Section 5.6 The Elements of Galois Theory
Given a polynomial in , the polynomial ring in over , we shall associate with a group, called the Galois group of . There is a very close relationship between the roots of a polynomial and its Galois group; in fact, the Galois group will turn out to be a certain permutation group of the roots of the polynomial. We shall make a study of these ideas in this, and in the next, section.
The means of introducing this group will be through the splitting field of over , the Galois group of being defined as a certain group of automorphisms of this splitting field. This accounts for our concern, in so many of the theorems to come, with the automorphisms of a field. A beautiful duality, expressed in the fundamental theorem of the Galois theory (Theorem 5.6.6), exists between the subgroups of the Galois group and the subfields of the splitting field. From this we shall eventually derive a condition for the solvability by means of radicals of the roots of a polynomial in terms of the algebraic structure of its Galois group. From this will follow the classical result of Abel that the general polynomial of degree 5 is not solvable by radicals. Along the way we shall also derive, as side results, theorems of great interest in their own right. One such will be the fundamental theorem on symmetric functions. Our approach to the subject is founded on the treatment given it by Artin.
Recall that we are assuming that all our fields are of characteristic , hence we can (and shall) make free use of Theorem 5.5.1 and its corollary.
By an automorphism of the field we shall mean, as usual, a mapping of onto itself such that
for all . Two automorphisms and of are said to be distinct if for some element in .
We begin the material with
THEOREM 5.6.1: If is a field and if are distinct automorphisms of , then it is impossible to find elements , not all in such that
for all .
Proof: Suppose we could find a set of elements in , not all , such that
for all . Then we could find such a relation having as few nonzero terms as possible; on renumbering we can assume that this minimal relation is
where are all different from . If were equal to 1 then for all , leading to , contrary to assumption. Thus we may assume that . Since the automorphisms are distinct there is an element such that . Since for all , relation (1) must also hold for , that is,
for all . Using the hypothesis that the 's are automorphisms of , this relation becomes
Multiplying relation (1) by and subtracting the result from (2) yields
If we put
for , then the are in ,
since , and yet
for all . This produces a shorter relation, contrary to the choice made; thus the theorem is proved.
DEFINITION: If is a group of automorphisms of , then the fixed field of is the set of all elements such that for all .
Note that this definition makes perfectly good sense even if is not a group but is merely a set of automorphisms of . However, the fixed field of a set of automorphisms and that of the group of automorphisms generated by this set (in the group of automorphisms of ) are equal (Problem 1), hence we lose nothing by defining the concept just for groups of automorphisms. Besides, we shall only be interested in the fixed fields of group of automorphisms.
Having called the set, in the definition above, the fixed field of , it would be nice if this terminology were accurate. That it is we see in
LEMMA 5.6.1: The fixed field of is a subfield of .
Proof: Let be in the fixed field of . Thus for all and . But then
hence and are again in the fixed field of . If , then
hence also falls in the fixed field of . Thus we have verified that the fixed field of is indeed a subfield of .
We shall be concerned with the automorphisms of a field which behave in a prescribed manner on a given subfield.
DEFINITION: Let be a field and let be a subfield of . Then the group of automorphisms of relative to , written , is the set of all automorphisms of leaving every element of fixed; that , the automorphisms of is in if and only if for every .
It is not surprising, and is quite easy to prove
LEMMA 5.6.2: is a subgroup of the group of all automorphisms of .
We leave the proof of this lemma to the reader. One remark: contains the field of rational numbers , since is of characteristic , and it is easy to see that the fixed field of any group of automorphisms of , being a field, must contain . Hence, every rational number is left fixed by every automorphisms of .
We pause to examine a few examples of the concepts just introduced.
EXAMPLE 5.6.1: Let be the field of complex numbers and let be the field of real numbers. We compute . If is any automorphisms of , since ,
hence . If, in addition, leaves every real number fixed, then for any where are real,
Each of these possibilities, namely the mapping
defines an automorphisms of being the identity automorphisms and complex-conjugation. Thus is a group of order 2.
What is the fixed field of ? It certainly must contain , but does it contain more? If is in the fixed field of then
whence and . In this case we see that the fixed field of is precisely itself.
