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Abstract Algebra: An Interactive Approach, 2e
©2015 This notebook is provided with the textbook, "Abstract Algebra: An Interactive Approach, 2nd Ed." by William Paulsen. Users of this notebook are encouraged to buy the textbook.
Chapter 15
Galois Theory
Initialization: This cell MUST be evaluated first:
The Galois Group of an Extension Field
In the last chapter, we explored the extensions of a field, and found that any finite extension could be entered into Sage fairly easily. In particular, we explored the splitting fields of several polynomials. In this chapter, we will explore the automorphisms on the field extensions, and discover that the group of automorphisms contains much information about the polynomial. For example, it will tell us if the roots of the polynomial can be expressed in terms of square roots and cube roots.
DEFINITION 15.1
Let K be a finite extension of the field F. An F-automorphism of K is a ring automorphism ϕ on the field K that fixes every element of F. That is, ϕ(x) = x whenever x is in F.
Note that there is at least one F-automorphism of K, the identity automorphism. Since we have seen that the set of group automorphisms of a group forms another group, it is not surprising that the same thing happens for F-automorphisms of a field.
DEFINITION 15.2
The set of all F-automorphisms of K is denoted GalF(K), and is called the Galois group of K
over F.
EXAMPLE:
Find the Galois group of the set of complex numbers ℂ over the real numbers.
According to Proposition 11.4, ℂ has two automorphisms that fix the real numbers: the identity automorphism, and the automorphism that sends each number to its complex conjugate. So there are exactly two elements of Galℝ(ℂ). In other words, Galℝ(ℂ) is isomorphic to Z2.
We want to find a way to compute the Galois group of any finite extension of a field F. Since we can define finite extensions in terms of polynomials, it is natural to ask what must happen to the roots of a polynomial.
EXAMPLE:
Let us use this lemma to find the Galois group of the splitting field of x3 − 2.
The splitting field is defined by letting a3 = 2, and b2 = −a2 − a b.
The three factors of x3 − 2 are a, b, and −a − b. Thus, Lemma 15.1 tells us that if ƒ(x) is an automorphism on ℚ(a, b), then ƒ(a) is either a, b, or −a − b, and ƒ(b) is either a, b, or −a − b. Let us try to find an automorphism such that ƒ(a) = b and ƒ(b) = a.
We can check to see if this is indeed a homomorphism.
This shows us that we have successfully defined one automorphism of the Galois group. (Any nonzero homomorphism on a field must be an automorphism in light of Proposition 10.5, and the fact that the kernel is always an ideal.)
EXPERIMENT:
Using the above example as a guide, define an automorphism G(x) on ℚ(a, b) such that G(a) = b, and G(b) = −a − b. Have Sage check that this is a homomorphism for you.
Since these two automorphisms fix the elements of ℚ, they are ℚ-automorphisms. From these two ℚ-automorphisms, we can produce more. Note that
shows that G(G(a)) = −a − b, and G(G(b)) = a.
Shows that F(G(a)) = a, and F(G(a)) = −a − b. Finally,
Shows that G(F(a)) = −a − b, and G(F(a)) = b.
Thus, we have found 6 ℚ-automorphisms on the field ℚ(a, b), the identity automorphism, F(x), G(x), G(G(x)), G(F(x)), and F(G(x)). Since G(F(x)) is not F(G(x)), the Galois group is not commutative.
It seems as though we must have found all of the automorphisms at this point, but this still needs to be proved. We begin by showing that there will always be an automorphism that moves one root of an irreducible polynomial to another.
The next lemma will be important in determining the subgroups of the Galois group.
Next we want to work on finding an upper bound on the number of elements in GalF(K).
PROPOSITION 15.3
Let K = F(u1, u2, u3, … un) be a finite extension field
of F. If ϕ1 and ϕ2 are two F-automorphisms in GalF(K), and
ϕ1(u1) = ϕ2(u1), ϕ1(u2) = ϕ2(u2), … ϕ1(un) = ϕ2(un),
then ϕ1(x) = ϕ2(x) for all x in K. In other words, an F-automorphism in GalF(K) is completely determined by its action on u1, u2, u3, … un.
We can now apply this proposition to the field ℚ(a, b). Any ℚ-automorphism is determined by where it sends the elements a and
b. By Lemma 15.1, these elements can only be sent to a, b, or −a − b. Yet an automorphism cannot send two elements to the same element. Thus, there are at most six ℚ-automorphisms on the field Yet we have found precisely
six ℚ-automorphisms
of ℚ(a, b). Thus, we have found all of the ℚ-automorphisms, so the Galois group of ℚ(a, b) contains exactly
six elements. We observed before that Galℚ(ℚ(a, b)) was non-commutative,
so Galℚ(ℚ(a, b)) ≈ S3.
We can similarly find an upper bound for the number of ℚ-automorphisms in any splitting field.
We immediately see from this corollary that the Galois group of a finite extension must be a finite group.
Let us look at another example of a Galois group of a
field. Consider the field ℚ(∛2). This is obviously a subfield of the field ℚ(a, b) we
have been studying, but in this subfield all of the elements are real. Thus, in this field ℚ(∛2) there
is only one root to the polynomial x3 − 2. Hence, if ƒ(x) is a ℚ-automorphism
of ℚ(∛2) then ƒ(∛2)must
be ∛2. By Proposition 15.3, the ℚ-automorphism is completely determined by where ƒ
sends ∛2. Thus, Galℚ(ℚ(∛2)) is merely the
trivial group.
In order to find the Galois group of a field, it is very helpful to know ahead of time the exact size of the Galois group. The next proposition allows us to compute the size of the Galois group for an important class of field extensions.
We are ready to try a more complicated example.
EXAMPLE:
Find the Galois group for the splitting field of the polynomial x4 − 2 x3 + x2 + 1.
First we verify that this polynomial is irreducible.
So this polynomial is irreducible over ℚ. Let us define a to be one root of this polynomial, and see how this polynomial factors over ℚ(a).
This tells us that if a is a root, then 1 − a is another root. However, it didn't factor completely, so we have to define b to be a root of the irreducible quadratic.
The polynomial should now factor completely.
So the four roots are a, 1 − a, b, and 1 − b. Any ℚ-automorphism will map each of these roots to another root, and so the Galois group will be a subgroup of S4. But which permutations will give rise to a ℚ-automorphism? A little trial and error will help.
Proposition 15.2 says that there will be some ℚ-automorphism that sends any one of these four roots to any other of the four roots. So there is a ℚ-automorphism that sends a to 1 − a. But where would it send the other three roots? Note that if ƒ(a) = 1 − a, then ƒ(1 − a) = ƒ(1) − ƒ(a) = 1 − (1 − a) = a. So we only have to determine if ƒ(b) is b or [removed]1 − b. Let's try both combinations.
They both work! We can now draw a picture of how these ℚ-automorphism act on the four roots of the polynomial.
If we number the roots as
| 1) | a |
| 2) | 1 − a |
| 3) | b |
| 4) | 1 − b |
then we can express the circle graph of F as P(2,1), while the circle graph of G represents the permutation P(2,1,4,3). These generate a subgroup of S4.
So far, we have 4 ℚ-automorphisms of the splitting field. Are there any others? Proposition 15.4 shows that there should be eight ℚ-automorphisms. Also Proposition 15.2 says we can find a ℚ-automorphism that send a to any of the other three roots! Suppose we send a to b.
If a is sent to b, then 1 − a will be sent to 1 − b. Thus, we find that b must be sent to either a or 1 − a. Try sending b to a:
This circle graph is like the permutation P(3,4,1,2). Thus, we can extend our subgroup of S4 to:
Now we have eight ℚ-automorphisms. Since this is the number predicted by Proposition 15.4, we are done. Hence, we found the Galois group as a subgroup of S4 of order 8.
