Galois Theory

In this text, we will introduce the Galois theory, which is a tool to correspond fields with groups under certain circumstances. This will empower us to study fields by studying their corresponding groups.

Field automorphisms

A field automorphism of is a field isomorphism whose domain and image are both . We know there's , where the complex number conjugation defined by is a field automorphism. Another little complicated example is defined by , which is a field automorphism by:

With analogous method, one can also easily verify that defined by and defined by are also field automorphisms.

Since field is both the domain and image of its field automorphisms , their composition is obviously also a field automorphism of . One would can simply verify that the identity map defined by is a trivial field automorphism, the inverse is a field automorphism, and the composition of maps has the nature of associativity. In this way, all field automorphisms of forms a group under the operation of composition with being the identity. Such a group is called the automorphism group of field , and is conventionally denoted as .

More specifically, in the field extension , it won't be hard to verify the automorphisms of that fix the base field pointwise (that is, when its domain is trimmed to , it collapse to identity map on ), forms a subgroup of , Such a subgroup is called the automorphism group of field extension , and is denoted as .

Field automorphisms of the smallest fields

We know all characteristic fields must have a subfield isomorphic to , and all characteristic fields must have a subfield isomorphic to , but we would like to show there's no field automorphism except for the identity map of and .

The proof is simple: just let there be any field automorphism , then we have , so , and then , since is the field of fraction of . The proof for field automorphism is even more simpler, just by and we are done. In this way, the only possible field automorphism of and must be the identity map.

We can conclude there're and , by simply trimming the domain of each field automorphisms to their own base fields.

Automorphisms as permutation groups of roots

One may have heard the Galois theory is a theory about the symmetry or permutation of roots, but what does it really mean?

We would like to prove that for field extension and polynomial who has a root , then for any , is also a root of .

The proof is done by simply applying to the expression of , that:

In case of one's not knowing what does it mean by "permutating roots", recall that any algebraic element is fundamentally a root of its minimal polynomial , we can clear see that there's . So a field automorphism must always send algebraic element to its conjugate element(s) over under its minimal polynomial .

Specially, for a finite extension , which has been shown to be equivalent to some finitely generated algebraic extension , as it can realized as tower of consecutive simple extensions , any automorphisms (or any homomorphism from ) are defined by the image of each . Since in each intermediate extension fields , by applying to its element we can see , with some mathematical induction we can see is defined as and . Since each must be mapped to one of its conjugate element over , which is no more than where is the minimal polynomial of over , obviously is finite then.

Actually the the choice of image can be even more scarce: consider finite extension , the elements conjugate to over are , however they are not in , so any automorphism must send to itself, and and is trivial.

Handling coincidence of simple extensions

It's also very common that adjoining different algebraic elements results in the same field, e.g. , with their minimal polynomial being and . However, let , there ought to be otherwise the automorphism group is not well defined. In fact, for any defined by and , and such that , , and (due to ), we would like to show it's the case of .

In fact, let the minimal polynomial of over be , it suffices to show is also a root of . Consider applying to the expression of , on the outer layer there's clearly , then by substituting and applying the trait of to fix the base field we have , finally by replacing we have , and is another root of in .

For example, for defined as and , it's , and there's , so the it's the case of when is taken as the member of .

This tiny little lemma is a direct application of the permutation group trait on roots, but help us beteer understand the connection between the same field obtained by adjoining different algebraic elements to the base field.

Automorphism groups of finite extensions

Acutally the case of finite extensions are the specific cases we care most about in the Galois theory, and in this section we would like to reveal some facts about them.

Lifting lemma

Fix a tower of finite extension , we would like to show by applying any field automorphism we have another tower , in which is kept all along the way.

Actually, we would like to prove in a more abstract way, since we can clip any subsequence in the tower , versus , when is trimed to each intermediate field serves as an isomorphism between two genally distinct fields, and thinking in an abstract way would be useful in the following topics of this text. We would like to show the lifting lemma, which states for a field isomorphism , the field isomorphism lifted to its finite extension with can be and only be implemented as where the minimal polynomial of is and the minimal polynomial of is . It can be visualized as below:

To prove, first we would like to show for two isomorphic base fields whose isomorphism is defined by , the two statements are equivalent:

1. defined by and is an isomorphism.
2. The minimal polynomial of is , and the minimal polynomial of is .

