This is one of my favorite topics, and an area of ongoing research: given a finite group G, is G the galois group for some number field?
There are three conjectures associated with finite galois groups, each stronger than the previous.
Every finite group g is the galois group of an extention of the rationals.
Every finite group g is the galois group of infinitely many mutually disjoint extentions for any number field K. Set K to the rationals and select any extention to show 2 implies 1.
Every finite group g is gal_t over the rationals. I'll describe gal_t below.
Let's warm up with some examples. If a number field F has galois group G, and E is the subfield fixed by the subgroup H, and H is normal in G, then E is galois over Q with galois group G/H. With this in mind, let's look for a number field whose galois group is cyclic of order 80. A theorem from analytic number theory, not included in this book, says there are infinitely many primes in any arithmetic series. So there is a prime p that is 1 mod 80, such as 241. Build the cyclotomic extension of order 241. The galois group is isomorphic to the multiplicative group of units mod 241, which is cyclic of order 240. G has a subgroup H of order 3, generated by y → y15, where y is the primitive 241 root of 1. H is normal in G, as one would expect since G is abelian, and the quotient group is cyclic of order 80. The subfield fixed by H is the desired number field with galois group Z/80. This can be done for any cycle length, hence every cyclic group is galois for some number field, in particular, a subfield of a cyclotomic extension. Here are a couple of cycles of order 3: x3 + x2 - 2x - 1, based on cyclo 7, and x3 - 3x + 1, based on cyclo 9.
The following is a tedious construction of A4. Since A4 has a kernel of Z/2 cross Z/2 and a quotient group of Z/3, we can expect a galois subfield whose galois group is cyclic of order 3. Two such fields were discussed in the previous paragraph; focus on the latter, based on cyclo 9. Its cubic equation is x3 - 3x + 1.
Review the quartic formula, and try to stumble upon the cubic z3 - 3z + 1, which has three roots derived from cyclo 9. Think of z as -2y, scaling the roots and giving the cubic 8y3 - 6y - 1. To obtain this, the coefficient denoted p in the quartic formula has to be 0, giving 8y3 - 8ry - q2. Set q = 1 and r = ¾. This comes from the quartic x4 + x + ¾, or 4x4 + 4x + 3.
Look for a rational root whose numerator divides 3 and denominator divides 4. ±1 and ±3 don't work, and if you plug in any fraction with 2 or 4 in the denominator the lead term gives a fraction and the linear and constant terms are integers. If this polynomial factors it is the product of two quadratics. Let the lead terms be positive, 2 and 2 for example, giving 2x2+vx+1 times 2x2-vx+3. (The constant terms could be negative.) Thinking about the middle coefficient on the product, 0 = -v2±8. There is no such integer v, so we have 4x2+4v+1 times x2-v+3. (Constants could be negative and/or swapped, and linear signs could be swapped.) 4v2 has to cancel ±13 or ±7. again there is no such v, hence our quartic is irreducible and generates an extension of dimension 4 over the rationals. Our quartic has 4 distinct roots.
The only zero derivative occurs at the cube root of -1/4, -0.63, and that yields a minimum above 0, hence all 4 roots are complex. Complex conjugation performs a double transposition on these roots, as we'll see below.
Let w be the primitive ninth root of 1, and let y = -(w+w8)/2, or perhaps -(w2+w7)/2, or perhaps -(w4+w5)/2. These are all real, the first two negative and the last one positive. The quartic formula says we can use any of these values of y and get the same 4 solutions, the same 4 values of x. Shift the values of y about, and the roots of the quartic shift as well. As per the formula, x is the root of the following quadratic equation.
x2 + y = ± sqrt(2y) × (x+1/4y)
The linear term changes with the square root of y, which changes with y, hence the quadratic, and the two roots selected, depends on y and the sign of the square root. One of the values of y is positive, giving two quadratics with real coefficients. In this case the root pairs are complex conjugates.
build a tower of extensions by adjoining y0 to Q, then the square root of y0, then x0. These have dimensions 3, at most 2, and at most 2. Yet the entire extension includes x0 with dimension 4, hence each quadratic step is a new field. The composite dimension is 12. We could achieve the same extension by adjoining y0 and then x0, since sqrt(y0) can be calculated from x0 and y0 and rational numbers. Thus our quartic remains irreducible over Q(y). With y0 comes y1 and y2, and their square roots thanks to x0. The conjugate root x0 is brought in, according to the quadratic that owns x0. At least two different quadratics are implied, thus two other roots besides x0 are present. That's 3 roots of the quartic. Each root implies its complex conjugate, hence all 4 roots are brought in. Adjoining x0 to Q(y) causes the quartic to split fully, giving the 4 roots.