The multiplication table
shows that this group is non-abelian, and has five elements of order 2. Thus, the Galois group is isomorphic to D4.
This example shows the usefulness of Proposition 15.4 in finding the Galois group. In fact, sometimes the Galois group can be determined using only Corollary 15.1 and Proposition 15.4.
One of the tools we will use for finding the ℚ-automorphisms is the close connection between the subgroups of the Galois group, and the subfield of the field extension. We begin by showing a way to produce subfields of a field extension.
DEFINITION 15.3
The field fix(H) is called the fixed field of the subgroup H.
Let us use the example of the Galois group of ℚ(a, b) to illustrate the fixed fields of the subgroups.
The Galois group was described as
{I(x), F(x), G(x), G(G(x)), F(G(x)), G(F(x))},
where I(x) represents the identity automorphism that sends every element to itself. The subgroups of Galℚ(ℚ(a, b)) are as follows:| H1 = | {I(x)} |
| H2 = | {I(x), F(x)} |
| H3 = | {I(x), F(G(x))} |
| H4 = | {I(x), G(F(x))} |
| H5 = | {I(x), G(x), G(G(x))} |
| H6 = | {I(x), F(x), G(x), G(G(x)), F(G(x)), G(F(x))} |
EXAMPLE:
Let us find the six fixed fields of ℚ(a, b) = ℚ(∛2, ω3∛2).
The field fix(H1) is the set of elements fixed by the identity mapping, which is of course all of ℚ(a, b). The field fix(H2) contains the elements fixed by the mapping F, which maps a to b, and b to a. Notice that the third root, −a − b, is fixed be the automorphicm F. Thus, fix(H2) = ℚ(−a − b). By a similar arguement, we see that fix(H3) = ℚ(a), and fix(H4) = ℚ(b). The field fix(H5) is a little bit trickier, since G moves a, b, and −a − b. Let us have Sage help us find the fixed field.
Let us try to find an element that is fixed by G. The basis for ℚ(a, b) is {1, a, b, a2, a·b, a2·b}. Of course G(1) = 1, but let us try the other six elements in the basis.
Only the last element, a2·b, is fixed by the automorphism G. What elementis this? We can find an approximation for a2·b by plugging in what a and b represent—two of the roots to the polynomial x3 − 2.
Here are the numerical approximations to the roots:
If we let a = 1.25992, and b = (-0.629961 + 1.09112 i), we find that a2·b =
The real part of the answer is obviously trying to be −1, but the complex part is also familar. Comparing this to
we see that a2·b = − 1 + √−3. This agrees with our previous obvervation that √−3 is in the field ℚ(a, b). Since −1 is already rational, we can write the fixed field fix(H5) as ℚ(√−3).
Finally, the only elements of ℚ(a, b) that are fixed by all ℚ-automorphisms are the elements of ℚ. Hence fix(H6) = ℚ.
Here is another example, although a bit easier.
EXAMPLE:
Find the fixed fields for the field ℚ(∛2).
Since the only ℚ-automorphism is the identity automorphism, which fixes the whole group, the only fixed field of ℚ(∛2) is ℚ(∛2), even though there is the obvious subfield ℚ within this field.
EXPERIMENT:
Consider the field ℚ(√2, √3). Note that every element of this field can be written as
a + b √2 + c √3 + d √6,
where a, b, c, and d are rational numbers. How many subfields does this field have? We will learn that since this field is a splitting field of x4 − 5 x2 + 6, every subfield is produced as the fixed field of some subgroup of the Galois group. Can you use this information to see what the elements of the Galois group are?We were hoping to be able to find all subfields of a field by looking at the fixed fields, but in the last example we failed. We will understand why the field ℚ(∛2) is not as well behaved as ℚ(∛2, ω3∛2) in the next section.
The Galois Group of a Polynomial in ℚ
To demonstrate Galois groups, let us concentrate on polynomials with rational coefficients. By working with rational numbers, we will avoid the problem of a splitting field having multiple roots. Recall that earier we saw that, even though x2 − t is irreducible in Z2(t), the factorization in the splitting field had multiple roots. This situation will never happen if we work in the field of rational numbers.
One advantage of working with a familar field is that we can borrow a result from calculus. It isn't often that we will use a calculus result in algebra, but in this case it greatly simplifies the proof.
Because of this lemma, we know from Proposition 14.10 that any splitting field can be expressed as a simple extension ℚ(w), and also we will be able to
use Proposition 15.4 to predict the size of the Galois group of the splitting field. We can relate the Galois group of the splitting field directly to the
polynomial.
DEFINITION 15.4
Let f(x) be a polynomial in ℚ. The Galois group of f(x) is the Galois group of the splitting field of f(x) over ℚ.
We have already seen some examples of Galois groups of splitting fields. The splitting field of x3 − 2 was isomorphic to S3. We also computed the Galois group of the splitting field of x4 − 2 x3 + x2 + 1, and found that the Galois group is isomorphic to D4.
EXAMPLE:
Find, up to isomorphism, the Galois group for the polynomial
x3 + x2 − 2 x − 1.
This polynomial is irreducible, as Sage can verify:Thus, we can let a denote one of the roots, and try to factor this in ℚ(a).
Note that this factors completely. Thus, the splitting field of x3 + x2 − 2 x − 1 is ℚ(a). This is a 3-dimensional extension of ℚ, so by Proposition 15.4, the Galois group has three elements, hence is isomorphic to Z3.
EXPERIMENT:
Using a similar argument, determine which familar group the Galois group of x5 + x4 − 4 x3 − 3 x2 + 3 x + 1 is isomorphic to.
EXAMPLE:
Find, up to isomorphism, the Galois group for the polynomial x5 − 5 x + 12.
In the last chapter, we were able to find a splitting field by making two extensions, one of dimension 5, and one of dimension 2. Here is the factorization that was produced:
If we define
we see that the product
simplifies to the original polynomial x5 − 5 x + 12. Thus, the five roots are a, b, c, d, and e. Thus, any ℚ-automorphism on the splitting field must send a and b to one of these five elements. Let us try to define a homomorphism ƒ that sends ƒ(a) = b, and ƒ(b) = a.
Thus homomophism works! Not only does Sage verify that this is a homomorphism, but it can drow a circle graph describing how this homomorphism acts on the five roots.
Note that we put the elements c, d, and e in quotes. This way, the single letter notations for these elements appear in the circle graph. Otherwise, the expanded values would be displayed, making the graph very messy.
From this, we see that ƒ(c) = d, ƒ(d) = c, and ƒ(e) = e.
EXPERIMENT:
What if we tried to define a homomorphism in Sage for which ƒ(a) = b, and ƒ(b) = c? Try changing the above commands to see if this indeed produces a homomorphism.
Not every possible way of mapping a and b to the roots a, b, c, d, and e will produce a homomorphism. Thus, finding other homomorphisms requires a little trial and error. In fact, exactly half of the attempts will produce a homomorphism in this example.
Let us try one more time to find a different homomorphism. This time, we will try to have ƒ(a) = a, and ƒ(b) = c.
Thus homomophism also works. The circle graph is given by
by which we see that ƒ(c) = b, ƒ(d) = e, and ƒ(e) = d.
