This chapter connects solvable groups with solvable extensions and solvable polynomials. A polynomial p(x) is solvable if its roots can be expressed as a formula in the coefficients of p and other elements of the base field F, using field operators (+*/) and nth roots. Think of the quadratic formula (b±sqrt(b24ac))/2a.
Square the elements of F, and you are still in F; but the square root may force a field extension of F. The same holds for cube roots, fourth roots, etc. That's ok; it's still a formula.
A formula exists for an arbitrary polynomial of degree 4 or less. Move to degree 5, and there may be a formula, such as x52 = 0, wherein the roots are the fifth roots of 2; but in general there is no formula. Quintics and above are unsolvable. This chapter will explain why.
The prefix quad has come to mean 4, as in quadrant (one of 4 regions), quadruple (times 4), and quadruped (an animal with 4 feet). So you might think a quadratic equation has degree 4. However, quad is latin for square, and the semantics of 4 has evolved over the years because a square has 4 corners and 4 sides. An example of the original meaning is a quadrangle, buildings arranged in a square. So a quadratic equation starts with x2, and has degree 2. In contrast, a quartic equation has degree 4, since quart is a latin descendent of 4, as in quarter, quartile, quartet, or even quarantine, wherein a ship was held in isolation from the shore for 40 days to ensure it did not carry the black death.
A quintic equation descends from the latin word for 5, which also gives us words like quintuplets and quintet. A cubic equation however is not connected to the number 3. It is so named because x to the third gives the volume of a cube. This is in line with quadratic, connected to the square.
In this section, and the next two, F is a field that does not have characteristic 2 or 3. Why? Because characteristic 2 is a bit like binary, 0 and 1, even and odd. If 1+1 = 0, then 2 is just like 0. The quadratic formula has 2a in the denominator, but you can't divide by 0. So 2 and 3 have to be legitimate numbers in the base field.
Given a polynomial p(x), divide through by its lead coefficient, so that p is monic. In other words, p begins with a power of x. If p use to be ax2 + bx + c, p is now x2 + (b/a)x + c/a, or simpley x2 + bx + c for notational convenience.
Let p be a monic polynomial of degree ≤ 4. Let b be the second coefficient, e.g. x2 + bx + c. Replace x with yb/n, where n is the degree of p. This creates a new polynomial q(y) that starts with yn, and has no term yn1. If there is a formula for the roots of q based on its coefficients, apply this formula to get y, then subtract b/n to get x. I will do this for the quadratic formula, but after that it is left as an exercise for the reader.
Consider any polynomial of degree 2, and divide through by its lead coefficient, so that p is monic. This leaves x2 + bx + c.
Substitute x = yb/2 to get rid of the second term, and get y2 + cb2/4 = 0.
Solve for y and get ±sqrt(b2/4c), or ±sqrt(b24c)/2. Subtract b/2, and x = (b±sqrt(b24c))/2. Finally, allow for a lead coefficient of a, by replacing b with b/a and c with c/a. After all, the second coefficient would be b/a if the polynomial had been ax2+bx+c, and we had divided through by a at the start. This gives the familiar quadratic formula.
x = (b ± sqrt(b24ac)) / 2a
This formula, involving field operations and radicals, proves quadratic polynomials are solvable.
The expression b24ac is sometimes called the discriminant. It is proportional to the square of the difference between the two roots. Thus the discriminant is 0 iff the two roots are identical.
If F is a subfield of the reals, the roots are complex iff the discriminant is negative, thus taking the square root of a negative number.
Consider the normalized polynomial x3 + px + q = 0. If p = 0 then x is the cube root of q. If q = 0 then x = 0 and the square root of p. So assume p and q are nonzero, else the solution is straightforward.
For a given solution x, let a and b be numbers satisfying ab = x, and ab = p/3. Combine these equations to give a quadratic.
a = (x+b)
(x+b)b = p/3
b2 + xb  p/3 = 0
Hence a and b can be derived from x and p. Since a quadratic equation has two solutions, there are two values of b associated with x and p. Once a and b are found, the "other" ab pair associated with x and p is produced by swapping and negating a and b.
a → b
b → a
Substitute x = ab in the original polynomial to get:
x3 + px + q = 0
(ab)3 + p(ab) + q = 0
a3  3a2b + 3ab2  b3 + p(ab) + q = 0
a3  3ab(ab)  b3 + p(ab) + q = 0
a3  3(p/3)(ab)  b3 + p(ab) + q = 0
a3  b3 + q = 0.
