Galois Theory Overview/Example Part 2: Galois Group and Fixed Fields I ll repeat the overview because it explains what I m doing with the example. Then I ll move on the second part of the example where we find the Galois group for the polynomial, and look at the relationship between its subgroups and the subfields of the splitting field for the polynomial. Overview: The main problem that motivated the development of Galois theory was the (un)solvability of the Quintic. Solvability by radicals: The quadratic formula x = b ± b2 4ac demonstrates that any 2 nd 2a degree polynomial can be solved by radicals in the sense that the roots can be expressed in terms of sums, products, and radicals involving the coefficients. Question: Can every 5 th degree polynomial be solved by radicals? The short answer is no because there are 5 th degree polynomials whose Galois group is isomorphic to S5 which is not a solvable group. This question of course raises other questions: What is a Galois group? What does the solvability of a Galois group have to do with whether you can solve a polynomial by radicals? The way it works (basically) is that given a polynomial, p(x), with coefficients in Q, you can find a bigger field containing Q that contains all the roots of p(x). If the smallest such extension of Q (the splitting field of p(x)) is the right kind of extension (an extension by radicals) then p(x) can be solved by radicals. Now, for every extension field, E, of Q we can consider the group of automorphisms of E that hold the elements of Q fixed. This is the Galois group of E over Q, denoted Gal(E/Q). If p(x) doesn t have any multiple roots and E is the splitting field of p(x), then Gal(E/Q) is called the Galois group of p(x). It turns out that there is a very close relationship between the subgroups of the Galois group of p(x) and the subfields of the splitting field E that contain Q. In fact, the subfield lattice is just the subgroup lattice turned upside- down. The solvability of a group is related to the structure of the subgroup lattice, so it is also related to the structure of the subfield lattice, which in turn is related to whether the splitting field is an extension by radicals. The big conclusion, then is that a polynomial is solvable by radicals if and only if its Galois group is solvable. The polynomial x 5-6x + 3 has Galois group S5. S5 is not solvable because A5 is simple. The End!
Example (54.3): Galois Group Now let s talk automorphisms. First, let s consider our first field extension. Let s see how many automorphisms of Q(α) there are that fix Q. First I want to show that such an automorphism must take a root of p(x) to another root of p(x). Suppose that ϕ is an automorphism of Q(α) that fixes Q. Let β be a root of p(x). Then p(β) = 0. This implies that ϕ(p(β)) = 0 since homomorphisms always map 0 to 0. Then ϕ(β 4 2) = 0, which implies that ϕ(β) 4 ϕ(2) = 0 by the homomorphism property. But ϕ(2) = 2 since ϕ fixes Q. So, we have ϕ(β) 4 2 = 0 and ϕ(β) is a root of x 4 2 = p(x). Actually this works in general. Any automorphism of an extension E of a field F that fixes F will take any zeros of a polynomial in F[x] to another (or the same) zero of that polynomial. Any such automorphism will also be completely determined by where it maps the elements of a basis of the extension (as a vector space). Recall that a basis for Q(α) over Q is {1, α, α 2, α 3 } So, we just need to see where we can send α, because that will determine where α 2 and α 3 are mapped. There are only two choices, α and - α because the other two roots are not elements of Q(α). This gives us two automorphisms. Now, let s look at the automorphisms of Q(α, i) that fix Q(α). Since the basis is {1, i} all we have to do is say were to send i. Again, it must go to another root of x 2-1. So, it can be mapped to either i or i. This gives us two automorphisms. Finally, let s consider the automorphisms of Q(α, i) that fix Q. Now our basis is {1, α, α 2, α 3, i, iα, iα 2, iα 3 }. We can completely define any automorphism by deciding where it sends i and α. α is a root of p(x), so an automorphism on Q(α, i) that fixes Q must take α to α, - α, iα, or - iα. And since i is a root of x 2 + 1, any such automorphism must take i to i or i.