Move y0 to y1, move sqrt(y0) to ±sqrt(y1), and move x0 solving the first quadratic to x1 solving the second quadratic. Both values of x solve the quartic. The automorphism on the subfield Q(y) has been extended up to x. The order of this automorphism is 3, 6, or 12; ratchet it up to an automorphism of order 3 that still moves y about. The galois group includes a 3 cycle moving y0 to y1 to y2.
fix y at its positive value, so that the square root of y is real, either positive or negative. These two quadratics with real coefficients have roots that are complex conjugates. You can swap roots by complex conjugation, or you can swap the two quadratics and trade one pair of roots for the other. This builds the subgroup Z/2 cross Z/2. These subgroups of order 3 and 4 imply a group of order 12, matching the dimension, hence the extension is galois.
The galois group G acts on the 4 roots of the quartic. We already saw the 4 sylow subgroup of G moving the roots about. Move y0 to y1 and the coefficients on the quadratic change from real to complex. The two roots of the quadratic cannot be the same. At least one of the two roots moves. The 3 cycle of G acts effectively on the 4 roots of the quartic. It moves three roots and leaves the fourth alone. The 3 cycle and the double transpositions are even permutations. Remember that G has size 12, hence G, captured as permutations on 4 elements, is A4. x4 + x + ¾ has galois group A4.
How about an arbitrary direct product of 2 cycles? Given k different prime numbers, each 1 mod 4, adjoin the square root of each prime to the integers. By induction this builds an extension of dimension 2k. Let p be a prime that is 1 mod each of these k primes. In other words, p is 1 mod their product. How does p split in the integral ring?
Bring in u1 the square root of the first prime q1, creating an extension R1 of dimension 2 over Q. Since p is 1 mod q1 it is a square mod q1, and by quadratic reciprocity, q1 is a square mod p. Review quadratic number fields; the ideal generated by p splits into two ideals, each having a residue degree of 1. The residue field looks just like the integers mod p.
Adjoin u2 the square root of q2, building R2 above R1. Any prime ideal in R1 lying over p looks like the integers mod p, and is 1 mod q2. Let x be the square root of q2 mod p, and the norm of x+u2 is a multiple of p. After localization, x+u2 and x-u2 generate the two conjugate prime ideals, lying over the given prime ideal, lying over p. There are now 4 prime ideals in R2 lying over p.
Adjoin the square root of q3, q4, etc, and p splits into 2k prime ideals. The residue degree is 1. Suppose this extension includes the square root of the next prime qk+1. Adjust p to be congruent to any nonsquare mod qk+1. If we then adjoin uk+1 first, p is inert at this step, and the residue degree is 2. This is a contradiction, hence uk+1 is not part of Rk. Each square root is new, and not contained in the prior field.
Since each qi is an integer, it does not move under any of the prior automorphisms. All automorphisms are independent, and the galois group is the direct product of K copies of Z/2. You could even build an infinite field extension with an infinite direct product of involutions, but this chapter focuses on finite extensions and finite galois groups.
One can join these side groups of order 2 to any cycle. If the cycle has length 80 for example, under the auspices of cyclo 241, then choose p = 1 mod 241 and apply the previous proof. p splits fully in cyclo 241, and in the fixed field inside this cyclotomic, and in each of the square roots. the next square root is new, and independent of the others, thus bringing in another Z/2. Z/80 * (Z/2)4 is possible, etc.
Actually, any abelian group G is galois, under the auspices of a cyclotomic extension. Write G as a direct product of cycles c1 through cn. Find a prime p1 that is 1 mod c1, and similarly for the other cycles. Make sure the n primes are distinct. Let m be the product of p1 through pn. Cyclo m has automorphisms according to the units mod m. The result is a direct product of cycles of length pi-1. Each ci is a quotient group of the corresponding cycle of length pi-1. Put this all together and a fixed field inside cyclo m has the galois group G.
Turning this around, every number field with an abelian galois group is a subfield of a cyclotomic extension. This is the Kronecker Weber theorem, and I may get round to proving it later on in this book.
We have encountered symmetric galois groups before: S3 here, and S5 here. The latter suggests Sp for any prime p, if we can build an irreducible polynomial of degree p that crosses the x axis p-1 times. In fact every symmetric group is the galois group of a number field, as we shall see below.
Next let's build a certain semidirect product. Adjoin the nth roots of 1, building a cyclotomic extension. This has dimension φ(n), with galois group according to the units mod n. On top of this, adjoin t, the nth root of some prime number p. Suppose t is already part of the cyclotomic. Adjoin t first; since the polynomial is irreducible by eisenstein's criterion, the dimension is n, which is larger than φ(n). Therefore t is new, and so are all the other roots of the polynomial: ty, ty2, ty3, etc. Furthermore, these roots move about in a cycle as t maps to ty, and ty maps to ty2, etc. The extension by t is galois, with cyclic galois group of order n. The order of the composite extension is n×φ(n).