Multiply through by 27a3, remembering that 3ab = p. The result is a quadratic in a3:
27a6 + 27qa3  p3 = 0
Apply the quadratic formula to get a3 in terms of p and q, then take the cube root of the result. The 6 values obtained from the two square roots and the three cube roots can be grouped into three pairs, each of whose product is p/3 and sum is x. Of course x is what we seek.
All this can be expanded into a closed formula with cube roots and square roots, perhaps incorporating lead and secondary coefficients, like the quadratic formula, but the result would be rather messy.
Consider the normalized polynomial x4 + px2 + qx + r = 0. If q or r is 0 the solution is straightforward.
Add px2+p2 to each side, and move qx and r to the right, so that the left side becomes the square of x2+p. Then introduce a new variable y into this square as shown below.
(x2+p)2 = px2  qx  r + p2
(x2+p+y)2 = (p+2y)x2  qx  r + p2 + y2 + 2py
This is a polynomial in two variables, x and y. If x was a solution before, it is still a solution, along with every value of y. Instead of four solution points (x), there are four solution lines (x cross y). Any value of y will do.
Adjust y so that the right side becomes a square. In other words, the right side should be (ux+v)2, for some u and v. This happens iff the discriminant comes out 0. The lead coefficient is p+2y, the linear coefficient is q, and the constant term is r+p2+y2+2py. The discriminant b24ac becomes a cubic polynomial in y.
8y3 + 20py2 + (16p28r)y + 4p34prq2
Solve this using the cubic procedure, and then use any of the three solutions for y. Use y to compute u and v, and write
x2+p+y = ±(ux+v)
Solve this by the quadratic formula, giving up to four solutions for x.
Should we try an example? How about a polynomial with small integer solutions.
x425x2+60x36 = 0
(x2  25)2 = 25x2 60x + 661
(x2  25 + y)2 = (2y25)x2 60x + 661 + y2  50y
8y3 + 500y2  10288y + 69700 = 0
y = 17 (a solution for y, by testing the factors of 69700, as I'm too lazy to employ the cubic procedure).
x28 = ±(3x10)
x23x+2 = 0 and x2+3x18 = 0
x = {1, 2, 3, 6}
We might have selected y = 25 instead, as that is another solution to the cubic in y. The left side becomes (x2+0)2, and the right side becomes 25x2  60x + 36, or (5x6)2. This produces the same 4 solutions for x.
The last option for y is 41/2, and this yields the same four values of x, as it must.
A more complicated example, building the galois group A4, is shown here.
Think about the tower of extensions implied by the quartic formula. First solve the cubic equation in y, which entails a quadratic equation (taking the square root of the discriminant), then the cube root of the result. With y in hand, solve one quadratic equation for two values of x, and another quadratic equation for the other two values of x. The entire extension could have a dimension as high as 24. 

At first it seems like you should be able to extend this line of reasoning to quintics and above, and for centuries the most brilliant mathematicians thought so as well, including Lagrange, who worked on the problem for over 30 years. But the process does not generalize, because S2, S3, and S4 are solvable groups, while S5 and beyond are not.
Let p(x) be an irreducible polynomial of degree n over K, and let F be the splitting field of p(x). Thus F contains all the roots of p(x). Adjoin one root, then another, then another, to show the dimension of F/K is at most n!.
The galois group G, of p(x), is, by definition, the galois group of F/K. Each automorphism in G permutes the roots of p(x), and since the roots generate F, each automorphism is defined by its action on these roots. Therefore G is faithfully represented as a subgroup of Sn.