Actually we have shown in the proof of uniqueness of splitting fields up to isomorphism, and we will not repeat it here. So all remained is to show .

We know whenever there's a field isomorphism , there will be an induced ring isomorphism defined by . For convenience, we denote the image of as directly.

Consider the expression , all elements taken into operation are already in and the operations are field operations. So we can apply on it, and we have:

Noted that is minimal polynomial and thus monic and irreducible, and obviously is also monic and irreducible, and thus minimal polynomial of over , and clearly it's again the case of .

Now back to the case. To prove the lifted isomorphism of can be realized as in the way we specified, just by applying from to from the bottom to the top and we are done. To prove the lifted isomorphism of can only be realized as the way we specified, first trim the domain of to , obviously its image is , and defines an isomorphism between them. From we've proved above we know can only be implemented in the way that is the minimal polynomial of over , and is the minimal polynomial of over , and now we have . Recursively doing in such way we can prove it's all along the way, with the minimal polynomial of being and the minimal polynomial of being . So the field isomorphism can only be implemented in the way we specified.

Finally, for the initial property about automorphism group of finite extension , noted that it's a field isomorphism onto itself with base field fixed by each automorphisms, which can be seen as lifted from , clearly for a fixed tower there's corresponding tower , with kept all along the way.

Counting in the automorphism groups

We know the automorphism group of a finite extension is also finite, and in this section we would like to show there's .

Fix a tower , we can construct an automorphism by lifting from , determining the image of one at a time. Each must be chosen to be the root of that is in , where is the minimal polynomial of so far. And can hold at most elements, since can have at most roots in the same field, in which splits into linear factors. On the other hand, even if splits , in which has exactly roots, they cannot be asserted to be distinct, since might not be separable (and we will come up with example later). Climbing up this fixed tower, you can find there're at most ways to build automorphisms in this fixed tower, while there's all along the way. By applying the Tower law we have , which is the maximum possible number of distinct automorphisms we can find in this fixed tower.

On the other hand, even if we are able to count all automorphism in the tower of , is it possible that some field automorphisms that only appears in some specific tower? Let there be with (it's possible that , consider that finite fields are splitting field of but also a simple extension adjoining primitive root to ), which specifies another tower Since they specify different basis for the same vector space over , we can write the transition matrix between them, which is invertible, and with which we can solve the relationship between and for any . If would like to be a valid automorphism, when we extract each values of from those of , we must see is the root of minimal polynomial over . And this implies choosing such an by the sufficiency of lifting lemma, and must have been counted in the tower of (assuming we've counted them exhaustively in this fixed tower). Actually with transition matrix we are able to transit the representation of in terms of into the one in terms of bijectively. In this way, we can conclude there's .

Let's try with real world example: splitting field of . For convenience, we let , and consider any field automorphism , which we are going to construct it starting from , in the fixed tower of .

When we let or , the intermediate field is still , and in which there's . The minimal polynomial still holds the roots and , so we can let or . There're automorphisms matching this setup (represented in cycle notation): , , and .

Alternatively we can let or , the intermediate field becomes and in which there's . The minimal polynomial holds the roots and , so we can let or . Noted that if we choose and , there'll automatically be and , which forms the loop . And there're automorphisms matching this setup (represented in cycle notation): , , , and .

Let , and we can see all these automorphisms can be represented as , which has clearly the structure of the dihedral group .

In this real example, we can see there's exactly , and we are specially interested in those finite extensions with , so we are going to explore the equality condition of .

Fixed fields

Before our exploring the equality condition of , we would like to adapt the concept of fixed field first, that instead of specifying to be fixed by the automorphisms in the group, we find out what's the exactly field fixed by a subset or subgroup of automorphisms.

For a subset of automorphisms in , the elements in fixed by forms a subfield of , since there're automatically and . Such a subfield is called the fixed field of . Conventionally, for a subgroup , the fixed field of is denoted as .