We have found two ℚ-automorphisms, but of course we know that there are more. Instead of trying to find homomorphisms, why don't we take advantage of Corollary 15.1? Since the Galois group is a subgroup of S5, we can rewrite the homomorphisms as permutations. All we need to do is number the roots of the polynomial in the same order as in the circle graphs:
| 1) | a |
| 2) | b |
| 3) | c |
| 4) | d |
| 5) | e |
Since the first ℚ-automorphism swapped the first two roots, and the third and forth roots, this can be represented by the permutation P(2,1,4,3). Likewise, the last ℚ-automorphism discovered swaps the last two roots, and the second and third roots. This becomes the permutation P(1,3,2,5,4).
Since the Galois group is a subgroup of S5, we can ask Sage to find the subgroup generated by these two permutations.
Thus we have 10 permutations. Proposition 15.4 states that the size of the Galois group is equal to the dimension of the splitting field. Since the splitting field is a 2-dimensional extension of a 5-dimensional extension, the Galois group contains exactly 10 elements. Thus, we have found all of the ℚ-automorphisms of the splitting field. Furthermore, we can display a multiplication table of the Galois group:
This reveals that the Galois group is non-abelian, and hence must be isomorphic to D5.
EXPERIMENT:
Can you see how to use Proposition 15.4 and Corollary 15.1 to prove directly that the Galois group of the splitting field of x5 − 5 x + 12 must be isomorphic to D5?
Here is another example that illustrates the variety of groups that can be produced by a Galois group of a polynomial.
EXAMPLE:
Find, up to isomorphism, the Galois group for the eighth degree polynomial
x8 − 12 x6 + 36 x4 − 36 x2 + 9.
Sage can verify that this is irreducible.Thus, we can define a to be one root of this equation.
Can we factor this in the field ℚ(a)?
We can instead verify the factorization:
Using either computation, Sage shows us that the splitting field for this polynomial is simply ℚ(a). Since this is an 8-dimensional extension of ℚ, the Galois group will have eight elements. But which group is this isomorphic to?
Note that the negative of each root is also a root. We will use the following order of the roots for the circle graphs:
| 1) | a |
| 2) | b |
| 3) | c |
| 4) | d |
| 5) | −a |
| 6) | −b |
| 7) | −c |
| 8) | −d |
By Proposition 15.2, there is a ℚ-automorphism ƒ for which ƒ(a) = b.
We can have Sage find where the other seven roots are mapped to.
We see that ƒ(b) = −a, ƒ(c) = −d, and ƒ(d) = c. Any of these results surprising? Using the numbering scheme given above, we can espress this element of the Galois group as P(2, 5, 8, 3, 6, 1, 4, 7).
Let us try a ℚ-automorphism which sends a to c.
Once again, we will use Sage's circle graph to convert this to a permutation.
This element of the Galois group behaves like the permutation P(3, 4, 5, 6, 7, 8, 1, 2). With these two permutations, we can determine whether we can generate the whole Galois group.
This gives eight elements, so this is the entire Galois group. Here is the multiplication table.
Notice that this group is non-commutative, and that there is only one element of order 2. Since D4 has five elements of order 2, this group must be isomorphic to the quaternionic group Q.
EXPERIMENT:
Use a similar approach to find the Galois group of the polynomial x4 − 2.
Here is one more example that at first seems difficult, but it is in fact easy!
EXAMPLE:
Find, up to isomorphism, the Galois group for the polynomial x4 − x + 1.
In the last chapter we saw that the splitting field was 24 dimensional over ℚ. We know from Corollary 15.1 that the Galois group is a subgroup of S4. But S4 has 24 elements, so the Galois group of x4 − x + 1 must be isomorphic to S4.
Sage has a way of determining the Galois group, up to isomorphism, for irreducible polynomials up to degree around 15 (although some polynomials of degree 14 or 15 take an unreasonable amount of time).
We first use InitDomain to define ℚ[x], and then use GaloisType on the polynomial.
In this way, we quickly verified the last two examples. However, this only gives an isomorphic group to the Galois group, instead of explicitly showing the elements of the group. Here is one last example.
Thus, Sage says that the Galois group for the polynomial x5 − x + 1 is S5.
For the final example, we wish to explore a whole class of polynomials at the same time. In the last chapter, we computed the splitting field of the cyclotomic polynomials, and determined that K = ℚ(ωn), where
ωn = e(2πi/n) = cos(2π⁄n) + i sin(2π⁄n).
We can use Proposition 14.9, along with some of the facts observed from §13.3 to find the Galios group of the nth cyclotomic polynomial Φn(x).
From all of these examples, we have seen a host of different groups produced as Galois groups of
polynomials: S3, Z3, D5, Z5, Q, D4, S4, and
all groups of the form Zn*. It is natural to ask whether all finite groups can be expressed as a Galois group of some polynomial
in ℚ[x]. This is still an open problem, known as the inverse Galois problem. There has been much progress made on this problem, and it is
very likely to be solved soon.
While we are working with cyclotomic polynomials and nth roots of unity, let us prove one more proposition that will be used later on.
To introduce the problem of whether a fifth degree polynomial can, in general, be solved in terms of square roots, cube roots, or fifth roots, we will
have Sage try to solve some polynomial equations for us. Sage can solve polynomials with the command
Note that the "double equals" == is Sage's way of expressing an equation. Obviously, Sage is using the quadratic equation to find the roots. Let us try changing the x^2 to x^3:
Sage was still able to solve this, but what a mess! The answer involves the square root of 78. Apparently Sage is using a formula that finds the roots of any cubic equation.
Let us try a fourth degree equation:
Once again, Sage was able to express the answer in terms of square roots and cube roots, yet this seems even more of a mess.
EXPERIMENT:
Replace the fourth degree equation x4 − x + 2 with any other fourth degree equation in x. Is Sage able to find the roots?
The equations for the cubic equation and the fourth degree equation were discovered in 1539 and 1545. The natural question is whether there is a similar formula for fifth degree polynomials. Let us try to solve a fifth degree polynomial in Sage.
What happened? Sage does not know of any formula for the fifth degree polynomial. As a result, Sage keeps the equation intact. Here is how we can get a list of approximate roots:
The CC is Sage's notation for floating point complex numbers. We see that there are indeed five complex roots to the equation.
The problem is not that Sage is not smart enough to find a formula for the fifth degree polynomials. In fact, it is impossible to find a formula for the roots of a fifth degree polynomial in terms of square roots, cube roots, or any other roots. The reason why is based on the properties of the Galois groups. The next section will reveal how the Galois groups are related to the splitting field.
The Fundamental Theorem of Galois Theory
In this section we will clarify the relationship between subgroups of the Galois group, and the subfields of the extension field. The natural correlation is to map to each subgroup of GalF(K) the fixed field of the subgroup. However, we ended §15.1 with what seemed to be a bad example—ℚ(∛2). The only fixed field was ℚ(∛2), even though there was the obvious subfield. The way we will deal with exceptions like this one is to consider only field extensions for which the original field appears as one of the fixed fields.
DEFINITION 15.5
Let K be a finite extension of F. We say that K is a Galois extension if the fixed field of GalF(K) is the field F.
Although this definition successfully rules out ℚ(∛2) from being a Galois extension, we need to find a simple test for determining whether a finite extension is a Galois extension. The following proposition takes us one step in that direction.
This proposition allows us to immediately rule out certain field extensions from being a Galois
extension. Clearly ℚ(∛2) is ruled out
because ℚ(∛2) is not a splitting field. But there are even some splitting fields that are not Galois
fields according to this proposition.
EXAMPLE:
Let Z2(t) be the field of rational functions in t, with coefficients in Z2. This field can be defined in Sage simply by the command
Note this is different than the AddFieldVar command, which only defines the polynomial ring Z2[t]. Here are some examples of elements in Z2(t):
We can now consider polynomials over this field.