The roots of p(x) are indistinguishable, insofar as an automorphism of F carries each root to any other root. This means G acts transitively on the roots of p(x). G has but one orbit, moving a designated root onto all the others. Since the size of the orbit, times the order of the stabilizing subgroup, gives the order of G, G is divisible by n, where n is the number of roots in p(x). For our purposes, this is almost always the degree of p. If K has characteristic 0, or K is finite, and p has a repeated root u, use formal derivatives to extract a smaller polynomial over K with root u, whence p was reducible after all. For all the usual fields, an irreducible polynomial p(x) of degree n has n distinct roots, and G is a transitive subgroup of Sn, and G is divisible by n.
As an exercise, what is the galois group of x32? We've seen this before; adjoin the real cube root of 2, but that doesn't bring in the complex cube roots of 2, so adjoin the cube roots of 1 to get the other cube roots of 2. The galois group of this extension has already been analyzed, and found to be S3.
A field extension is cyclic of order n if it is galois, and has dimension n, and its galois group is cyclic. For example, let F and K be finite fields. The extension F/K is galois, and the galois group is cyclic, generated by the frobenius automorphism. Every finite extension of a finite field is cyclic.
A field extension is solvable if it is finite, and galois, and its galois group is solvable. For example, a cyclotomic extension of Q has an abelian galois group, and is solvable. In many cases the extension is cyclic, e.g. adjoining the pth root of 1 where p is prime.
Intuitively, a solvable extension is a tower of cyclic extensions. This makes sense, because a solvable group decomposes into cyclic factor groups. Let's look at this more closely.
Let F/K be galois with solvable group G, and let H be a normal subgroup of G, having fixed field L between F and K. Examine F/L, with galois group H, and L/K, with galois group G/H. Both H and G/H are solvable.
If G is finite, create a tower of intermediate field extensions, according to the composition series of G. Each extension is galois with a cyclic galois group. Therefore an extension F/K is solvable iff it is a tower of intermediate cyclic extensions, each having prime order.
Let E and F be two fields with the common subfield K. Recall that E and F are both vector spaces over K. Give E a basis b over K, and give F a basis c over K.
Let S be the set of K linear maps from E into F. If g is a function in S, then g maps each basis element from b into a linear combination of basis elements of c. By linearity, this defines g on all of E. For instance, g(a1b1+a2b2) = a1g(b1) + a2g(b2).
Each function in S can be scaled by an element of K, and two functions in S can be added together to get another function in S. In other words, S is itself a K vector space.
Let T be the functions in S that also happen to be field homomorphisms. In other words, the functions in T preserve multiplication. Thus g(b1b2) = g(b1) * g(b2). Since g maps 1 to 1, and g is K linear, g fixes K.
You can add or scale the functions in T, but you probably won't get another function in T. If K is the real numbers and g is complex conjugation, 2g is not a field homomorphism, because 2g(1) = 2, whereas a field homomorphism always maps 1 to 1.
Suppose there is a linear combination of field homomorphisms in T that produces 0. We usually take coefficients from K, but let's generalize this to include coefficients from F. Since there is some linear combination that produces 0 with coefficients in K, there are linear combinations that lead to 0 with coefficients in F. Select the shortest example.
Let the sum of aigi give the 0 homomorphism, where each gi is a field homomorphism in T, and each ai is a coefficient taken from F. This is equation 1.
1: ∑ aigi(x) = 0
There are at least two terms in equation 1, for if there is only one, it drops out. Apply a1g1 to 1 and get a1*1 = 0, hence a1 = 0.
If g1 and g2 agree everywhere, they are in fact the same function, so select y in E such that g1(y) ≠ g2(y). If g1(y) is 0, swap the roles of g1 and g2. Now g1(y) is nonzero. In the following, y is held constant.
For any x in E, do the following. Apply equation 1 to xy. Thus the sum of aigi(xy) = 0. Pull each gi(y) out and make it part of the coefficient. This gives another linear equation of homomorphisms, equation 2, that still maps every x to 0.
2: ∑ aigi(y) gi(x) = 0
Next, multiply equation 1 through by g1(y). This is equation 3, a scaled version of equation 1.
3: ∑ aig1(y) gi(x) = 0
Subtract equation 2 from equation 3, giving equation 4.