Let be the subfield of and the fixed field of , then one can easily see there's , otherwise one can put into the field operation of , resulting in , which is a contradiction.

Please notice it's in general: consider , since the only automorphism is , the whole is fixed, not just the base field .

Artin's lemma

This lemma is due to Artin, who shows that for a finite automorphism group , is finite with .

Let , this proof is done by showing for any , pick up any distinct , there's always a set of not all zero such that , rendering them as linear dependent and thus must be finite with degree no more than .

Let be distinct members of . Consider the equation , where are expected to be in , by applying every to both sides of the equation we have:

Obviously the null space of is non-trivial. However, since all coefficients of are in and cannot be naively expected to be over , we still need to show there's some solution over .

Let the solution be the one with maximal possible number of zeroes and non-trivial, where is assumed to be non-zero without losing generality. And we assert to be since otherwise we can normalize by multiplying to every components of the solution, without affecting the number of zeroes inside. Assume there's a component , then . We know is a permutation of since it's a bijection on finite set. Substitute the solution into the system of equations and apply on both sides of each row equations, we can see it just permutate the rows in and thus is also a solution in 's null space. By subtracting these two solutions we have , which is still non-trivial by , and since this solution has no zeroes than the first solution. But the first component is while the first solution is , so this solution has more zeroes than the maximal possible one, which is a contradiction. So we cannot assume there's . On the other hand, if we assume every component , after our permutating and subtracting, we will just get a zero vector, which is a trivial solution, so the maximality of the first solution is preserved then.

In this way, the equation must have a non-trivial solution whose components are in , and thus we have , which means any are linearly dependent when .

Finally, to show is finite, assume it's the contrary. Obviously only algebraic element over is allowed in . So for any sufficiently large integer , we can pickup , in which each extension in the tower is not trivial. By the Tower law, we have , however this is impossible, since requires no less than linearly independent basis. In this way, we've shown is finite with , when is finite.

Consequence of Artin's lemma

Although we have in general, but if is finite, we have with , which can be easily tweaked into .

By Artin's lemma we know is finite extension with , and since is finite we have . Finally, since any must fix every element in , and they are from , clearly there's , and thus . When we put these inequalities together, we have

The only possibility is and .

On the other hand, for a finite extension , we obviously have , and , so we have the tower of . By the Tower law, it won't be hard to find , which means we can explore the equality condition of by finding the condition for to hold, and vice versa.

Normal and separable extensions

An algebraic extension (not necessarily finite) can have two special properties: it's said to be a normal extension if a irreducible polynomial over has a root in , then it splits in completely; it's said to be a separable extension if any irreducible polynomial over of algebraic element in is separable.

In this section, for a finite extension , we will focus on the relationship between the condition for to hold, and the normality and separability of .

Group actions of the automorphism group

We will start with inspecting the group actions of the automorphism group on any first. Clearly the orbit of under will be , with . Let the minimal polynomial of be over , of course there'll be so are the conjugate elements of over .

Let's multiply the linear factors of the points in togerther, resulting in the polynomial . For any , we have . Since is just a permutation of , given it's a subset of and bijective with by multiplying its inverse on the left, there must be . So is a symmetric polynomial under the group action of , and its coefficients must be in .

To show is irreducible, assume it's reducible, and the minimal polynomial of is over . Since there will be , and contains distinct linear factors, some factor is not contained in , so . In the same time, there's , which means , and is not fixed by . However must fix a polynomial over , which is a contradiction, so must be irreducible. Since is monic and irreducible, it's the minimal polynomial of and also all of the points in the orbit, over .

Finally, since there's , when is viewed as polynomial over the extension field, there must be .

Assume there's , then both and are minimal polynomials over and thus . Noted that the choice of is arbitrary, and each renders the minimal polynomial of over splits in , so must be normal. To show is separable, assume it's the contrary and let be the instance whose minimal polynomial over is inseparable. Then is impossible since there's no repeated factor in while there's some in . In this way, the extension satisfying must be normal and separable.