Note that there is no element whose square is equal to −t, since this polynomial is irreducible.
Suppose we define a new element a that solves the equation x2 + t = 0.
We can test this extension by computing a2:
Now x2 + t factor should factor in Z2(t)(a).
The factorization gives a "Not Implemented Error" because this field has become too complicated for Sage to have a factoring algorithm. Nonetheless, we can determine what the factorization is.
This simplified to x2 + t, so this is the factorization.
Note, however, that there is a double root in this factorization! Thus, by Proposition 15.8, Z2(t)(a) is not a Galois extension of Z2(t).
One immediate consequence from Proposition 15.8 is that a Galois extension can be written as a simple extension.
In order to introduce the correlation between the subgroups of the Galois group and the subfields of the Galois extension, let us consider the familiar splitting
field of x3 − 2. Since ∛2 and
ω3∛2 are two roots, we can express the splitting field
as ℚ(∛2, ω3∛2). The
subfields of this Galois extension are ℚ, ℚ(∛2),
ℚ(ω3∛2),
ℚ(ω32∛2),
ℚ(√−3), and the whole field
ℚ(∛2, ω3∛2). We
can draw a diagram of these subfields, showing which subfields are subfields of other subfields.
The green lines in this diagram indicate which subfields are Galois extensions of the subfield above it. Also, whenever we have a Galois extension, the corresponding Galois group is shown in blue. This diagram indicates that the splitting field of x3 − 2 is a Galois extension of ℚ(√−3). This can be verified with Sage.
First, we define ℚ(√−3) by defining a2 = −3.
Now, x3 − 2 is still irreducible in this field.
Let us define b to be the cube root of 2.
Now, will the polynomial x3 − 2 factor completely over the field ℚ(√−3, ∛2)?
This factors completely without multiple roots. In fact, this is predicted by Proposition 15.7, since ℚ(√−3) contains the cube roots of unity.
Now let us look at the subgroups of the Galois group S3:
Once again, we draw lines connecting two subgroups if one subgroup is contained in the other subgroup. We draw a green line to indicate that the smaller subgroup is a normal subgroup of the larger. Whenever the subgroup is a normal subgroup, the quotient group is indicated in blue.
The pattern is now obvious. The two pictures are the same, except that the subfields are replaced by a subgroup of S3. This feature of Galois extensions is the heart of Galois theory. In fact, there is a natural way that the subfields of K, and the subgroups of GalF(K) are related: for each subfield E of K, we can consider GalE(K), the set of automorphisms of K that fix E. This is a subgroup of GalF(K). On the other hand, given a subgroup H of GalF(K), we can consider the fixed field fix(H), which is a subfield of K. To show that, indeed, the two pictures will be essentially the same, we need four steps.
(1) Show that if we start with a subfield E, then form the Galois group GalE(K), and find the fixed field of this subgroup, we get back E.
(2) Show that if we start with a subgroup H of GalF(K), find the fixed field, and then find the Galois group of the fixed field, we get back H. These first two steps establish a one-to-one correspondence between the subfields and the subgroups of the Galois group.
(3) Show that if one subgroup N is a normal subgroup of another subgroup H, then the corresponding subfields form a Galois extension. That is, a green line on the second picture corresponds to a green line on the first.
(4) Show that if one subfield E is a Galois extension of another, L, then the corresponding Galois groups will have a normal subgroup relation. Furthermore, the quotient group of the Galois groups will be isomorphic to the Galois group of the Galois extension. Thus, a green line on the first picture corresponds to a green line on the second, and the blue groups in the pictures will be isomorphic.
Let us begin by proving the first step.
Lemmas 15.4 and 15.5 show that there is a one-to-one corresponence between the subgroups of GalF(K) and the subfields of K containing F. We now consider the special significance of the normal subgroups of GalF(K).
Lemmas 15.4 through 15.7 explain the amazing similarity in the diagrams of the subfields, and the subgroups of the Galois group.By putting these four pieces together, we get the fundamental theorem of Galois theory.
THEOREM 15.1: The Fundamental Theorem of Galois Theory
Let K be a Galois extension of the field F. Then there is a one-to-one correspondence between the subfields of K containing F and
the subgroups of GalF(K), given by mapping E to the
subgroup GalE(K). The dimension of K over
the subfield E is |GalE(K)|. Furthermore, a subfield E is a Galois extension of another subfield L if, and only
if, GalE(K) is a normal subgroup of GalL(K) in which case GalL(E) is isomorphic
to
The fundemental theorem of Galois theory has many different applications. With this theorem one can prove that it is impossible to trisect an angle with only a straight
edge and a compass. This finally puts to rest a problem introduced by the ancient Greeks. A similar problem is that of producing a
line ∛2 times the length of a given line with only a straight edge and compass. Both of these Greek problems involves
a field extension of order 3, while any compass and straight edge contruction involve a series of field extensions of order 2. Of course 3 does not divide any power of
2, so a field extension of dimension 3 cannot be a subfield of a field created by a sequence of extensions of order 2. Another important application shows
that a fifth degree equation
cannot be solved in terms of radicals. We will explore both of these applications in the next section.
Applications of Galois Theory
There are two main results of Galois theory. One result is that one can demonstrate that it is impossible to find a formula for the solutions to a fifth degree polynomial in terms of square roots, cube roots, or fifth roots. This finally closed the door on a centuries old problem, ever since the solutions for a third and fourth degree equations were discovered in the 16th century. But the other application of Galois theory applies to problems thousands of years old. The three great construction problems of antiquity are trisecting an angle, duplicating the cube, and squaring the circle. Galois theory proved once and for all that these problems are impossible to construct with only a straight edge and compass.
We will begin with the problem of determining whether or not a polynomial can be solved in terms of radicals. The first step is to show that, in ℚ, a Galois extension is the same thing as a splitting field.
The next step is to give a clear definition of what it means for a polynomial to be solvable by radicals.
DEFINITION 15.6
A field K is called a radical extension of F if K = F(u1, u2, u3, … un), where a power of each ui is contained in F(u1, u2, … ui−1).
EXAMPLE:
Express the splitting field of x4 − 8 x2 − 8 x − 2 as a radical extention.
We can have Sage solve for the roots explicity.
We see the solutions are
| x = − √2 ± | √ 2 − √2 | or x = √2 ± | √ 2 + √2 |
How would we express the splitting field as a radical extension? It is apparent that we first must include √2 in this field. But then it seems we need to include
| √ 2 + √2 | and | √ 2 − √2 |
in our field. Note, however, that the product of these two numbers is √2. Thus, all four roots are in the field
| ℚ(√2, | √ 2 + √2 | ). |
This is a radical extension of ℚ of dimension 4, and the splitting field of x4 − 8 x2 − 8 x − 2 must be at least 4. Hence, we have found that the splitting field is a radical extension of ℚ.
DEFINITION 15.7
The polynomial equation f(x) = 0 is said to be solvable by radicals if there is a radical extension of ℚ that contains the splitting field of f(x).
This definition agrees with our intuitive understanding of what it means for a polynomial to be solved in terms of radicals. For example, Sage's solution to the equation
reveals that the splitting field is contained in the following radical extension:
| ℚ(√26·3, | ∛ 1 − √26·3/9 | ,√−3 ). |
This is in fact overkill, since the splitting field is at most a 6-dimensional extension of ℚ, while the above radical extension may be up to a 12-dimensional extension of ℚ. Yet the point is that there is some radical extension of ℚ that contains the roots of x3 − x + 2, because the roots can be solved in terms of square roots and cube roots.