As mentioned above, the roots of p move about in a cycle, but they also move via the automorphisms of the cyclotomic. Complex conjugation is one such automorphism. It maps yj to yφ(n)-j, and in so doing it maps tyj to tyφ(n)-j. The automorphisms of the cyclotomic extend to t, and the entire ring is galois.
The group is faithfully represented by its actions on the roots of p. If these roots are numbered by the exponent on y, 0 to n-1, then apply an automorphism from the cyclotomic, thus multiplying by j, then shift by c, then multiply by k, where jk = 1 mod n. A root with index b maps to (bj+c)k, or b+ck. This is another shift, showing the cyclic subgroup is normal. That's not a surprise; the subfield fixed by this subgroup, wherein t is fixed, is the cyclotomic, and that is galois, hence the subgroup has to be normal within the larger galois group. It's just nice to confirm these things.
Call the galois group J, where G is the normal cyclic subgroup of order n, and H is the cyclotomic subgroup of order φ(n). Members of H induce automorphisms on the group G, and J is the semidirect product of G by H. If you like, represent a member of J as an integer mod n, for the shift, followed by a unit mod n, for the automorphism on G.
When n = 4, G is a 4 cycle, like spinning a square about: t → ty → ty2 → ty3 → t. H is complex conjugation, which swaps ty and ty3. That is like reflecting the square through its diagonal. The result is the dihedral group D4. Thus D4 is represented by a number field.
For larger values of n, J includes Dn as a subgroup, through the circular shift and complex conjugation, but this is not a quotient group, so doesn't become galois over Q. We do however get one more dihedral group for free, D6. This because complex conjugation is the only nontrivial automorphism in cyclo 6, just as it was the only nontrivial automorphism in cyclo 4.
An extension F/K is regular if each element in F that is algebraic over K already lies in K. F could be K of course, or a transcendental extension of K, or a transcendental extension followed by an algebraic extension, like some leaves atop some very tall trees.
Assume v is algebraic atop a transcendental extension. The transcendental ring is a ufd, and integrally closed, thus the minimum polynomial p satisfying p(v) = 0 is monic and lies in this ring. There is no need for denominators. If this polynomial has at least one coefficient with at least one transcendental element in it, and if q is some other polynomial that shows v is algebraic over K, then p divides into Q, and p lives in the polynomial ring K[x], which is a contradiction. Conversely, if all coefficients live in K then v is algebraic over K.
Even if v is not algebraic over K, riding on top of the indeterminant t, the field K(t,v) might not be regular. The field could bring in algebraic elements of K. Let v = sqrt(t) + sqrt(7). I'm merging two quadratic extensions into one via the primitive element theorem. Verify that v4 = (2t+14)v2 - t2 + 14t - 49. This is degree 4 as expected. v2 implies sqrt(7t), and when this is multiplied by v you get 7*sqrt(t) + t*sqrt(7). this is a different linear combination of sqrt(t) and sqrt(7), different from v, so both sqrt(t) and sqrt(7) are implied. The algebraic extension K(sqrt(7)) of dimension 2 is contained in this extension of dimension 4. A regular extension based on K(t) is more than just an algebraic extension that sits on top of K(t). It is entirely atop K(t), with no algebraic elements brought in.
Verify that F/K is regular iff it is linearly disjoint from every finite extension E/K. F would intersect the normal closure of E iff there was something algebraic in F.
Hereinafter, all fields are assumed to be perfect, with only separable extensions. For any E/K, take the normal closure, so that E/K is galois. Thus the dimension of E+F over F is the dimension of E, and this holds for each subfield of E.
Any intermediate extension of F is regular over K. It still has nothing algebraic.
If L is algebraic over K, then F is regular over L. Anything algebraic over L would be algebraic over K.
The composition of regular extensions is regular. If v in F2 is algebraic over K then v is certainly algebraic over F1. This puts v in F1, and with v algebraic over K, v lies in K.
A finite group G is gal_t with respect to K if there exists a galois extension L/K(t) (t an indeterminant) with galois group G, such that L is a regular extension of K. We aren't just extending K, we're extending K(t). If K is not specified the rationals are assumed.
As mentioned in the previous section, a galois extension with galois group G implies another galois extension over the same field with galois group H, where H is a quotient group of G. Use the intermediate subfield fixed by the normal subgroup having quotient group H. Thus conjectures 1 2 and 3 descend from G down to H. Since the subfield of a regular extension is regular, there is no trouble with (3).