Please notice, we haven't tested whether the inverse that being normal and separable implies by now, nor whether the normality implies separability, or vice versa.

Normality and splitting fields

The normality of finite extension is a special case, we will show that a finite extension is normal iff is a splitting field of some polynomial over .

First, we will show finite extension is normal when is a splitting field, which means minimal polynomials of every algebraic element will eventually split in . Let be a splitting field of , and the minimal polynomial of be . Since is already in , we have and , and assume is another root of , if would like to be outside , adjoining to have to be a proper extension, where we must have .

But this will not happen, consider versus , where we reorder them so that and are to be adjoined first. Clearly there's isomorphism defined by and . The polynomial can be viewed as polynomial over , and there's under , so it's the case of isomorphic base fields with isomorphic polynomial to split. Noted that the way to split over is identical to the way to split over and , since the decomposition over is also a decomposition of over and , and all of the polynomial rings over are UFDs. So clearly is the splitting field of over , and is the splitting field of over , as they only contains nothing besides the base fields and the roots of . By our theory of splitting field, the splitting field of isomorphic polynomial over isomorphic base field must be unique up to isomorphism. So it's the case of , and the only possibility is and thus . Applying this to any root of will render it as in , so splits in .

Please notice this is not naively the case lifting lemma is applicable to, that is all along the way, try and you will see. In fact, we will count on the and 's uniquely splitting , in some step of our splitting field constructing algorithm, if is lifted to where are the actual roots we've grown into, the minimal polynomial of we are about to adjoin is over and over , we don't know what the image of is, but clearly when viewing from thus must contain some elements conjugate to , available to be chosen as image of and lifting into , this will hold all along the way, as we grow into and into concurrently.

So in this way, the proof also give us a construction that for every conjugate elements where is a splitting field, we can always lift the field isomorphism defined by and into an field automorphism in . This is useful later.

Then, we will show when finite extension is normal, it's a splitting field of some polynomial. Consider its equivalent finitely generated extension , for every , all of its conjugate elements over are also in by normality, and its minimal polynomial over splits in . When we multiply these into , with or without deduplication of , we will see is the smallest field containing only base field and all roots of , and thus the splitting field of .

In this way, we've shown the equivalence of being normal and extension field's being splitting field in a finite extension.

Separability and characteristic

In this section, we will discuss the connection between separability of an extension and the characteristic of the field. We will first have a look at judging the separability of irreducible polynomials, and then discuss the separability by characteristic of fields.

Actually judging the separability of polynomial is easy for the speciality of irreducible polynomial: for irreducible polynomial , we would like to show there's , where both greatest common divisor and formal derivative are evaluated in the field of . When , it's simply the case of . When , we can assume in the splitting field of polynomial , there's some such that . Since is irreducible and is a root of , the minimal polynomial of is associate with , and thus by the theory of simple extensions. But this is impossible due to when , so it can only be the case of . Now we are done.

In the field of characteristic , we can conclude that every irreducible polynomial is separable, since any non-constant polynomial in has non-zero formal derivative. In this way, every algebraic extension of characteristic field must be a separable.

However, such guarantee of non-zero formal derivative does not hold for irreducible polynomials over fields of characteristic . In the field where is transcendental over , consider the polynomial , since is the field of fractions of and is polynomial over , we just need to verify the (ir)reducibility of over in order to establish its (ir)reducibility over . Since , and the latter one is a PID, must also be a PID. And since is irreducible, must also be irreducible and thus prime in . Then by Eisenstein's criterion is irreducible over , and thus irreducible over . However, given that and , is not separable, simple extension is not separable.

In fact, it won't be hard to find there's , so is the splitting field, with the linear factor being only . Specially, Such kind of polynomial and extension is said to be purely inseparable.

So cautious analysis must be taken when handling the separability of extensions over field of characteristic .

On one hand, in the field of characteristic , it's simple to model the irreducible polynomial such that : for the term , its derivative is , and as long as the term will not be . In this way, as long as contains a term whose degree is not divisible by , it's formal derivative is non-zero, and thus it's separable.