EXAMPLE 5.6.2: Let be the field of rational numbers and let where is the real cube root of 2. Every element in is of the form
where are rational numbers. If is an automorphisms of , then
hence must also be a cube root of 2 lying in . However, there is only one real cube root of 2, and since is a subfield of the real field, we must have that . But then
that is, is the identity automorphisms of . We thus see that consists only of the identity map, and in this case the fixed field of is not but is, in , larger, being all of .
EXAMPLE 5.6.3: Let be the field of rational numbers and let thus and satisfies the polynomial over . By the Eisenstein criterion one can show that is irreducible over (see Problem 3). Thus is of degree 4 over and every element in is of the form where all of , and are in . Now, for any automorphisms of , since , and , whence is also a 5th root of unity. In consequence, can only be one of , or . We claim that each of these possibilities actually occurs, for let us define the four mappings , and by
for , and 4. Each of these defines an automorphisms of (Problem 4). Therefore, since is completely determined by is a group of order 4, with as its unit element. In light of
is a cyclic group of order 4. One can easily prove that the fixed field of is itself (Problem 5). The subgroup of has as its fixed field the set of all elements
which is an extension of of degree 2.
The examples, although illustrative, are still too special, for note that in each of them turned out to be a cyclic group. This is highly atypical for, in general, need not even be abelian (see Theorem 5.6.3). However, despite their specialty, they do bring certain important things to light. For one thing they show that we must study the effect of the automorphisms on the roots of polynomials and, for another, they point out that need not be equal to all of the fixed field of . The cases in which this does happen are highly desirable ones and are situations with which we shall soon spend much time and effort.
THEOREM 5.6.2: If is a finite extension of , then is a finite group and its order, satisfies .
Proof: Let and suppose that is a basis of over . Suppose that we can find distinct automorphisms in . By the corollary to Theorem 4.3.3 the system of homogeneous linear equations in the unknowns :
has a nontrivial solution (not all ) in . Thus
Since every element in is left fixed by each and since an arbitrary element in is of the form with in , then from the system of equations (4) we get
for all . But this contradicts the result of Theorem 5.6.1. Thus Theorem 5.6.2 has been proved.
Theorem 5.6.2 is of central importance in the Galois theory. However, aside from its key role there, it serves us well in proving a classic result concerned with symmetric rational functions. This result on symmetric functions in its turn will play an important part in the Galois theory.
First a few remarks on the field of rational functions in -variables over a field . Let us recall that in Section 3.11 we defined the ring of polynomials in the -variables over and from this defined the field of rational functions in , over as the ring of all quotients of such polynomials.
Let be the symmetric group of degree considered to be acting on the set ; for and an integer with , let be the image of under . We can make act on in the following natural way: for and , define the mapping which takes onto . We shall write this mapping of onto itself also as . It is obvious that these mappings define automorphisms of . What is the fixed field of with respect to ? It consists of all rational functions such that
for all . But these are precisely those elements in which are known as the symmetric rational functions. Being the fixed field of they form a subfield of , called the field of symmetric rational functions which we shall denote by . We shall be concerned with three questions:
1. What is ?
2. What is )?
3. Can we describe in terms of some particularly simple extension of ?
We shall answer these three questions simultaneously.
We can explicitly produce in some particularly simple functions constructed from known as the elementary symmetric functions in . These are defined as follows:
That these are symmetric functions is left as an exercise. For and 4 we write them out explicitly below.
Note that when and are the roots of the polynomial
that when , and are roots of
and that when , and are all roots of
Since are all in , the field obtained by adjoining to must lie in . Our objective is now twofold, namely, to prove
Since the group is a group of automorphisms of leaving fixed,
Thus, by Theorem 5.6.2,
If we could show that
then, since is a subfield of , we would have
But then we would get that
and so , and, finally,
(this latter from the second sentence of this paragraph). These are precisely the conclusions we seek.