Not all radical extensions of ℚ are Galois extensions. For example, ℚ(∛2) is not a Galois extension, since this extension is not the splitting field of a polynomial. In order to utilize Galois theory, we need to show that a radical extension is contained in some extension that is both a radical extension and a Galois extension.
Lemma 15.8, when combined with the definition of a polynomial solvable by radicals, tells us that if a polynomial is solvable by radicals, then the splitting field of
the polynomial is contained in a field extension of ℚ that is both a radical extension and a Galois extension.
What can we say about such an extension? The answer is rather startling, since we begin to see a connection with the Jordan-Hölder theorem (8.2).
The light is beginning to appear at the end of the tunnel. We know that any subgroup of a solvable group must be solvable. Thus, we can immediately tell whether a polynomial is solvable by radicals from its Galois group.
Galois' criterion theorem is able to show us that there are some polynomials whose roots cannot be expressed in terms of square roots, cube roots, and other roots. In
fact we can find one of them, namely x5 − x + 1.
At this point, it is clear that we will need yet another second degree extension to finish the splitting field. Since it took a second degree extension of a third degree extension of a fourth degree extension of a fifth degree extension to form the splitting field, the splitting field has dimension 120, and hence the Galois group is S5.
By the way, Sage is able to finish the splitting field, but it takes a VERY long time.
Galois' criterion theorem ended the long search for a formula that finds the roots of a fifth degree polynomial. In fact, Galois' criterion theorem works the other
direction as well—if the Galois group is solvable, then the polynomial is solvable by radicals. Since a fourth degree equation is a subgroup
of S4, which is solvable, there must be a formula for the roots of a fourth degree polynomial. The change of the structure
between S4 and S5 is what changes the behavior of fifth degree polynomials from fourth degree polynomials.
Galois theory also can be used to analyze construction problems. The ancient Greeks had a very rigorous definition of what it meant for a certain geometrical object to be constructed using an unmarked straightedge and compass.
• Given two points that have already been constructed, we can construct the straight line going through the points.
• Given two points that have already been constructed, we can construct a circle with a center at one point and passing through the other.
• Given two lines that have been constructed, the point of intersection becomes a constructed point.
• Given a circle and a line, or two circles, that have been constructed, the points of intersection, if they exist, become constructed points.
Starting with a unit segment, what other lengths can be constructed?
DEFINITION 15.8
We say that a real number x is constructible if we can create a line segment whose length is |x| times the length of the unit segment in a finite number of steps, using only the allowed procedures using the straightedge and compass.
Using plane geometry, it is not hard to see that if x and y ≠ 0 are constructible numbers, then x + y, x − y, x·y, and x/y are constructible. (See Problems 9 through 11.) Thus, the set of constructible numbers forms a subfield F of ℝ. We can characterize the field F by the following theorem.
We can now show that two of the three great construction problems of antiquity are impossible.
The last construction problem involves knowledge that π is transcendental, so it requires a more difficult proof to show that π it not a constructible number.
There are many other applications of Galois theory. For example, one can use the fundemental theorem of Galois theory to prove that any polynomial in the complex numbers has a root in the complex numbers. This is known as the fundemental theorem of algebra.
Proofs:
Proof of Proposition 15.1:
By Lemma 11.5, the set of all ring automorphisms of a ring forms a group. So we only need to show that the set of F-automorphisms of K is a subgroup of the group of all automorphisms. If ϕ1 and ϕ2 are two F-automorphisms of K, then ϕ1(x) = ϕ2(x) = x for all x in F. Thus, (ϕ1·ϕ2)(x) = ϕ1(ϕ2(x)) = x for all x in F. Thus, ϕ1·ϕ2 is an F-automorphism of K. Note also that ϕ1-1(x) = x for all x in F, so ϕ1-1 is also an F-automorphism of K. Since the set of all F-automorphisms of K is closed under multiplications and inverses, the set is a subgroup of the group of ring automorphisms of K. Thus, the set of F-automorphisms of K is a group.
Proof of Lemma 15.1:
Let f(x) = c0 + c1 x + c2 x2 + c3 x3 + … + cn xn. Since u is a root of f(x), we have that
c0 + c1 u + c2 u2 + c3 u3 + … + cn un = 0.
Since ϕ is a ring homomorphism, we have that0 = ϕ(0) = ϕ(c0 + c1 u + c2 u2 + c3 u3 + … + cn un)
= ϕ(c0) + ϕ(c1) ϕ(u) + ϕ(c2) ϕ(u2) + ϕ(c3) ϕ(u3) + … + ϕ(cn) ϕ(un)
Since c0, c1, c2, … cn are in F, we have0 = c0 + c1 ϕ(u) + c2 ϕ(u)2 + c3 ϕ(u)3 + … + cn ϕ(u)n.
Therefore, ϕ(u) is also a root of f(x).Proof of Proposition 15.2:
If there is some ϕ such that ϕ(u) = v, we can let g(x) = IrrF(u, x) and h(x) = IrrF(v, x). Then u is a root of g(x), and v is a root of h(x). By Lemma 15.1, u is a root of h(x) and v is a root of g(x), since v = ϕ-1(u). So g(x) is a multiple of h(x), and vice versa. Since both have a leading coefficient of 1, we have that g(x) = h(x).
Now suppose that IrrF(u, x) = IrrF(v, x). Then by Corollary 14.2 there is an isomorphism ϕ from F(u) to F(v) such that ϕ(u) = v, and ϕ(x) = x for all x in F. Since K is a splitting field of f(x) over F, it is a splitting field of f(x) over both F(u) and F(v). Therefore ϕ extends to an F-automorphism of K (which we will also denote ϕ) by Proposition 14.8. Therefore, ϕ is in GalF(K), and ϕ(u)= v.
Proof of Lemma 15.2:
Let E be the set of all elements x such that ϕ(x) = x. Since ϕ is an F-automorphism, by definition E must contain the elements of F. If x and y are in E, note that
ϕ(x + y) = ϕ(x) + ϕ(y) = x + y,
ϕ(x·y) = ϕ(x)·ϕ(y) = x·y,
ϕ(−x) = −ϕ(x) = −x,
ϕ(x-1) = ϕ(x)-1 = x-1 if x ≠ 0.
Thus, x + y, x·y, and −x are in E whenever x and y are, and x-1 is in E whenever x ≠ 0 is in E. Thus, E is a subfield of K.Proof of Proposition 15.3:
Consider the F-automorphism ϕ2-1(ϕ1(x)). It is clear that this automorphism fixes u1, u2, u3, … un, as well as the elements of F. By Lemma 15.2, the set E of all elements x such that ϕ2-1(ϕ1(x)) = x forms a subfield of K. But K is by Lemma 14.3 the smallest field containing u1, u2, u3, … un, and F. Thus, K = E, and so ϕ1(x) = ϕ2(x) for all x in K.
Proof of Corollary 15.1:
Since f(x) has degree n in F[x], there are at most n roots of f(x) in K. Call these roots u1, u2, u3, … um. Since K is the splitting field of f(x) over F, we can write K = F(u1, u2, u3, … um). If ϕ is in GalF(K), then ϕ(u1), ϕ(u2), ϕ(u3), … ϕ(um) will be distinct roots of f(x) by Lemma 15.1. Hence, ϕ will act as a permutation on the roots of f(x). By Proposition 15.3, ϕ is completely determined by this permutation on the roots of f(x). Thus, GalF(K) is isomorphic to a subgroup of Sm, and since m is not larger than n, GalF(K) is isomorphic to a subgroup of Sn.