One the other hand, there're still separable extensions over fields of characteristic . We claim that algebraic extensions must be separable. To see, consider any algebraic element with its minimal polynomial , is a simple extension, and thus finite. So is also a finite field, let it be isomorphic to , in which is the root of in . Since is also a polynomial over , there is , and if is not separable, is not separable too, which is a contradiction. So in this way, algebraic extension of a finite field is separable.

And in this way, it won't be hard to find normality does not neccesarily indicate separability, nor vice versa: is separable but not normal, while is normal but not separable.

Primitive element theorem

This topic arises from the separability of finite extensions.

Consider the field , which is a finite extension field over . Although it seens to be little common at the first glance, but actually such a field is just the same as . First by and , the minimal polynomial of over must be a factor of . Simply by substituting in we know it has no factor of degree . If it's factorizable by polynomial of degree , then the two factors are , and , and thus , which has no solution for any . In this way, is the minimal polynomial of , and is of degree with basis . And we can transform the basis using:

By Gauss's elimination the transition matrix is of full rank and thus invertible, so .

In a simple extension , the element is said to be the primitive element. And the primitive element theorem states that all finite separable extensions are simple extensions.

If is finite and is a finite separable extension, then is a also a finite field. Let the primitive root of be , obviously is a subfield of covering every point of , so it can only be , and thus is a simple extension.

For the case of is infinite, if in a finite separable extension we are able to prove , then given that is fundamentally a finitely generated algebraic extension adjoining the basis , we can iteratively replace , and then we are done. Actually, we would like to claim and prove .

Let minimal polynomials of over be , and continue to "grow" over into such that splits . Given that formal derivatives can be purely evaluated in and they are known to be non-zero, are also separable polynomials over . Let the roots conjugate with over be and the roots conjugate with over be . We claim to be any element as long as . If the point to evade is not in , we can happily ignore it. And since the points to evade are finite while there're infinite elements in , we must be able to find such a value .

Next, let there be and consturct the polynomial . In the field of we can evaluate and . Let the minimal polynomial of over be , if then . Otherwise it's the case of , and by the theory of simple extension we have . And since and is separable, there must be some . But is not a root of , let and we have when evaluated in , which is a contradiction. So the only possibility is , and since every element on the right hand side are from .

Finally, since , we have ; on the other hand, since , we have . There must be by mutual set inclusion, and we are done.

Epilogue of this chapter

In the beginning of this chapter, in the finite extension we've shown the connection between and the group actions of the , with which we've shown there's only if is normal and separable; in the middle of this chapter, we've shown that normality and separability can be independent from each other, as well as discussing their own internal mechanisms; and finally in this section we are going to show there's iff is normal and separable, rendering it as a sufficient and essential condition.

First, we are going to show for any algebraic element , its minimal polynomials and coincides with each other when finite extension is normal and separable. Since is normal, must split in , and since , the roots of must be non-repetitively , and is among them. If we still want to have , the only possibility is the group action of would partition the roots of into two or more disjoint orbits, or briefly the group action of would be non-transitive. However, the group action of on the roots of is transitive, since for any two roots of , by normality of there must be that is lifted from defined by and . In this way, we will find coincides with for every algebraic element .

Then, since is finite and separable, by the primitive element theorem it's a simple extension , where is the primitive element and is the minimal polynomial of over . Noted that there's , and is formed by 's group action on , which requires there're at least group elements to mark every points in the orbit . So it is the case of , and by combining with the general inequality for finite extension , we will know it's only possible that and thus .

Alternatively, although the normality of finite extension does not necessarily imply its separability, but when is the splitting field of a separable polynomial over , is also separable.

In fact, we want to show there's iff is a splitting field of a separable polynomial over . The essentiality is simple, when , then is normal and separable, by normality it's a splitting field of polynomial by separability we can deduplicate irreducible factors of and each factor must be separable, and so is . The sufficiency is rather different, given that is already a splitting field of separable polynomial , we already have normality, and it requires separability to go. But instead of striving to prove its separability, based on our analysis on the normality of splitting fields, we can fix the path we used to grow into the splitting field of by splitting , and count the number of automorphisms directly over there. Clearly on the intermediate extension corresponding to splitting minimal polynomial of , has distinct roots by the separability of and , and each root can be selected as image of , which will be lifted into a field automorphism when the splitting field construction is complete. Noted that on this step there's , and by induction in such way there's , implying is finite normal and separable.

