radicals. With these theory as building blocks, we clarify what does it truly mean by "solvable by radicals", and construct the model to characterize what Galois group will look like if it's solvable by radicals. The characterization turns out to be a sufficient and essential condition eventually.

We will focus on the solvability by radicals of equations over characteristic fields. And to make it more useful in practise, we also come up with some methods of calculation for Galois groups.

Cyclotomic extensions

Let's first take a look at the splitting field of where is any field. The equation in the form of is called a cyclotomic equation, it got its name as when , the roots will divide the unit circle on complex plane, in -sect way. The "cyclo-" means the circle, and the "-tomic" means to cut something.

Galois group of cyclotomic extensions

First, when is separable, is Galois, and all of its roots are distinct in such way. While clearly the primitive -th root generates a cyclic group of all roots of , we need to prove this still holds in general field , where all distinct roots of form a cyclic group under multiplication.

Let be the set of all distinct roots of . And since , and , is an finite abelian group under multiplication. The group has no duplicate cyclic -group factor of the same order, otherwise the polynomial will have solutions. So the group is cyclic, and we shall denote one of its generator as , and call it a primitive -th root in . The theory below must be independent of the choice of generator.

To ensure is separable, clearly we can evaluate over . Whenever by doing division we have . It's obvious that when there's or , and then will be separable.

Contrarily, if is inseparable, then it must be the case of , let . When is odd prime, there's ; when , there's . Given that , the roots of are this case.

Now consider the extension , is the splitting field of , which is separable, thus is a finite Galois extension, and usually called a cyclotomic extension. For any valid field automorphism , if there's , then , so every valid can be represented as and . And for any valid , we have , so is abelian.

But clearly there's where is the -th cyclotomic polynomial and embeds into the base field if , or perform a modulo- reduction and then embed it into the base field if , so there may be many that are not conjugate to over . And since , is valid only if is another primitive -th root. We know every such that is a member of . So there's an injective group homomorphism from to , mapping such that to modulo-.

Please notice that it's not saying that all primitive -th roots are conjugate under any circumstance, nor the group homomorphism to is surjective! A prominent example is , given that is normal and , there's no irreducible polynomial of degree greater than . And for every there's paired with it, multiplying into an irreducible polynomial over . One can easily find the image of in is , and is not surjective when is sufficiently large.

Cyclotomic extensions over rational numbers

However, under certain circumstance the homomorphism from the Galois group to can still be surjective and thus a group isomorphism. An classical and important case is . The extension field is sometimes called the cyclotomic field.

Remember that -th cyclotomic polynomial is defined as , whose coefficients are integers. It contains the primitive -th root of unity, and if it's irreducible over , the Galois group must be of size in order to cover every distinct primitive -th root of unity, and thus the group homomorphism is surjective given that it's already injective and .

The following proof is due to Dedekind. Assume is reducible, given that it's monic, we can write such that is the minimal polynomial of over and is a monic primitive polynomial.

We claim that is also a root of and thus conjugate to over , where is prime and . To prove, we assume it's the contrary that is a root of . Let , clearly is a root of when evaluated in , so we have over . Noted all coefficients of are integers by Gauss's lemma, so we can perform the modulo- reduction on them, and there's:

Let be an irreducible factor of , then since is also a UFD. In this way, we have over and is not separable over . But consider , and when , is actually separable over , which is a contradiction. In this way, where is prime and is also a root of , and conjugate to over .

Consider any primitive root of unity , which has , factorize and we can find , so starting with we can see all of the are conjugate to over , and this rationale is applicable to any primitive -th root . So in this way all primitive -th root are conjugate over , and is irreducible. Noted that we've prove all cyclotomic polynomial over are irreducible by now.

In this way, the group homomorphism from to is surjective and thus a group isomorphism.