Proof of Proposition 15.4:
Let d be the dimension of K over F. Then if g(x) = IrrF(w, x), then g(x) has degree d. Since K is a splitting field and contains one root of g(x), by Lemma 14.5 g(x) splits completely in K. Since there are no double roots of g(x) in K, then there are d roots w = w1, w2, w3, … wd. Since g(x) is irreducible, IrrF(wi, x) = IrrF(w, x) so Proposition 15.2 states that there is an F-automorphism that sends w to wi for 1 ≤ i ≤ d. Hence, there are at least d F-automorphisms. But by Proposition 15.3, the F-automorphism of F(w) is determined by where it sends w, which must be one of the d roots. So |GalF(K)| = d.
Proof of Proposition 15.5:
For each ϕ in H, let Eϕ be the set of elements that are fixed by ϕ. By Lemma 15.2, Eϕ is a subfield of K containing F. By taking the intersection of all Eϕ with ϕ in H, we obtain a subfield of K containing F.
Proof of Lemma 15.3:
Since we are working in ℚ[x], we can use the familiar tools of calculus. Suppose that K is the splitting field of f(x), and u is a multiple root of f(x) in K. Then
f(x) = (x − u)2·g(x).
Since we are working in a field extension of ℚ, we can take the derivative of both sides to getf ′(x) = 2 (x − u)·g(x) + (x − u)2·g′(x).
Thus, u is a root of f ′(x), which has lower degree than f(x). Note that f ′(x) is not 0, since it has degree of at least one.Since f ′(x) is also in ℚ[x], we see that Irrℚ(u, x) has degree less than the degree of f(x), and so Irrℚ(u, x) is a divisor of f(x). But this contradicts the fact that f(x) is irreducible. Therefore, f(x) cannot have multiple roots in its splitting field.
Proof of Proposition 15.6:
Let g(x) = Φn(x), which by Gauss' theorem on cyclotomic polynomials (13.2), is irreducible. The roots of g(x) are of the form ωnk where k is coprime to n. Hence, K is the splitting field of g(x). Since the degree of Φn(x) is ϕ(n), we know from Proposition 15.4 that the size of Galℚ(K) is ϕ(n).
To show that Galℚ(K) is isomorphic to Zn*, note that every ϕ in Galℚ(K) is determined by where it sends ωn, and that it must send it to one of the roots ωnk for some k coprime to n. Thus, there is a natural mapping
ƒ : Galℚ(K) → Zn*
defined byƒ(ϕ) = ( the value k for which ϕ(ωn) = ωnk).
This mapping is well defined since ωnn = 1. This mapping is a homomorphism, for if ƒ(ϕ) = k and ƒ(μ) = m, then(ϕ·μ)(ωn) = ϕ(μ(ωn)) = ϕ(ωnm) = (ωn)k·m,
soƒ(ϕ·μ) = k·m = ƒ(ϕ)·ƒ(μ).
Finally, an element in the kernel of this homomorphism sends ωn to ωn, so Ker(ƒ) is just the identity element of Galℚ(K). Thus, ƒ is an isomorphism from Galℚ(K) to Zn*. Since we know that both Galℚ(K) and Zn* have ϕ(n) elements, the mapping is onto, so Galℚ(K) ≈ Zn*.Proof of Proposition 15.7:
Since u is a root of xn − c, we have that un = c. But (ωn)k·u is also a root of this polynomial for integers k = 0, 1, 2, … n − 1, since
(ωnk·u)n = (ωn)k·n·un = 1·c = c.
Since there are n distinct roots of the polynomial xn − c in K, the polynomial factors completely in K[x]. Thus, K is the splitting field of f(x).To show that GalF(K) is abelian, note that any F-automorphism is determined by where u is sent, which must be of the form ωnk·u. Thus, if ϕ1 and ϕ2 are two F-automorphisms of K, where
ϕ1(u) = ωnk·u and ϕ2(u) = ωnm·u,
then(ϕ1·ϕ2)(u) = ϕ1(ϕ2(u)) = ϕ1(ωnm·u) = (ϕ1(ωn))m·ϕ1(u) = (ωn)m·(ωn)k·u,
while(ϕ2·ϕ1)(u) = ϕ2(ϕ1(u)) = ϕ2(ωnk·u) = (ϕ2(ωn))k·ϕ2(u) = (ωn)k·(ωn)m·u.
Thus, ϕ1·ϕ2 = ϕ2·ϕ1, and so the Galois group is abelian.Proof of Proposition 15.8:
Since f(x) has at least one root in the field K, we will let u1, u2, u3, … un be the set of all roots of f(x) in K. Consider the polynomial
g(x) = (x − u1)·(x − u2)·(x − u3) ⋯ (x − un).
By Lemma 13.3, any automorphism in GalF(K) extends to an automorphism on K[x] with ϕ(x) = x. Thus,ϕ(g(x)) = (x − ϕ(u1))·(x − ϕ(u2))·(x − ϕ(u3)) ⋯ (x − ϕ(un)).
By Lemma 15.1, ϕ(u1), ϕ(u2), ϕ(u3), … ϕ(un) will all be roots of f(x), and so this list is a permutation of the list u1, u2, u3, … un. Therefore, ϕ(g(x)) = g(x) for all ϕ in GalF(K).Now, since K is a Galois extension of F, the fixed field of GalF(K) is the field F. Thus, g(x) is a polynomial in F[x]. Since g(x) certainly divides the polynomial f(x), and f(x) is irreducible in F[x], we have that f(x) and g(x) have the same degree. Thus, n is the degree of f(x), and so f(x) factors completely in the field K. Furthermore, f(x) has no multiple roots in the field K.
Proof of Corollary 15.2:
Since K is a Galois extension of F, K is finite dimensional over F. Thus, K = F(u1, u2, u3, … un) for elements u1, u2, u3, … un in K. But the polynomials [removed]IrrF(ui, x) all have a root in K, and so factor completely in the field K without multiple roots. Then we can use Corollary 14.4 to show that there is an element w in K such that F(w) = K.
Proof of Lemma 15.4:
Let H = GalE(K), which is a subgroup of GalF(K). Let E0 be the field fixed by H. Certainly E0 contains the field E, since every automorphism in H fixes E. Suppose that u is an element of K which is not in E. Let f(x) = IrrE(u, x). Since u is not in E, f(x) has degree at least 2. Note that g(x) = IrrF(u, x) is a polynomial in F[x] for which f(x) is a factor in the domain E[x]. Since F is a Galois field over F, g(x) factors completely in K with no repeated factors. Thus, f(x) also factors completely in K with no repeated factors, so there are at least two solutions to the equation f(x) = 0 in K. One solution is of course u, so let v be another solution. By Proposition 15.2, there is an E-automorphism in H such that ϕ(u) = v. Thus, u is not in E0. Therefore, E0 = E, and so K is a Galois extension of E.
Proof of Lemma 15.5:
Let n be the dimension of the field K over E. By Lemma 15.4, K is a Galois extension of E. Thus, by Corollary 15.2, there exists an element w in K such that K = E(w). If f(x) = IrrE(w, x), then the degree of f(x) is n by Corollary 14.1. Since K is a Galois extension of E, by Proposition 15.8, the polynomial f(x) factors completely in the field K, and there are no multiple roots. Thus, by Proposition 15.4, the number of E-automorphisms of K is the dimension of K over E, which is n.
Suppose that H contains m E-automorphisms. Let v1, v2, v3, … vm be the images of w under the automorphisms in the subgroup H. That is, for each vi there is an ƒ in H such that vi = ƒ(w).
Consider the polynomial
g(x) = (x − v1)·(x − v2)·(x − v3) ⋯ (x − vm).