Fundamental theorem of Galois theory

Those finite normal and separable extensions we've done hard work about are the central object of the Galois theory, and are usually referred as finite Galois extensions, whose automorphism groups are usually referred as the Galois groups, and conventionally denoted as .

The Galois theory, which arose from Galois's characterization of solvable equations and has been improved by numerous mathematicians to improve its generality and intrisiciality, builds up a bridge between field theory and group theory.

The fundamental theorem

In finite Galois extension , the fundamental theorem claims there's a bijection with its inverse , between the intermediate field from every tower of , and every subgroup .

First, we need to show is Galois. First for any algebraic element , whose minimal polynomials are and , noted that when is viewed as polynomial over there must be . For the normality, assume the does not split in , then will not split in either, which is a contradiction to is normal. Similarily for the separability, assume is not separable, then will not be separable either, which is a contradiction to is separable. In this way, is Galois and we can safely use our notation of .

Then, we need to show is a subgroup of . Since , the the member of should fix , and by , they fix the member of automatically, so we have .

Conversely, for every subgroup , by the theory of fixed fields we know there's subfield such that with . Noted that this is already the expression of composition of and then , by we know it's injective. And since every subgroup can be used as input of , it's surjective. In this way, we've proved our claim of the existence of bijection alongside with its inverse .

Noted that we always have , and by the Tower law we have , which can be easily tweaked into .

And it won't be hard to find (be cautious about the direction of inclusion!): from left to right, we first clamp at the Galois extension , then by we know there's ; from right to left, clearly is Galois, with in it, so it must be .

Such correspondence relationship between subgroup of and fixed field by each subgroups is usually referred as Galois correspondence by some mathematicians.

Normal subgroup and normal extensions

While is always Galois for , it's generally not the case that is Galois, since it's very likely not to be normal in smaller fields. But specially, we have is normal (and thus Galois) iff , this brings accordance to the normality of extensions with subgroups.

Consider any field automorphism , we can trim its domain to the field of , so it's a field isomorphism whose image is a field and a subfield of , while might be or not be . Denote the image as , and if , then clearly we can pick up some ¹ , leaving evidence of 's not being normal. Conversely, if there's , we take any algebraic element , we know every element conjugate to is the form of for some , and it's . In this way, we have is normal iff .

Noted that this can also be interpreted as the group action of on the subfields , but by the Galois correspondence we know such subfield corresponds to , and we can also intepret it as the group action of on the subgroup , and the whole group must fix it under such group action iff is to be true.

And we know for every subfield the corresponding Galois group is , in order to know what group action is it on , we need to reveal the connection between and . Noted that despite and 's not necessarily being normal, they are finite and separable and thus primitive element theorem is applicable, where we have and , and by the lifting lemma we know it can only be implemented as and . Let's take any , we know there's and , when we replace we get , which can be fixed into so that and , so we have . By the knowledge of group action we know the mapping in the form of defines field isomorphism between and , so it's the case of when considering only set inclusions. And by their finiteness it's only the case of , and thus . So the group action of on is the conjugation action on subgroups, and by the conclusions in group theory we know the stabilizer subgroup of this action on is the normalizer of in which . And if the whole fixes under such group action, it's only the case that .

Finally, when the quotient group of is also take into consideration, we want to show there's:

Where we denote so that it's more unified in form. To show, we trim the domain of every to , given that there's only single point in the orbit , they are all automorphisms on , and thus members of . The process of trimming can be represented as a map , and it won't be hard to find it's a surjective group homomorphism by testing their group operations and . Obviously all elements in are trimmed to the identity element , conversely any automorphism in trimmed to must fix and thus in by definition. In this way, we have , and thus .

Some instance of Galois correspondence

Let's have a look at some real world instances of Galois correspondence.