Cyclotomic extensions over finite fields

On the other hand, let's look at the case of where . The rationale applied to is not applicable to since it requires feasibility of modulo- at arbitrary prime distinct from . So will the group homomorphsm from the Galois group to also be surjective?

Clearly is a finite field, let the minimal polynomial of be , and there's where . We have analyzed such case in the discussion of automorphisms on finite fields, that , so all elements conjugate to are , and the Galois group is . Please notice we have , which implies it's only possible the case when we consider the group operations of . So the image of is , which is cyclic with equal to the order.

And this is the problem, since is not cyclic in general. An easy example to verify is , whose members are , while , , , so it can't be cyclic. In this way, the group homomorphism from to is not surjective in general.

Meantime, we can also derive is not irreducible in general, where is the modulo- reduction: inspired by the structure of , we can see there's . In fact, we can furtherly derive where is irreducible iff . It's actually quite intuitive when we compare it with the case of , where is of degree , clearly the group action of on will not cover all of the roots if we don't have .

The same case also holds for , since we have , while any normal subgroup of is cyclic.

Simple radical extensions

Then let's take a look at the splitting field of . We claim that must also split in .

The proof is simple, let splits as , pick up arbitrary , clearly it must fulfil . And by dividing both sides by we have . By Substituting we have . Noted that there's , and thus must split in .

Let the splitting field of be , clearly there's a tower and is an intermediate normal extension. And in order to apply the Galois theory, we can assume to be separable, which is clearly the case that or . In this case, is cyclotomic extension with . By studying the structure of , we can also derive the structure of .

On the other hand, on the way of growing up to the splitting field , we are inevitably adjoining the roots of to every intermediate field . The extension with is called a simple radical extension, which we are also going to cover later, alongside with some criteria.

Galois group of its splitting field extension

Actually we want to explicitly place the splitting field of to make our discussion clear. In the tower where is the splitting field of , the characteristic of fulfils or , is arbitrary field that splits and maybe larger than , is the splitting field of . Clearly is cyclotomic and Galois, and is Galois given that is splitting field of a separable polynomial. We claim that Galois group is cyclic group of order .

First, it's obvious that , and thus must split as where is any element such that . For any element , it's clear that , so we have , thus means every automorphism is defined as .

Then, consider two automorphisms in defined as , we have , so we can see there's a group homomorphism from to defined by mapping to modulo-.

Again, the homomorphism is clearly injective, but it's generally not surjective, since need not to be conjugate to over in general. Consider the roots of , where , we can decompose into and both factors are irreducible over . So is conjugate to while is conjugate to over .

Finally, even though the group homomorphism need not to be surjective, but any subgroup of are cyclic, so can only be cyclic of order in this case.

And if the group homomorphism wants to be surjective, it must be the case that , where ought to be irreducible. And if irreducible over , there must be and it's only possible the case that .

Let be one of the root of , then we can see , thus and is a simple radical extension now ¹.

Please notice and be cautious that although and are Galois in our specified tower of extensions, need not to be Galois.

Cyclic extensions and simple radical extensions

An Galois extension is said to be cyclic when is cyclic group. Clearly by our discussion in previous section, by adjoining the root of separable to a field that is a sufficiently large field to split , we obtain a simple radical extension that is cyclic.

And here comes potentially the most important part of this text. Is any cyclic extension a simple radical extension? And this is the case: we would like to show in the tower where ² is cyclic with , the characteristic of fulfils or , the splitting field of , then can be realized as a simple radical extension by adjoining that is a root of to .

First, find out any such that . We are guaranteed to find one of them, since is finite Galois, we can choose the primitive element of over to be .

Then, we can let , the proof is done by considering a delicately constructed polynomial called the Lagrange resolvent, in the form of:

If we let any to act on , it becomes:

Noted that for each , we have (for the right hand side, just consider any how we expand ). If we are able to show there's some such that , then by and the orbit of under the action of must have points, it's only the case and is a simple radical extension by letting .