Thus, we merely must prove that
To see how this settles the whole affair, note that the polynomial
which has coefficients in , factors over as
(This is in fact the origin of the elementary symmetric functions.) Thus , of degree over , splits as a product of linear factors over . It cannot split over a proper subfield of which contains for this subfield would then have to contain both and each of the roots of , namely, ; but then this subfield would be all of . Thus we see that is the splitting field of the polynomial
over . Since is of degree , by Theorem 5.3.2 we get
Thus all our claims are established. We summarize the whole discussion in the basic and important result
THEOREM 5.6.3: Let be a field and let be the field of rational functions in over . Suppose that is the field of symmetric rational functions; then
2. , the symmetric group of degree .
3. If are the elementary symmetric functions in , then
4. is the splitting field over of the polynomial
We mentioned earlier that given any integer it is possible to construct a field and a polynomial of degree over this field whose splitting field is of maximal possible degree, , over this field. Theorem 5.6.3 explicitly provides us with such an example for if we put , the rational function field in variables and consider the splitting field of the polynomial over then it is of degree ! over .
Part 3 of Theorem 5.6.3 is a very classical theorem. It asserts that a symmetric rational function in variables is a rational function in the elementary symmetric functions of these variables. This result can even be sharpened to: A symmetric polynomial in variables is a polynomial in their elementary symmetric functions (see Problem 7). This result is known as the theorem on symmetric polynomials.
In the examples we discussed of groups of automorphisms of fields and of fixed fields under such groups, we saw that it might very well happen that is actually smaller than the whole fixed field of . Certainly is always contained in this field but need not fill it out. Thus to impose the condition on an extension of that be precisely the fixed field of is a genuine limitation on the type of extension of that we are considering. It is in this kind of extension that we shall be most interested.
DEFINITION: is a normal extension of if is a finite extension of such that is the fixed field of .
Another way of saying the same thing: If is a normal extension of , then every element in which is outside is moved by some element in . In the examples discussed, Examples 5.6.1 and 5.6.3 were normal extensions whereas Example 5.6.2 was not.
An immediate consequence of the assumption of normality is that it allows us to calculate with great accuracy the size of the fixed field of any subgroup of and, in particular, to sharpen Theorem 5.6.2 from an inequality to an equality.
THEOREM 5.6.4: Let be a normal extension of and let be a subgroup of ; let
be the fixed field of H. Then
(In particular, when .)
Proof: Since every element in leaves elementwise fixed, certainly By Theorem 5.6.2 we know that
and since we have the inequalities
If we could show that , it would immediately follow that and as a subgroup of having order that of , we would obtain that . So we must merely show that to prove everything.
By Theorem 5.5.1 there exists an such that ; this must therefore satisfy an irreducible polynomial over of degree and no nontrivial polynomial of lower degree (Theorem 5.1.3). Let the elements of be , where is the identity of and where . Consider the elementary symmetric functions of , namely,
Each is invariant under every . (Prove!) Thus, by the definition of are all elements of . However, (as well as is a root of the polynomial
having coefficients in . By the nature of , this forces
whence . Since we already know that we obtain , the desired conclusion.
When , by the normality of over ; consequently for this particular case we read off the result .
We are rapidly nearing the central theorem of the Galois theory. What we still lack is the relationship between splitting fields and normal extensions. This gap is filled by
THEOREM 5.6.5: is a normal extension of if and only if is the splitting field of some polynomial over .
Proof: In one direction the proof will be highly reminiscent of that of Theorem 5.6.4.
Suppose that is a normal extension of ; by Theorem 5.5.1, . Consider the polynomial
over , where are all the elements of . Expanding we see that
where are the elementary symmetric functions in . But then are each invariant with respect to every , whence by the normality of over , must all be in . Therefore, splits the polynomial into a product of linear factors. Since is a root of and since generates over can be in no proper subfield of which contains . Thus is the splitting field of over .
Now for the other direction; it is a little more complicated. We separate off one piece of its proof in
LEMMA 5.6.3: Let be the splitting field of in and let be an irreducible factor of in . If the roots of are , then for each there exists an automorphisms in such that .