The finite extension , whose extension field is splitting field of , is clearly Galois. Since we are going to use them a lot later, denote for convenience. On splitting over , the choices are clearly , fix a path , on this layer there's , The is irreducible over , and on splitting it we have choice , and we've chosen to reach to the splitting field, fixing the path . Now consider the field automorphisms on it, when we've chosen any root out of the three roots on the first layer, then on the next layer both the remaining two roots are available to be chosen. So the Galois group by our knowledge of symmetric groups.

Now consider the subgroups of the Galois group. All transpositions generate a cyclic subgroup, and all three cycles generate the alternating group . Noted that the three cyclic subgroups are not the subgroup of .

Where every subgroup corresponds to a subfield.

For the cyclic groups of transpositions, like , clearly are mapped into and vice versa, but since , it's fixed by the transposition, thus is fixed by the group. Using this method we can derive , and .

The thing is interesting when we come to the alternating group . By definition it's generated by , and it's an normal extension, but what is it? Consider any element , let the generator automorphism act on it and we have:

This will require , and can be solved into . So in order to be fixed, the element must be in the form of , which is an element from . The field is identical to . In this way, we have . By now, we've found all subfields in finite Galois extension :

Another instance we want to have a look is we've discussed above, here we will inherit the notation for convenience.

Again, we will first have a look at the subgroup structure of dihedral group . Noted that there's held, the distinct cyclic subgroups generated by the elements are , , , , , . Then we try to make pairwise combination of these generators, and we can see this will result in , , , , , , , , , , , , , , , noted that once there's in the same group, it will generate the whole through their group operations. When we try to add gnerators to the newly generated groups, we have , , , , , , and no new group is generated. By now, we've enumerated all subgroups of , and by their inclusion relationship we can show in in the diagram as above:

And we are going to find out subfield occupying these question marks:

First, we know the element of are represented as , and we will try to apply the group elements to them. Try first, which requires , and we have , and to be fixed the element can only be in the form of , so it's contained in the field . Similarily let's apply , which requires , and then we have , and to be fixed it must be in the form of , which is contained in . And let's apply , which requires , and then we have , and to be fixed it must be in the form of , which is in the field of .

Now consider the groups containing , in which we just need to let more group elements act on . To be fixed by , we just need apply , resutling in , which must be in the form of , and is thus in the field . To be fixed by , we need to apply , which requires by , apply it and we have , which must be in the form of , and thus in the field . To be fixed by , we ned to apply , which requires and thus , apply it and we have , which must be in the form of and in the field .

Finally, for the remaining two fields, we need to apply the generator element to , and see how it can be fixed. To be fixed by , apply and we have , which can be fixed into . In order to be fixed we have and thus , noted that there's , and thus the element is in the field . To be fixed by , apply , which requires and thus , and we have , which can be fixed into , so this time it's , and thus , noted that there's , and thus the element is in the field . By now we've found all subfields in Galois extension :

These instances can be used as reference when we are for calculating subfields corresponding to subgroups in given finite Galois extensions.

Automorphisms on finite fields

The finite group is the splitting field of , and the extension has been proved to be finite, normal and separable, and thus a finite Galois extension, which can be studied by our developed tools. Noted that we have , thus the Galois group of such extension is identical to the automorphism group of finite fields. We would like to show is cyclic with order .

First, for commutative rings of characteristic , by the Freshmen's dream we've know defined by is a ring homomorphism onto itself and is called the Frobenius endomorphism. When it comes to the case of finite fields, since clearly fixes the elements in the base fields and does not collapse the whole field into zero ring, thus and it can only be injective. By the finiteness ² of we know it can only be surjective. In this way, can only be a field automorphism, and an element in as elements in are fixed.

Then, clearly its self composition defined as is also a field automorphism. Clearly by , and we would like to show there's . Assume there's some , then we have , all elements from are the solution of , while can only hold roots, which is a contradiction. In this way, we have while .

Finally, noted that there's while and , so it can only be the case that , and we are done.