This gives us 8 automorphisms as shown in the following table: Automorphism α i e α i r 2 - α i r iα i r 3 - iα i f α - i for 2 - α - i for 3 iα - i for - iα - i I got the names by composing the maps called r and f to generate the rest of them. Note that f 2 = e and r 4 = e. And rof = for 3 so the group of these automorphisms is isomorphic to D8. Since Q(α, i) is the splitting field for p(x), this group of automorphisms is the Galois group for p(x). Since it is isomorphic to D8, we can easily list off its subgroups: Gal(Q(α, i)/q) = r, f, r, r 2, f, f r, f r 2, f r 3, r 2, f, r 2, f r, and {e}. Let s draw a cool subgroup lattice!
Example (54.3): Fixed Fields Now, let s try and identify all of the fields between Q and Q(α, i). More specifically, let s find the fixed field for each of these subgroups of the Galois group. First, let s think about how big each of these fixed fields should be. The fixed field for {e} leaves all of Q(α, i) fixed. That is the biggest fixed field we can expect (it is of degree 8 over Q). The fixed field for the whole Galois group is Q. The Galois group is the full set of automorphisms that fix Q. It is clear that no basis element stays fixed under all of these maps. It s less clear that no combination is fixed (but still true). This is a degree 1 extension over Q. We ll expect to find degree 2 and degree 4 extensions between these. The subgroups of order 4 contain more maps that the subgroups of order 2, so they should have smaller fixed fields (more chances for things to move). The subgroups of order 2 contain more than the trivial map, so should have smaller fixed fields than {e}. So we should expect the order- 4 subgroups to have degree- 2 fixed fields and the order- 2 subgroups to have degree- 4 fixed fields. Now, we ll consider all the proper subgroups in turn:
First, r 2. Recall that our basis is: {1, α, α 2, α 3, i, iα, iα 2, iα 3 }. The map r 2, fixes i, and it also fixes α 2 (= 2 ) and hence iα 2. So, it leaves all of Q(α 2, i) fixed. Since this subgroup has index 2, we expect the fixed field to be a fourth degree extension. This one, is because it can be created in two steps (each of degree 2). First, extend to Q(α 2 ) by forming Q[α 2 ]/<x 2-2>. This is a second- degree extension since x 2-2 is an irreducible second degree polynomial (Eisenstein with p = 2). Since this new field has no imaginary numbers in it, the polynomial x 2 + 1 is irreducible and so Q(α 2, i) Q(α 2 )[x]/< x 2 + 1 > is an extension of degree 4 over Q. Now, r. r fixes i, but sends α to iα. This means that α 2 goes to - α 2, and α 3 goes to - iα 3. The map will take iα to - α, iα 2 to - iα 2, and iα 3 to - α 3. This means that of the basis elements, r only fixes 1 and i. Note that both r 2 and r 3 also fix i. This is a root of the irreducible 2 nd degree polynomial x 2 + 1. Thus the fixed field for r is Q(i) Q[x]/< x 2 + 1 >. Now, f. f fixes α, and hence α 2 and α 3. It does not fix i, so doesn t fix any other basis element (other than 1). This is a subgroup of index 4, so we are looking for a fourth degree extension. Since α is a root of our original irreducible 4 th degree polynomial, our fixed field is the first extension field we constructed: Q(α). Now, f r 2. f r 2 sends α to - α. So, it fixes α 2, but sends α 3 to - α 3. Since it sends i to i, iα is fixed! Note that iα one of the roots of our irreducible polynomial x 4 2, so we know that Q(iα) Q[x]/< x 4-2> is a field extension of 4 th degree. This has to be our fixed field because any bigger field would have to be all of Q(α, i). Now, r 2, f. The basis elements fixed by this subgroup are those fixed by f, r 2, and f r 2. Looking at our previous work, the only thing fixed by all of these is α 2. This element is a root of the irreducible 2 nd degree polynomial x 2 2, so the fixed field for r 2, f is Q(α 2 ) Q[x]/< x 2-2>. Now, r 2, f r. The map f r takes α to - iα, and i to - i. So, it takes iα 2 to i(- iα)(- iα) = iα 2. The map f r 3 takes α to iα, and i to i, so it also leaves iα 2 fixed. We saw above that r 2 left i and α fixed, so it also leaves iα 2 fixed. Since this subgroup has index 2 in the group Gal(Q(α, i)/q), we are looking for a field extension of degree 2 over Q. That works out because iα 2 is a root of the irreducible (by Eisenstein with p = 2) polynomial x 2 + 2. So the fixed field for r 2, f r is Q(iα 2 ) Q[x]/< x 2 +2>.