If G is gal_t over K it is gal_t over K(v).
Let L/K(t) be the regular extension with galois group G. It does not matter whether v is adjoined before or after the indeterminant t. Let E = K(t) for notational convenience. Remember that v is linearly disjoint from L. Every automorphism of L/E extends uniquely to an automorphism of L(v)/E(v) by fixing v; and conversely, since L is normal over E, each such automorphism pulls back to an automorphism of L/E. The galois groups are the same. G fixes E in L, and at least E(v) in L(v).
Assume v is algebraic over K. Let L have dimension m over E and let K(v) have dimension n over K. Thus E(v) has dimension n over E, and L(v) has dimension n over L. L(v) has dimension mn over E, and it must have dimension m over E(v). The dimension agrees with |G|. L(v) is galois over E(v) with galois group G.
Let x ∈ L(v) be algebraic over K(v), hence algebraic over K. Suppose x is not in K(v). K(v,x) is larger than K(v), and being linearly disjoint from L, L(v,x) would be larger than L(v). This is a contradiction, hence x is already in K(v), and L(v) is regular over K(v). L(v) proves G is gal_t over K(v).
If G is gal_t over Q then extend Q by v and G is gal_t over any number field. This is why conjecture 3 specifies gal_t over Q. We can bump up to any number field from there, as indicated by conjecture 2.
Now let v be an indeterminant that is algebraically independent of t. Recall that v is an indeterminant over K(t) iff t and v are algebraically independent over K. With this in mind, suppose v is not an indeterminant over L. In other words, v is algebraic over L. Let p be the polynomial that makes this so. Multiply p by all its conjugates under G to get a polynomial q. Each coefficient of q is an expression in the coefficients of p, that is fixed by G. The coefficients of q lie in E. Thus t and v are no longer independent, and this is a contradiction. Therefore v is an indeterminant over L, or if you prefer, v and L are independent over K. This same proof works if you adjoin any number of indeterminants v1 v2 v3 … to L. Assume they are independent of each other and of t, and suppose a polynomial p connects some of these indeterminants using coefficients in L. Build q as before, and connect t with some of the indeterminants from the set V, using coefficients from K, which is a contradiction. Therefore it is enough to say the adjoined indeterminants are independent of t.
Again, start with one indeterminant v. Let u ∈ L(v) be algebraic over K(v). u is a quotient of two polynomials in L[v], in lowest terms, and at the same time u is the root of a polynomial with coefficients in K(v), each coefficient a quotient of polynomials in K[v]. Multiply through by a common denominator so that the coefficients are all in K[v]. Then multiply through by the denominator of u, to an appropriate power, giving a polynomial in the numerator of u (a member of L[v]), with coefficients in K[v]. this makes v the root of a polynomial in L, which is a contradiction. Thus L(v) is regular over K(v). Since K(v) is regular over K, L(v) is regular over K. This is a bonus we get because v is transcendental.
Fix v as we did before, so that G becomes a group of automorphisms of L(v) over E(v). L is K(t) adjoin some element x, giving a dimension of m. Adjoin x to K(v,t), wherein x has the same minimum polynomial, and the dimension is still m. Therefore G is gal_t over K(v).
Assume finitely many indeterminants are added to K and L. Bring them in one at a time as above. G becomes gal_t over K(V) for a set of indeterminants V. Also, the new extension is regular over K. This leads to an equivalent definition of gal_t, in which the extension L/K(T) may have finitely many indeterminants rather than just one. We just stepped up from one to many; reducing finitely many indeterminants to one is proved later.
In fact one can extend K by arbitrarily many indeterminants. The same isomorphism exists between galois groups, and a similar proof shows L(V)/K(V) is regular. If u in L(V) is algebraic over K(V), then u and the coefficients of its polynomial entail finitely many indeterminants from the set V. Clear denominators and find a polynomial in finitely many indeterminants that is equal to 0. This contradicts algebraic independence.
If G is gal_t over K, you can push K up to any transcendental extension, and then a finite algebraic extension on top.
If G and H are gal_t over K then so is G*H. Let L and M be the corresponding extensions over K(t1) and K(t2). Extend M by adjoining t1 to get a new M over K(t1,t2) with galois group H, regular over K(t1). This means L and M are linearly disjoint. The compositum L+M has the desired galois group G*H over K(t1,t2). Since M is regular over K(t1), L+M is regular over L by propagation, and regular over K by composition. Therefore G*H is gal_t. I'm using the second definition of gal_t here, wherein the algebraic extension sits atop two indeterminants t1 and t2, and that's good enough.