One interesting fact about finite field is that for any minimal polynomial with root , then we have . First, by adjoining to we have , this field is a finite field so that we have an field isomorphism ³ . We know is Galois so minimal polynomial of over splits in , and apply the inverse to that decomposition we can see a decomposition of a minimal polynomial over , which is undoubtedly the . And clearly we have of order , by applying to we have , they must be distinct by the separability of a Galois extension.

Conversely, every primitive root ⁴ of generate a minimal polynomial over of degree under the group action of . Let's fix one of the primitive root , and every element in multiplicative group is in the form of . In order for to be a primitive root, it must generate the whole multiplicative group, so we must have and thus . So there're primitive roots in the field. Then consider the orbit of , which is , we have already, and , so both and are elements in . And clearly we have so all of the are distinct modulo-, and thus are distinct modulo-, so are the elements in the orbit of . In this way, every primitive root in generates a minimal polynomial over under the group action of . And we can see there're distinct minimal polynomials ⁵ generated by them.

However, besides primitive root, minimal polynomial of degree can also be generated by elements that are not primitive root in . An instance is where is any primitive root of , by applying the rationale above we can see its orbit has distinct elements as , multiplying to a minimal polynomial of degree over . So some extra verification must be taken if we are to guarantee a primitive root to be found instead of any common primitive element of , among irreducible polynomials of degree over , which is mandatory for certain use cases of finite fields.

Diamond rule

We also have some second isomorphism style rule or diamond rule for Galois extensions.

Let there be arbitrary field extension which is not necessarily finite, in the subfield of over there're and such that is finite Galois extension with primitive element , and is arbitrary extension (yes, including ), we claim there's whose isomorphism is defined by trimming the domain of each to .

Clearly and are both fields and lie in the position as is shown in the diagram on the left. We can always assume there's , otherwise it's just trivially the case that and , the trivial field extension is obviously finite Galois, but not worth discussing, as is shown on the right.

The extension is clearly finite and Galois as it's an intermediate field of Galois extension . It's the still the case that , since it's the mutual inclusion that and , so is a simple extension by adjoining . Let the minimal polynomial of over be , clearly is the splitting field of .

Noted that there's , so is irreducible over , by adjoining to we have the finite extension . Clearly the extension is normal since now is the splitting field of polynomial , and separable since is separable. In this way, is also finite Galois, and simple extension by adjoining .

Finally, let's build the connection between and . Noted that their group members are both defined by fixing the base field and mapping to its conjugates, while there're of them in both Galois groups. So we can make correspondence , which trims the automorphism defined by and and , to the automorphism defined by and . One can easily verify is a group homomorphism, and since it's obviously bijective over finite groups, it's a field isomorphism, connecting .

Conclusion

In this text, starting with the concept of field automorphism, we've established multiple foundation of behavioural nature about field extensions, and built the bridge between field extensions and groups, the Fundamental theorem of Galois theory.

First, we've established the nature of field automorphisms on finite fields, which forms a finite group whose order is capped at the degree of extension, and we're eager to know under which condition will the order of group equal to the degree of extension. On the other hand, we've inspected the fixed field, and the automorphism group of the extension field over the fixed field has the desirable property of order equal to the degree of extension, but the fixed field is potentially an intermediate extension field over the base field, and seeking the equality condition of group order and degree of extension can also be intepreted as the condition in which the fixed field coincides with the base field.

Then, to establish the condition of equality and coincidence, we've inspected the normality and separability of finite extensions. The normality guarantees us to find all of the conjugate roots in the extension field, and the separability guarantees us that all of the roots are unique. Both of them are essential and sufficient condition for automorphism group order equal to degree of extension, and the fixed field's coincidence with the base field.

Finally, those finite extensions fulfiling the equality and coincidence condition have desirable properties that we want to study them in Galois theory, and we specially nominate them as finite Galois extensions, as well as nominating their automorphism groups as Galois groups. By the Fundamental theorem of Galois theory, we create a bijective correspondence relationship between subfields in Galois extensions and subgroups of the Galois group, allowing us to compare and inspect their structures as well as importing group theory tools into studying fields.

One who has truly understood the Galois theory will realize its beauty and usefulness. And it's helpful to leverage Galois theory for studying algebraic equations, finite fields and number theory problems.