To show, just consider the summation:

Obviously it's impossible for all to be in , and by now we've proved is a simple radical extension under such settings.

This allow us to decompose cyclic extension , where (prime can duplicate) and splits , into a chain of intermediate simple radical extension with prime degree.

To prove, first we know must have unique cyclic subgroup of order , we may denote each of them as . By fundamental theorem of Galois theory, we know there're a chain of intermediate subfields as with . Clearly each is normal by with , which is a cyclic of order . And we know by . So we can apply the theorem we've just proved above to conclude that is a simple radical extension with prime degree with for some .

Radical extensions and solvable groups

With the knowledge of cyclotomic extensions and simple radical extensions, we are able to model the solvability by radicals.

A radical extension is every pair of extensions in the tower of:

Where every intermediate extension fulfils for some . That is, the extension field is obtained from adjoining a radical to the base field, which might be a cyclotomic extension (when ) or a simple radical extension (when ). The tower itself is conventionally known as the root tower.

An polynomial is said to be solvable by radicals if over its splitting field there's a finite extension , so that is a radical extension. Clearly all roots of will be expressible in terms of radicals appeared in the root tower of then.

On the other hand, a solvable group is a group with the following subnormal series:

Where every intermediate quotient group is a cyclic group of prime order .

In this chapter, we will prove the Galois's theorem over the field of characteristic , that Galois extension has an extension field over so that is radical extension iff is solvable. Such sufficiency and essentiality creates a strong bound between Galois extensions that can be embedded into radical extensions and solvable groups.

Subnormal series and solvable groups

We will begin with discussing some basic properties about subnormal series, and applying them on solvable groups. These properties are essential to our proof of Galois's theorem later.

An alternative definition for solvable group is, in the subnormal series of , every intermediate quotient group is cyclic. However, we can prove such a group can be broken down to match our definition.

First, we would like to show a refinement in the form of exists when is not simple. By the correspondence theorem we know there's . To show , first by we know and thus . By the normality of , we have . Combining them together we have , and thus . With similar method we can derive , and thus . In this way, we've shown a refinement exists between and in the subnormal series when is not simple.

In fact, we would like to call the subnormal series such that every is simple the composition series. And back to our case, if is cyclic of order , then by the theory of cyclic group there exists a normal subgroup of order so that with being cyclic of order . Clearly this can go on until there's a composition series with being cyclic of prime order. In this way, we are okay with being cyclic and is solvable then is solvable, as it can be broken down to match our initial definition.

Then, we would like to show that any subgroup of a solvable group is also solvable.

Again, we would like to prove this is due to the property of subnormal series, that for any subgroup , by intersecting it with every component normal series of , we obtain the normal series of :

Where is isomorphic to a subgroup of all along the way.

First, it won't be hard to see normalizes , and by the second isomorphism theorem there's:

Then, since , we can see , and actually we have:

And since , we have , and we are done.

When we put it back to our case of solvable group and its subgroup , we can see has a normal series with being either trivial or cyclic, for they are isomorphic to a subgroup of , and thus is also solvable.

Next, we would like to show when is solvable then is also solvable.

Once again, this is due to the property of subnormal series, that for any group homomorphism , by applying to every every component of 's subnormal series, we have:

Where is isomorphic to a normal subgroup of .

First, let's ensure it's the case of . This can be done by simply checking .

Then, consider another group homomorphism defined by , it won't be hard to find the homomorphism is surjective by testing every point in the image, and by first isomorphism theorem is isomorphic to a normal subgroup of .

Now back to our case of solvable group and normal subgroup , which defines its canonical homomorphism . For any in the subnormal series , which can be recovered into and by the third isomorphism theorem, we know it's isomorphic to a normal subgroup of , which is must be trivial or cyclic. Thus we can yield is also solvable.