If ϕ is an automorphism in H, then ϕ(vi) = ϕ(ƒ(w)) = vj for some j. Also, since ϕ is one-to-one, the images of ϕ(v1), ϕ(v2), ϕ(v3), … ϕ(vm) must all be distinct. Thus, each ϕ in H is a permutation on the elements v1, v2, v3, … vm. Hence, ϕ(g(x)) = g(x) . Since E is the fixed field fix(H) of the subgroup H, we see that g(x) is in E[x]. Thus, f(x) = IrrE(u, x) divides g(x), so m is at least n. Thus,|H| ≤ |GalE(K)| = n ≤ m = |H|.
Therefore, H = GalE(K).Proof of Lemma 15.6:
First, we want to show that if u is in E, and ϕ is in GalL(K), then v = ϕ(u) is in E. Since GalE(K) is a normal subgroup of GalL(K), for any ƒ in GalE(K) we have that ψ = ϕ-1·ƒ·ϕ is in GalE(K). Then ƒ·ϕ = ϕ·ψ, or
ƒ(ϕ(u)) = ϕ(ψ(u)).
Since u is in E, ψ(u) = u, so
ƒ(v) = ƒ(ϕ(u)) = ϕ(ψ(u)) = ϕ(u) = v.
Thus, v is fixed by every automorphism ƒ in GalE(K). By Lemma 15.4, K is a Galois extension of E, so the fixed field of GalE(K) is E. Thus, v is in E.To show that the fixed field of GalL(E) is L, consider an element u in E that is not in L. By Lemma 15.4, K is a Galois extension of L. Since u is not in the fixed field of GalL(K), there is an L-automorphism ϕ that moves u to another element, v. But ϕ moves all elements of E to elements of E, so we can consider the restriction of ϕ on the field E, denoted ϕ′. This is an automorphism of E, since the inverse is (ϕ-1)′. Thus, there is an L-automorphism of E that moves the element u, so the fixed field of GalL(E) is only L. Therefore, E is a Galois extension of L.
Proof of Lemma 15.7:
By Lemma 15.4, K is a Galois extension of L. We begin by showing that if ϕ is an L-automorphism of K, and u is in E, then ϕ(u) is in E. Let g(x) = IrrL(u, x). Since E is a Galois extension of L, by Proposition 15.8, g(x) factors completely in E[x], which is of course the same factorization in K[x]. By Lemma 15.1, ϕ(u) is a root of g(x) in K, but all of the roots are also in E. Thus, ϕ(u) is in E.
Next, we define the mapping ƒ that sends an L-automorphism of K to its restriction on the field E. We denote the restriction of ϕ on the field E by ϕ′. Since ϕ maps elements of E to elements of E, we see that ϕ′ is an L-automorphism of E. However, (ϕ-1)′ is also an L-automorphism of E, and (ϕ-1)′·ϕ′ is clearly the identity mapping on E. Thus, ϕ′ is an element of GalL(E).
To show that ƒ is a homomorphism, note that
ƒ(ϕ1·ϕ2) = (ϕ1·ϕ2)′ = ϕ1′·ϕ2′ = ƒ(ϕ1)·ƒ(ϕ2).
The kernel of this homomorphism is simply the L-automorphisms of K that fix the elements of E, which is of course GalL(E).Finally, to show that this homomorphism is surjective, let ψ be an L-automorphism of E. Since K is a splitting field of E, we can use Proposition 14.8 to extend ψ to an L-automorphism of K, which we will call ϕ. Then ƒ(ϕ) = ψ, and we have shown that ƒ is surjective.
Proof of Theorem 15.1:
If GalE(K) = GalL(K) for two subfields E and L of K, then by Lemma 15.4, both E and L are the fixed field of the subgroup GalE(K) = GalL(K), so E = L. Thus, the mapping E → GalE(K) is one-to-one. But if H is any subgroup of GalF(K), then we can consider E to be the fixed field fix(H), and by Lemma 15.5 GalE(K) = H. Thus, the correspondence is also onto. Also by Proposition 15.4, the dimension of K over E is |GalE(K)|, since K is a Galois extension of E.
If E is also a Galois extension of another subfield L, then by Lemma 15.7 there is a surjective homomorphism from GalL(K) to GalL(E), whose kernel is GalE(K). Thus, GalE(K) is a normal subgroup of GalL(K), and by the first isomorphism theorem (4.1), GalL(E) is isomorphic to GalL(K) / GalE(K).
Finally, suppose that GalE(K) is a normal subgroup of GalL(K). By Lemma 15.6 E is a Galois extension of L.
Proof of Proposition 15.9:
Let K be the splitting field of f(x) in E[x]. If u is an element of K not in E, then g(x) = IrrE(u, x) has degree > 1. By Lemma 14.5, g(x) factors completely in the field K. Thus, the splitting field of g(x) is contained in the field K. However, g(x) is a factor of Irrℚ(u, x), which by Lemma 15.3 does not have multiple roots in K. Therefore, g(x) cannot have multiple roots in K, so there exists at least two roots of g(x) in K. Let v be a root of g(x) different from u. Then g(x) = IrrE(v, x), and so by Proposition 15.2 there exists a ϕ in GalE(K) such that ϕ(u) = v. Thus, u is not in the fixed field of GalE(K). Since E is obviously contained in the fixed field of GalE(K), we find that the fixed field is E, so K is a Galois extension of E.
Proof of Lemma 15.8:
Let E = ℚ(u1, u2, u3, … un) be a radical extension of ℚ. Then for every i = 1, 2, 3, … n, there is a ki for which
(ui)kⁱ = v, for which v ∈ ℚ(u1, u2, … ui−1).
Note that if n = 0, then E = ℚ, and the lemma is obviously true. We will prove this by induction on n. That is, we will assume that the lemma is true for the fieldℚ(u1, u2, u3, … un−1).
That is, this field is contained in a radical extension L of ℚ that is also a Galois extension of ℚ.By Corollary 15.2, there exists an element w of L such that L = ℚ(w).
Let g(x) = Irrℚ(w, x) and p(x) = Irrℚ(un, x). Let K be the splitting field of g(x)·p(x) over ℚ. By Proposition 15.9 K is a Galois extension of ℚ. Since w is in K, L is a subfield of K. The only thing left to show is that K is a radical extension of L.
Let v1, v2, v3, … vm be all of the roots of p(x) in K. Since p(x) is irreducible, by Proposition 15.2 there is a ℚ-automorphism ϕi that sends vi to un. Since (un)k = b is in L, we have
(vi)k = (ϕi(un))k = ϕi(unk) = ϕi(b).
Now, L is a Galois extension of ℚ, so by the fundamental theorem of Galois theory (15.1), GalL(K) is a normal subgroup of Galℚ(K). So by Lemma 15.6 ℚ-automorphisms of K map elements of L to elements of L. Thus, ϕi(b) is in L, and so K = L(v1, v2, v3, … vm) is a radical extension of L.Proof of Lemma 15.9:
Since K is a radical extension of ℚ, we can write
K = ℚ(u1, u2, u3, … un)
where some power of each ui, (ui)kⁱ, is in ℚ(u1, u2, u3, … ui−1).Let m be the least common multiple of all of the ki, and let u0 = ωm, the mth root of unity. We would like to add u0 in the front of the sequence of u's to get a larger field
M = ℚ(u0, u1, u2, u3, … un)
Since (u0)m = 1, we see that M is still a radical extension of ℚ. To show that M = K(u0) is a Galois extension of ℚ, note that by Corollary 15.2, K = ℚ(w) for some element w in K. If f(x) = Irrℚ(w, x), then M is the splitting field of the polynomial f(x)·(xm − 1). Thus, by Proposition 15.9, M is a Galois extension of ℚ.Consider the sequence of subfields
| E0 = | ℚ(u0), |
| E1 = | ℚ(u0, u1), |
| E2 = | ℚ(u0, u1, u2,) |
| E3 = | ℚ(u0, u1, u2, u3), |
| … | ……… |
| En = | ℚ(u0, u1, u2, u3, … un) = M. |
By Proposition 15.7, each of these fields is a Galois extension of the previous field, since the mth roots of unity were designed to be in all of these fields. Also, by Proposition 15.6, E0 is a Galois extension of ℚ.