Now for the two hard ones! The subgroup f r leaves iα 2 fixed. But the field Q(iα 2 ) Q[x]/< x 2 +2> isn t the fixed field that we are looking for. Since f r is a subgroup of r 2, f r which has fixed field Q(iα 2 ), it should have a larger fixed field (fewer maps, so they should leave more things fixed). So, we re looking for a field extension of degree 4 that contains Q(iα 2 ). I claim that for any β in Q(α, i), the element e(β) + f r (β) is fixed under both e and f r. This follows from f r being a group of order 2: e(e(β) + f r (β)) = e(e(β)) + e( f r (β)) = e(β) + f r (β) and f r [(e(β) + f r (β))] = ( f r) e (β) + ( f r) ( f r)(β) = ( f r) (β) + e(β) = e(β)+ ( f r) (β). Now, if we let β = α, we see that this means that e(α) + f r (α) = α - iα is fixed by both elements of f r. Just for fun, let s make sure this element really is fixed by f r. f r (α - iα) = f r (α) + f r ( - iα) = - iα - f r ( i) f r (α) = - iα - (- i)(- iα) = - iα +α. Now, do we know if Q(α - iα) is an extension of the correct degree? We re looking for a 4 th degree extension, so we should see if we can find a 4 th degree irreducible polynomial with α - iα as a root. The elements of Gal(Q(α, i)/q) must take roots of this polynomial to other roots of this polynomial. Even better, we don t need to check all 8 automorphisms, we can just choose one from each left coset of f r. Let s compute those cosets: ro f r ={r, f} r 2 o f r ={ r 2, for 3 } r 3 o f r ={ r 3, for 2 } So, we can find our roots by plugging α - iα into r, r 2, and r 3. r(α - iα) = r(α) - r(i) r(α) = iα - i(iα) = iα +α (good, we would expect to find the conjugate) r 2 (α - iα) = r 2 (α) - r 2 (i) r 2 (α) = - α - i(- α) = - α + iα (hopefully the next one will be the conjugate of this one) r 3 (α - iα) = r 3 (α) r 3 (i) r 3 (α) = - iα - i(- iα) = - α - iα (Yay!) So, let s multiply and see if we find a nice irreducible (over Q) polynomial. (x - (α - iα))(x - (α + iα))(x - (- α - iα))(x - (- α + iα)) = (x 2 [(α - iα) + (α + iα)]x + 2α 2 ) (x 2 [(- α - iα) + (- α + iα)]x + 2α 2 ) = (x 2 2αx + 2α 2 ) (x 2 +2αx + 2α 2 ) = x 4 2αx 3 + 2α 2 x 2 + 2αx 3-4α 2 x 2 + 4α 3 x + 2x 2 α 2-4α 3 x + 4α 4 = x 4 + 8.
Note that Eisenstein doesn t help us to prove that this polynomial is irreducible over Q. But looking at it in factored form, we can see that it clearly has no linear factors in Q[x], and any method of forming quadratic factors (from the linear factors) other than the one used above will result in complex coefficients. The factorization (into two quadratics) above also doesn t work over Q since the linear and constant coefficients of both terms are irrational. Anyhow, our fixed field is Q(α - iα) Q[x]/< x 4 +8> which is a 4 th degree extension of Q as desired. Let s use a similar method to figure out the fixed field for f r 3. Again for any β in Q(α, i), the element e(β) + f r 3 (β) is fixed by both e and f r 3. We let β = α to see that e(α) + f r 3 (α) = α + iα. This is actually another root of x 4 +8, so we know that Q(α + iα) is a 4 th degree extension of Q. (This one is clearly isomorphic to the previous one but is a different field). Note that the inverse of α - iα is α + iα = α + iα α + iα, so is in Q(α - iα), but since 2 2 is not in 2α 2 2 2 2 2 Q(α - iα), α + iα is not in Q(α - iα). Now that we have all of our fixed fields, we can make a subfield lattice!
Let s put these side by side (with the subfield lattice inverted), so that we can easily see both the inverse relationship and the correspondence between subgroups and subfields This illustrates the fundamental theorem of Galois theory, which basically says that this is what will always happen.