Finally, we would like to show when is solvable and is solvable then will also be solvable. Let there be subnormal series of annotated by subgroups of under correspondence theorem:

To show , since , it can be interpreted as , this can be used to derive that , and with similar method we can prove , and thus . By using the third isomorphism theorem we have . In this way, we build subnormal series over such that each is cyclic, and adding to is solvable, the whole is solvable.

We will soon apply these properties in our proof.

Galois's theorem

Here comes the part of proof of the Galois's theorem. Let the root tower of be:

Where is cyclotomic of degree or simple radical by adjoining root of .

First we would like to show the Galois group of is solvable group.

We clearly know the structure of cyclotomic extensions and the simple radical extensions with base fields splitting , but we don't know what the structure of a simple radical extension without its base field splitting . To solve, let , and we adjoin to each field in the tower so that we have:

All intermediate extension is either trivial or simple radical extension with degree . When is cyclotomic or simple radical extension with coinciding with , e.g. for , . Otherwise by our theory of simple radical extension, splits and thus is normal and simple radical extension, with cyclic of order .

One thing should be noticed is that is not necessarily normal. A good example is , noted that is the root of , another root of it is , but it's clearly not in , so the polynomial will not split there. We will see a Galois extension is mandatory for proving our problem later.

To solve, let , and we will create normal clousure of . First let the minimal polynomial of over be , then we evaluate . The least common multiple of is defined as and , where and . Finally, we grow into the splitting field of , which must cover the path of first. In this way, the extension is a normal extension and thus Galois for characteristic base field .

Then, let for . For with , we can see there's . Clearly we also have , by the normality of , and . In this way, we can extend into shown as below:

(Noted that many of these intermediate extensions are trivial, but it doesn't hurt.)

It's clear that it's just lingering the similar cases of , that each is either trivial or simple radical extension, when it's the latter case is cyclic. So let's descend down from , where is cyclic and thus solvable, and we can see there's all along the way, given that both and are solvable, is also solvable. The final case is , is isomorphic to a subgroup of , and is thus solvable, adding to that is already solvable, is a solvable group.

Finally, we can see on another path there's , where is a normal subextension. So we have , and since any quotient group of a solvable group is also solvable, we can derive that has to be solvable.

The proof of the converse that if is solvable then can be embedded in a radical extension is even simpler.

First, fix a subnormal series of that:

Let , according to the diamond rule we have:

And there's . Although we don't know what exactly is, but we do know it's a solvable group, and so is . Let its subnormal series be:

Descending from , we evaluate the fixed field for each Galois group, down to , in a tower as below:

Descending from , there's , where is cyclic of prime order . And since the corresponding primitive -th roots have already been in , cyclic extension is also a simple radical extension.

Finally, we can see is a radical extension: by adjoining the primitive -th roots first, and then through a series of simple radical extension it grows up to (again, need not to be normal). And is an extension field over .

In this way, we've proved the Galois's theorem, that there's an extension field over such that is a radical extension iff is a solvable group.

The Jordan-Hölder theorem

There's still one thing a little bit itchy, could be a group that is complicated enough and have multiple subnormal or composition series, that we can't say it's not solvable even if we witness a composition series that is not solvable.

And this is what role the Jordan-Hölder theorem comes into playing. The theorem states a general fact about composition series, that for a group in which every composition series is finite, for composition series among them:

It must be the case that and such that all along the way.

The theorem is done by mathematical induction on any subgroup that appears earlier in any of the composition series of , in which the Jordan-Hölder theorem has been held. This is completely feasible since every composition series of must also be finite, otherwise concatenate the portion of composition series appears after yields an infinite composition series of , which is a contradiction. And by recursively doing so reduces down to the case of simple group over , which has unique composition series as .

For the case of , given that the Jordan-Hölder theorem has been true for it, and the subsequence from up to and the subsequence from up to specify two composition series of , it must be the case of and thus . For the permutation group, there must be specifying the permutation of quotient groups of , and lifting it to by fixing the letter specifies the permutation associated with the composition series we came up with in this case.