We can now apply the fundamental theorem of Galois theory (15.1). We find that GalEₖ(M) is a normal subgroup of GalEₖ₋₁(M), and the quotient group
GalEₖ₋₁(M) / GalEₖ(M)
is isomorphic to GalEₖ₋₁(Ek).By Proposition 15.7, each of these quotient groups are abelian. Also, by Proposition 15.6, Galℚ(E0) is isomorphic to Zm*, which is abelian. Thus, the sequence of subgroups
Galℚ(M) ⊇ GalE₀(M) ⊇ GalE₁(M) ⊇ GalE₂(M) ⊇ GalE₃(M) ⊇ ⋯ ⊇ GalEₙ(M) = {e}
is a subnormal series for which all of the quotient groups are abelian. Therefore, the composition series of Galℚ(M) will consists of only prime, cyclic factors. By the solvability theorem (8.3), Galℚ(M) is a solvable group.To finish the theorem, we note that E is a Galois field of ℚ, so by the fundamental theorem of Galois theory (15.1), GalE(M) is a normal subgroup of Galℚ(M), and Galℚ(E) is isomorphic to Galℚ(M) / GalE(M). Using Proposition 8.3 we see that Galℚ(E) is solvable.
Proof of Theorem 15.2:
Suppose that f(x) is a polynomial that is solvable by radicals. Let E be the splitting field of f(x). By Lemma 15.8, there is a field K containing E which is a Galois extension of ℚ, and also is a radical extension of ℚ. By Proposition 15.9, E is a Galois extension of ℚ. Thus, we can use Lemma 15.9 to show that the Galois group of f(x), Galℚ(E), is a solvable group.
Proof of Corollary 15.3:
We have already shown that the Galois group of x5 − x + 1 is isomorphic to S5. But S5 is not solvable, since it contains the non-cyclic simple subgroup A5. Thus, by Galois' criterion theorem (15.2) this particular equation cannot be solved with a formula involving only field operations and extraction of roots, so certainly there can be no general formula.
Proof of Theorem 15.3:
Since it is possible to construct a line perpendicular to a given line and going through a particular point, we can use Cartesian coordinates to describe the constructible numbers. By first constructing two perpendicular lines, the x- and y-axes, given two constructible numbers a and b, we can construct the point (a, b) using two perpendicular lines. Likewise, if some point is constructible, we can drop perpendiculars to construct the numbers of the coordinates. Thus, we see that the point (a, b) is constructible if, and only if, a and b are constructible numbers.
We can likewise characterize the lines and circles that are constructible. A line is clearly constructible if and only if the x and y intercepts are constructible numbers. Thus, the line a x + b y + c = 0 is constructible if, and only if, a, b, and c can be made to be constructible numbers. Likewise, the circle x2 + y2 + a x + b y + c = 0 is constructible if and only if a, b, and c are constructible numbers.
Now, the intersection of two lines can be found using only field operations. But the intersection of a line and a circle, or two circles, will involve a quadratic equation. The solution of a quadratic equation involves taking a square root of an element in F. This means that a number x is a constructible number if, and only if, there is a sequence of fields ℚ = F1, F2, F3, … Fn with x ∈ Fn such that each Fk+1 = Fk(√aₖ), where ak > 0 is in Fk.
Thus, we see that whenever x is a constructible number, it is contained in a radical extension of ℚ. Furthermore, since only square roots are involved in the field extensions, the dimension of the radical extension will be 2n for some n.
Proof of Corollary 15.4:
Suppose it were possible to trisect any angle. Then in particular it would be possible to trisect a 60° angle, hence we would be able to construct a 20° angle, so x = cos(20°) would be a constructible number. But we saw in §11.4 that x is a root of the irreducible polynomial 8 x3 − 6 x − 1. Hence, x cannot be in an field extension of ℚ of dimension 2n. So we cannot trisect a 60° angle with a straightedge and compass, let alone the general angle.
Likewise, we can show that duplicating the cube is impossible. This would involve constructing a line of length ∛2, which is a root of x3 − 2. Again, since ℚ(∛2) is a 3 dimensional extension of ℚ, this cannot be contained in a field of dimension 2n, so ∛2 is not a constructible number.
Sage Interactive Problems
For Problems 17 through 22: Find the set of ℚ-automorphisms for the splitting fields of the following polynomials.
§15.1 #17) x3 − 3 x − 1
§15.1 #18) x3 − 3 x + 3
§15.1 #19) x4 − 5 x2 + 5
§15.1 #20) x4 − x2 + 1
§15.1 #21) x4 + x2 - 1
§15.1 #22) x5 − x4 − 4 x3 + 3 x2 + 3 x − 1
For Problems 16 through 21: Use Sage to find the Galois group of the polynomial. Determine the number of elements in the Galois group, and display a multiplication table of the subgroup of Sn isomorphic to the Galois group.
§15.2 #16) x4 − 2
§15.2 #17) x5 − 2
§15.2 #18) x5 + 15 x + 12
§15.2 #19) x5 + x4 − 4 x3 − 3 x2 + 3 x + 1
§15.2 #20) x4 − 10 x + 1
§15.2 #21) x8 − 108 x6 + 1548 x4 − 3888 x2 + 1296
§15.2 #22)
Use Sage to find the Galois group of x5 + 20 x + 16. How many elements are in the Galois group? (This may take longer than the above problems.)
§15.3 #16)
Let f(x) = Φ8(x) = x4 + 1, and let F be the splitting field of f(x) over ℚ. Use Sage to find the fixed fields for the 3 elements of Galℚ(F) of order 2. Express these fields in the form ℚ(√a), where a is rational.
§15.3 #17)
Let f(x) = Φ12(x) = x4 − x2 + 1, and let F be the splitting field of f(x) over ℚ. Use Sage to find the fixed fields for the 3 elements of Galℚ(F) of order 2. Express these fields in the form ℚ(√a), where a is rational.
§15.3 #18)
Let f(x) = Φ16(x) = x8 + 1, and let F be the splitting field of f(x) over ℚ. Use Sage to find the fixed fields for the 4 elements of Galℚ(F) of order 4. Express these fields in the form ℚ(√a), where a is rational.
§15.3 #19)
Let f(x) = Φ15(x) = x8 − x7 + x5 − x4 + x3 − x + 1, and let F be the splitting field of f(x) over ℚ. Use Sage to find the fixed fields for the 4 elements of Galℚ(F) of order 4. Express these fields in the form ℚ(√a), where a is rational.
§15.3 #20)
Let f(x) = Φ20(x) = x8 − x6 + x4 − x2 + 1, and let F be the splitting field of f(x) over ℚ. Use Sage to find the fixed fields for the 4 elements of Galℚ(F) of order 4. Express these fields in the form ℚ(√a), where a is rational.
§15.4 #19)
Express ℚ(ω5) as a radical extension using only square roots. This, along with Problem 14, can be used to show how to construct a regular pentagon.
Hint: Find a subnormal sequence for the Galois group, and then find generators for the fixed fields for each subgroup.
§15.4 #20)
Express ℚ(ω17) as a radical extension using only square roots. This, along with Problem 14, can be used to show how to construct a regular pentagon. See the hint for Problem 19.