4: ∑ ai (g1(y)  gi(y)) gi(x) = 0
The first term, with coefficient a1*(g1(y)g1(y)), drops out. The second term is a2(g1(y)g2(y))g2(x). Since g1 and g2 disagree on y, the coefficient is nonzero, and the second term remains. Equation 4 is a shorter linear combination of field homomorphisms that yields 0. This is a contradiction, therefore all field homomorphisms are linearly independent.
This theorem is often applied to the automorphisms of F, i.e. field homomorphisms from F into F.
The best way to understand solvable extensions is to understand cyclic extensions. After all, every solvable extension is a tower of cyclic extensions. I will describe cyclic extensions of order p (that's good enough), over fields of characteristic p, and notp.
Hilbert's theorem 90 deals with trace and norm in a cyclic extension. It is so named because it is the 90th theorem in David Hilbert's famous Zahlbericht, published in 1897.
Let a cyclic extension F/K have a galois group G, and let the automorphism c generate G. The trace of u is 0 iff u = vc(v) for some v in F, and similarly, the norm of u is 1 iff u = v/c(v) for some nonzero v in F. Remember that the trace is the sum of the conjugates, and the norm is the product of the conjugates.
One direction is easy, and holds whether G is cyclic or not. The trace of vc(v) is the automorphic images of v, minus the automorphic images of c(v). Yet the latter is merely a permutation of the former. The conjugates of v are subtracted from themselves, and the result is 0. A similar argument shows the norm of v/c(v) = 1.
Conversely, assume the trace of u is 0. The trace is a linear combination of field automorphisms of F, in fact the sum of the automorphisms (all coefficients are 1), and by the previous theorem, this linear combination cannot be 0 everywhere. There is some z whose trace is nonzero.
Let w = z/trace(z). F/K is galois, hence the trace of anything lies in K. In particular, the trace of z lies in K, and is unaffected by any of the group automorphisms. The trace of w is the trace of z over the trace of z, or 1.
As a warmup, let n = 2, so there is but one nontrivial automorphism c, and let v = uw. Now v  c(v) = uw  c(uw). Remember that c(w) = 1w. Substitute and get uw  c(u) + c(u)w = (u+c(u)w  c(u) = c(u). With trace(u) = 0, c(u) = u, so indeed, v  c(v) = u.
The construction for n > 2 is a little more complicated. With trace(u) = 0, and trace(w) = 1, set v equal to the following sum, up to an exponent of n2.
v = uw + (u+c(u))c(w) + (u+c(u)+c2(u))c2(w) +
(u+c(u)+c2(u)+c3(u))c3(w) … n1 terms
Consider c(v). What does c do to the above sum? Each term becomes the term that follows, almost. For instance, watch what happens to the second term, as it almost becomes the third.
(u+c(u))c(w) → (c(u)+c2(u))c2(w)
All we're missing is uc2(w).
Consider v  c(v). As mentioned above, we have, left over, uc(w) + uc2(w) + uc3(w) + … ucn2(w). And v starts with uw, so the previous sum begins with uw. Pull out u, and find all the conjugates of w except the last one. This last conjugate is d, where d is the inverse of c. Since trace(w) = 1, we have, so far, u*(1d(w)), or u  ud(w).
The last term of c(v) also stands alone:
(c(u) + c2(u) + c3(u)…d(u)) * d(w)
Subtract this from u ud(w) and get u  trace(u)d(w), or u.
Next assume norm(u) = 1. Set v to the following sum, containing n terms. Note, the value of w is not the same as it was above. I'll describe how to select w below.
v = uw + uc(u)c(w) + uc(u)c2(u)c2(w) + uc(u)c2(u)c3(u)c3(w) + … + ud(w)
Since norm(u) = 1, u is nonzero. Similarly for uc(u), uc(u)c2(u), and so on. The "coefficients" on the functions c0(), c1(), c2(), c3(), etc, are all nonzero. This gives a nontrivial linear combination of distinct Kautomorphisms of F. By linear independence, the resulting function cannot map all of F to 0. Select any w, such that the image of w is nonzero. In other words, v is nonzero.
Start with v, apply c, and multiply by u. Once again the first term becomes the second, the second becomes the third, and so on, until the last term becomes the first. (This because u = 1, and c(1) = 1. Therefore v = uc(v), or u = v/c(v).