For the case of , consider the subgroup . Noted that , which means is normal in the whole group, by the second isomorphism theorem there's . The has to be , otherwise given that , it won't be hard to find it's the case of . Combining with the fact that due to under set inclusion, we can see there's a subnormal series where every intermediate quotient group is non-trivial, but has been expected to be simple, which is a contradiction. Meantime, we know is also a simple group.

Let's expand the subnormal series of :

Since the Jordan-Hölder theorem has been true for , it has to be the case of .

With analogous rationale, we have , is also simple, and the subnormal series of can also be expaned into:

In which the Jordan-Hölder theorem has been true for , with . So it has to be the case of and thus .

For the permutation part, we can see there's permutation such that , similarily there's such that , combining with the fact that and , we can conclude that by lifting to by fixing the letters , the permutation is the permutation between the two composition series we came up with under this case.

In this way, we can see once all of a group's decomposition series has finite length, which is clearly true for finite groups, by the Jordan-Hölder theorem such decomposition is unique, up to the permutation of quotient groups.

And back to our case, if we are to see a group is actually a solvable group, then there is a composition series in which every intermediate quotient groups are cyclic, and by the Jordan-Hölder theorem, any other decomposition series should render it a solvable group. So once we've got to determine a group not being solvable by a composition series, it will never be solvable under any other composition series.

Calculation of Galois groups

All these efforts, profound visions by predecessor mathematicians, and beauty of algebra, without feasibility of putting into application, will forfeit. So in case of this, we should ramble some methods for calculating Galois group for the splitting field over base field a polynomial, or simply the Galois group of a polynomial, over characteristic fields, as far as we've discussed and proved.

Symmetric group as Galois group

We claim the Galois group of an irreducible polynomial with prime degree is isomorphic to symmetric group if it has non-real roots in . Clearly we need to excavate roots of in so that it's eventually.

For convenience, let be the roots of , and we let by denoting any as .

First, specially in the group of , the element of order must be a -cycle. Assume is composed of cycles of length so that (identity cycle included by letting ), by the theory of symmetric groups we know . And if there's any chance that , by the Euclid's lemma there has to be , which has to be and has to be a -cycle.

When we grow up to the splitting field , without losing generality we claim the first step is to adjoin a root to , which is a simple extension with , by the Tower law there's , and thus . By the Cauchy's theorem there has to be a subgroup of order of , which has to be cyclic and generated by a -cycle. We can write since can rearrange those to match up with the -cycle generator.

Then, without losing generality we can let be the non-real roots in , since again we can rearrange one of the non-real root to label , then for the other non-real root labelled at , noted that has adjacent to and is also a -cycle, generating the same group as . Once again, we can use some rearrangement if needed. By applying the diamond rule we have:

And we have ³. Let's inspect the extension first. The irreducible polynomial of must be of degree , by and . And clearly there's . Then splits in , and under the group action of we know the other root of must be the complex conjugation of , which is also the root of . And if is not the complex conjugation of , then may have non-real roots instead of , which is a contradiction. So in this way we know , where is defined as , and . When is mapped to , it becomes , and , which is when it's viewed as permutation on roots. In this way, we have and thus .

Finally, consider the group operations of and in . First , , where does not contain the letter (but may contain ). Then noted that the permutation are in are now in reversed order of , so it's , and thus . So . And finally . This is sufficient to build every permutation in , and thus .

As a real application, recall that we've mentioned that is not solvable by radicals this earlier, we would like to show it's exactly the case that the Galois group of is .

First, by Eisenstein's criterion with we know is irreducible over . Then, to determine the number of roots, noted that has only two real roots and , rendering it changes its monotonicity twice over . And around the extrema we see , so it has roots in the interval of , leaving roots to be non-real. In this way, the Galois group of is , which means is not solvable by radicals.