Proof: Since every root of is a root of , it must lie in . Let be any two roots of . By Theorem 5.3.3, there is an isomorphism of onto taking onto and leaving every element of fixed. Now is the splitting field of considered as a polynomial over ; likewise, is the splitting field of considered as a polynomial over . By Theorem 5.3.4 there is an isomorphism of onto (thus an automorphisms of ) coinciding with on . But then and leaves every element of fixed. This is, of course, exactly what Lemma 5.6.3 claims.
We return to the completion of the proof of Theorem 5.6.5. Assume that is the splitting field of the polynomial in . We want to show that is normal over . We proceed by induction on , assuming that for any pair of fields of degree less than that whenever is the splitting field over of a polynomial in , then is normal over .
If splits into linear factors over , then , which is certainly a normal extension of . So, assume that has an irreducible factor of degree . The distinct roots of all lie in and is the splitting field of considered as a polynomial over . Since
by our induction hypothesis is a normal extension of .
Let be left fixed by every automorphisms ; we would like to show that is in F. Now, any automorphisms in certainly leaves fixed, hence leaves fixed; by the normality of over , this implies that is in . Thus
By Lemma 5.6.3 there is an automorphisms of , such that ; since this leaves and each fixed, applying it to (5) we obtain
Thus the polynomial
in , of degree at most , has the distinct roots . This can only happen if all its coefficients are ; in particular, whence so is in . This completes the induction and proves that is a normal extension of . Theorem 5.6.5 is now completely proved.
DEFINITION: Let be a polynomial in and let be its splitting field over . The Galois group of is the group of all the automorphisms of , leaving every element of fixed.
Note that the of can be considered as a group of permutations of its roots, for if is a root of and if then is also a root of .
We now come to the result known as the fundamental theorem of Galois theory. It sets up a one-to-one correspondence between the subfields of the splitting field of and the subgroups of its Galois group. Moreover, it gives a criterion that a subfield of a normal extension itself be a normal extension of . This fundamental theorem will be used in the next section to derive conditions for the solvability by radicals of the roots of a polynomial.
THEOREM 5.6.6: Let be a polynomial in its splitting field over , and its Galois group. For any subfield of which contains let
and for any subgroup of let
Then the association of with sets up one-to-one correspondence of the set of subfields of which contain onto the set of subgroups of such that
3. = index of in .
4. is a normal extension of if and only if is a normal subgroup of .
5. When is a normal extension of , then is isomorphic to
Proof: Since is the splitting field of over it is also the splitting field of over any subfield which contains , therefore, by Theorem 5.6.5, is a normal extension of . Thus, by the definition of normality, is the fixed field of , that is, , proving 1.
Since is a normal extension of , by Theorem 5.6.4, given a subgroup of , then , which is the assertion of part 2. Moreover, this shows that any subgroup of arises in the form , whence the association of with maps the set of all subfields of containing onto the set of all subgroups of . That it is one-to-one is clear, for, if then, by part 1,
Since is normal over , again using Theorem 5.6.4, ; but then
in . This is part 3.
The only parts which remain to be proved are those which pertain to normality. We first make the following observation. is a normal extension of if and only if for every . Why? We know by Theorem 5.5.1 that ; thus if , then for all . But, as we saw in the proof of Theorem 5.6.5, this implies that is the splitting field of
which has coefficients in . As a splitting field, , by Theorem 5.6.5, is a normal extension of . Conversely, if is a normal extension of , then , where the minimal polynomial of , over has all its roots in (Theorem 5.6.5). However, for any is also a root of , whence must be in . Since is generated by over , we get that for every .
Thus is a normal extension of if and only if for any and and so ; that is, if and only if . But this says that is normal over if and only if for every . This last condition being precisely that which defines as a normal subgroup of , we see that part 4 is proved.
Finally, if is normal over , given , since induces an automorphisms of defined by for every . Because leaves every element of fixed, must be in . Also, as is evident, for any whence the mapping of into defined by is a homomorphism of into . What is the kernel of this homomorphism? It consists of all elements in such that is the identity map on . That is, the kernel is the set of all such that ; by the very definition, we get that the kernel is exactly . The image of in , by Theorem 2.7.1 is isomorphic to , whose order is
Thus the image of in is all of and so we have isomorphic to . This finishes the proof of part 5 and thereby completes the proof of Theorem 5.6.6.