Resolvent methods

Another way to evaluate the Galois group is to utilize the resolvents, whether it has rational roots reveals the group structure of the Galois group itself.

For a multivariate polynomial where is a UFD of characteristic , the group action of on it is defined by:

And for a polynomial , whose roots are where is a splitting field over , a resolvent relative to and is defined as:

Where is the orbit of under the group action of . Noted that every point in the orbit will be taken only once.

Of course, the resolvent method will not be useful if we will have to solve the equation prior to using them. In fact, The resolvent is expected to be a deliberatedly designed polynomial that exposes certain symmetry, so that we can use merely the coefficients of to calculate them.

To guarantee, we will need to show the theory of symmetric polynomials first.

A symmetric polynomial is a multivariate polynomial that is fixed by the group action of .

Consider the polynomial which has roots where is a splitting field over , there will be:

And by Vieta's formula we have:

Define , so that we can write . All of these must be fixed by the group action of , by observing that:

Those polynomials realizing coefficient as polynomial of the roots of are called the elementary symmetric polynomial.

It won't be hard to find all symmetric polynomials forms an integral domain and is subring of . And we would like to show the Fundamental Theorem of Symmetric Polynomials, that all symmetric polynomials in are generated by ring operations of elementary symmetric polynomials.

To prove, first let's sort the terms of the symmetric polynomial : for every term , sort them by the lexical order of . If is the term that is lexically maximum, there must be . Otherwise for any , consider the permutation , since is symmetric, . Noted that is lexically larger than , which is contradiction to its being lexically maximum. Similarily, we can derive that any term in must fulfil , otherwise just apply to and it's again a contradiction.

Then, we claim that for all symmetric polynomial, the lexically maximum term of is . This is due to every term , in must fulfil , while there's all the way (tips: consider maximizing the degree of by multiplying appropriate terms, but every term in contains for at most once, while degree must be distributed among these terms). So the lexically maximal configuration is to let , which can be realized by multiplying the leading for times.

Next, for with lexically maximum term and with lexically maximum term , the lexically maximum term of is . Since any term in is lexically surpassed by at and or (let then), and any term in is lexically surpassed by at and or (let then), their product is lexically surpassed by . at or .

Finally, consider the polynomial , it also has the same lexically maximum term as . By doing subtraction we can see the lexically maximum term of is strictly lower than 's. By iteratively doing such reduction we will find a way to represent using elementary symmetric polynomials. The process of reduction will eventually halt after at most steps, since all terms in and are capped at degree , and the iterative subtractions are strictly monotonically decreasing in lexical order.

And back to our resolvent, let , clearly there's:

As is required by the axiom of group actions. And consider:

Where are elementary symmetric functions. Let , we have:

So every is a symmetric polynomial, convertible from elementary symmetric polynomials of , and can be evaluated by using merely coefficients of by Vieta's formula when the roots are substituted in, so is .

So what does a resolvent , have to do with Galois group of ? Well, Let be the stabilizer subgroup of , and injective group homomorphism by . We would like to show:

1. If , then has a rational root.
2. If has a rational root that is not duplicate root, then .

On one hand, notice the stabilizer subgroup of is , thus must fix . While we can see there's:

Thus we know , and there's some such that , rendering it a rational root of .

On the other hand, assume the root of be , then there must be as it's in . And for any , we have:

But this will render a duplicate root, but we've asserted that there's no duplicate root. So it's only possible the case that , , and finally .

Please notice it's always the case that , which is an implication of 's capability to be embedded into .

And to search for rational root, we will sometimes need to multiply the least common multiple of denominators of the coefficients of in order to convert it into a polynomial over . Then by Gauss's lemma its reducibility over is the same as the one over , and we will just need to factorize the leading coefficient and the constant term, then make up possible combinations to test for rational roots.

By now we've proved why the resolvents are capable of excavating structure of Galois group, and we are going to have a look at some real world examples.

Subgroup of alternating group as Galois group

The first example we want to have a look at is determining whether the Galois group is a subgroup of alternating group.

First, we deliberatedly construct a polynomial as below:

When we apply to , we can see how factors in transform, as is listed below:

1. .
2. .
3. .
4. .
5. .

Where denotes they swap their position, while represents what they transform into.

So we have , and is fixed by all even permutations, so the stabilizer subgroup of is .

When we put it into resolvent, we can see:

Where is called the discriminant of , we may also denote it as for convenience. And we can clearly see if would like to have a rational root, it must be the square root of over .

The discriminant itself, can be used for detecting whether there's duplicate root in , as indicates there's some immediately.

The discriminant we maybe most familiar with is:

And since we've known the origin and principle of discriminant, the concrete calculation is just a brute force work and can be completely left to software. Or you can code one on your own, as all principles are clear by now.

The calculation of discriminant can also be done effectively by resultants, however this is completely technical and we will not have to cover it here.

Please notice that without rational root of does not necessarily imply the Galois group is , it just imply there're odd permutations. For example, among the subgroups of there's Frobenius group which is not a subgroup of , and is solvable. A careful case study of subgroups of is required prior to application.

Subgroup of dihedral group as Galois group

As another example of resolvent method, we want to create a resolvent for detecting if a polynomial of degree can possess a Galois group isomorphic to a subgroup of dihedral group . This is completely possible, since the Galois group of is a dihedral group.

Let's analyze the group structure first. The dihedral group is generated by rotation and mirror . Spend some time on calculating we can see the group has these members: .

Then, we will try to brainstorm some polynomials, which is fixed by a subgroup of , and can't be fixed by . And we can see are the cosets of in .

Let's try , this is fixed by , and moved by , similarily is the case of . Both of them are not fixed by , so they can be preserved. But if we add them together, we can see is fixed by the whole , and and does not fix them. So is the desired polynomial to generate resolvent.

I don't know whether there's some more intuitive and effective way to brainstorm a polynomial that has specified stabilizer subgroup under symmetric group. All what I can think of is to find all cosets of the stabilizer subgroup, then pick up a member of the subgroup and generate some polynomials that are fixed by the member, Check whether it's not fixed by the representative element of the coset (besides ), and discard if so. Check whether the whole stabilizer subgroup fix the poynomial, if so, we are done. Otherwise put it into a "pool" of polynomials. After we've visited all members of the stabilizer subgroup and still can't find a desirable polynomial, we should try to combine the members from the pool using ring operations, do some check, and put it to the pool, in some breadth first search manner, as the ones of lower degree are favored.

Whatever, let's evaluate the resolvent. Let , , , the evaluation is done using sympy in python:

import sympy

x1 = sympy.Symbol('x1')
x2 = sympy.Symbol('x2')
x3 = sympy.Symbol('x3')
x4 = sympy.Symbol('x4')

e1 = x1 + x2 + x3 + x4
e2 = x1 * x2 + x1 * x3 + x1 * x4 + x2 * x3 + x2 * x4 + x3 * x4
e3 = x1 * x2 * x3 + x1 * x2 * x4 + x1 * x3 * x4 + x2 * x3 * x4
e4 = x1 * x2 * x3 * x4

p1 = x1 * x3 + x2 * x4
p2 = x1 * x2 + x3 * x4
p3 = x1 * x4 + x2 * x3

r1 = sympy.expand(p1 + p2 + p3)
r2 = sympy.expand(p1 * p2 + p1 * p3 + p2 * p3)
r3 = sympy.expand(p1 * p2 * p3)

print(f'r1={r1}') # Simplify r1
s1 = e2
print(f'r1-s1={sympy.expand(r1-s1)}')
# r1 = e2

print(f'r2={r2}') # Simplify r2
s2 = sympy.expand(e1 * e3)
print(f'r2-s2={sympy.expand(r2-s2)}')
s2 = sympy.expand(s2 - 4 * e4)
print(f'r2-s2={sympy.expand(r2-s2)}')
# r2 = e1 * e3 - 4 * e4

print(f'r3={r3}') # Simplify r3
s3 = sympy.expand(e1 ** 2 * e4)
print(f'r3-s3={sympy.expand(r3-s3)}')
s3 = sympy.expand(s3 + e3 ** 2)
print(f'r3-s3={sympy.expand(r3-s3)}')
s3 = sympy.expand(s3 - 4 * e2 * e4)
print(f'r3-s3={sympy.expand(r3-s3)}')
# r3 = e1 ** 2 * e4 + e3 ** 2 - 4 * e2 * e4

By now, we can conclude the resolvent of is:

To test, we just need to try it against , which yields with rational root . So the Galois group of it is a subgroup of by our resolvent, matching our previous knowledge.

Conclusion

In this text, we've introduced and proved the Galois's theorem, along side with fundamentals required to derive it, and tools to apply it in practise.

First, we've inspected some extensions by adjoining "radicals" in our common cognitions to the base field. A cyclotomic extension is a Galois extension obtained by adjoining primitive -th root , which is the root of to the base field of characteristic or , and the Galois group can be embedded into , which is an abelian group. A simple radical extension is a Galois extension obtained by adjoining the root of to the the base field that has already splitted , and its Galois group can be embedded into , which is also an abelian group. I love they when they are Galois since their normality allows us to use them as building blocks to stack extensions over each other while each the intermediate extensions have structure that have been studied by us.

Conversely, a cyclic extension is a Galois extension whose Galois group is cyclic group. And cyclic extension extension of degree with splitting can be realized as simple radical extension, the radical can be recovered using Lagrange's resolvent. This is the foundation of recovering a cyclic extension back into simple radical extension under specific circumstance.

Then, we characterize what does it mean by "solvable by radicals", as when there's a radical extension comprising the Galois extension of specified polynomial, then all of the roots of the polynomial are expressibl by radicals in this radical extension, and is thus "solvable by radicals". Based on such characterization we've proved the Galois's theorem, that a Galois extension can be embedded into a radical extension iff the Galois group is a solvable group. The proof is done by first showing the Galois group of the normal closure of a radical extension is a solvable group, then by any quotient group of a solvable group is also solvable, and Galois group of our specified polynomial is a normal subgroup of the one of normal closure of the radical extension, it can only be solvable. Conversely, by elevating to Galois extension of our specified polynomial over a sufficiently large cyclotomic extension field, we have obtained an elevated Galois extension whose Galois group is a quotient group of solvable group, and thus also solvable. The intermediate cyclic extensions of elevated Galois extension can be then recovered back to simple radical extensions. Eventually the elevated Galois extension, annexing the cyclotomic extension on the bottom, forms a radical extension containing the Galois extension of our specified polynomial.

And in order to apply the Galois's theorem, we've come up with tools for calculating Galois group of polynomials. The most special case is when a polynomial of prime degree has non-real roots, which can be proved to possess a Galois group isomorphic to , and when it will not be solvable. For a more common scenario, we apply the resolvent methods, which aims at constructing some delicately constructed polynomial that have specific symmtries by stabilizer subgroup, and then evaluate the resolvent using this polynomial and the polynomial to solve. If the resolvent related to this polynomial has rational root that is not duplicate root, the Galois group is the subgroup of the stabilizer subgroup. We've also shown examples of testing whether the Galois group is subgroup of an alternating group or dihedral group , using the resolvent method.

(There're also other methods for calculate Galois group like the Dedekind's theorem, but its proof requires algebraic number theory, and I hate introducing things that I don't know how it works. As using this tool is not an urgent, I decide to leave it out.)

By now, we are imbued with the mind of Galois's theorem, and equipped with the tools to apply them. Hope one will find their usefulness in the future.