Let q(x) = xpxa over a field K of characteristic p. Either q is irreducible, or it splits in K, i.e. all its roots are in K.
Let r be a root of q(x). Evaluate q(r+j) for any integer j. By the frobenius homomorphism, the result is rp+jprja, which is rpra, or 0. If r is a root then every r+j is a root. That's p distinct roots, and there can be no more, hence q(x) splits.
Conversely suppose q(x) has an irreducible factor s(x), but has no roots in K. Let s have the root r, which lies in a field extension E/K. Within E, s has the roots r+j for every j, and these are distinct, so s has degree p, s divides into q, and s = q. Either q is irreducible or it splits.
Now let K have any characteristic other than p, perhaps 0, and let q(x) = xpa. As shown earlier, the roots are distinct. Start with a = 1, and assume some root r is not in K. Note that rj is a conjugate of r, another root of q(x), for any j. If rj = 1 then j divides p, which only happens when j = p. Similarly, two powers of r cannot be the same, for then their quotient would be a lesser power of r that is equal to 1. The p1 powers of r are all in the field K(r), and these are all the nontrivial roots of q(x). Hence q/(x1), 1 + x + x2 + x3 + … xp1, is irreducible, and generates an extension of order p1. This is galois, in fact cyclic, with galois group r → rb, where b is primitive mod p.
Next let a be something other than 1. Let q(x) have the root u. The conjugates are now u times the powers of r, where r is a pth root of 1 as above. If two conjugates are present then divide them to find a pth root of 1. This brings in all the other roots of 1, and all the other conjugates of u. The irreducible polynomial containing u has degree 1, in which case u is already in K, or it has degree > 1, in which case it has degree p and is all of q(x). Once again q splits or q is irreducible.
Let F be an extension of a field K having characteristic p. By the previous theorem, if F = K(u), where upua = 0, then either F = K, or F is an extension of dimension p, F includes all the conjugates of u, the automorphism group is cyclic (keep adding 1 to u), F is galois, and F is a cyclic extension of order p.
Now assume the converse. Let F be a cyclic extension of order p, and note that the trace of 1 = 0. If c generates the galois group G, then set 1 = vc(v) for some v in F by Hilbert's theorem 90.
Set u = v, and note that c(u) = u+1. Since c(u) ≠ u, u does not lie in K.
There are no proper subgroups of G, and no intermediate fields of F, so K(u) = F.
Finally, c fixes upu, so upu = a for some a in K. Therefore u is a root of xpxa, which is irreducible over K.
When K has characteristic p, F/K is cyclic of order p iff F = K(u) for some u that is a root of the irreducible polynomial xpxa.
Let the field K have a characteristic other than p, perhaps 0. Consider the polynomial q(x) = xpa, for some nonzero a. As shown earlier, either q splits, in which case all the roots are already in K, or q is irreducible. Let u be a root of q, and let F = K(u), having dimension p. This is a traditional pth root, consistent with the quadratic, cubic, and quartic formulas at the top of this chapter.
Assume there is an element r in K that is a primitive pth root of 1. Thus K contains all the pth roots of 1, and splits xp1. The elements u, ur, ur2, etc, are the p distinct roots of xpa, and the automorphism u → ur generates the cyclic galois group of order p. Hence F/K is a cyclic extension of order p.
Conversely, let F/K be a cyclic extension of order p. Let r be a primitive pth root of 1. If a field, any field, contains rj for some nonzero j, raise this to the m, where m is the inverse of j mod p. The result is r. Thus K, or F, or any field, contains r iff it contains all the powers of r. One conjugate brings in the rest.
If F contains r, and r is not in K, then K(r) is an intermediate extension of dimension p1 inside F. Since F/K has dimension p, this is impossible. So if r is in F, r lies in K.
If r is not in F then adjoin r to create a larger field E. All the conjugates of r are present, E is a splitting field, and E is normal over F.
Think of E as K adjoin F and the powers of r. Remember that F is galois by assumption, hence a splitting field for some polynomial of degree p. Adjoining r is also a splitting field over K, hence E is a splitting field over K. In other words, E/K is a normal extension. F/K is separable by assumption, and the powers of r are seprable, hence E/K is separable. Since E is normal and separable it is galois.
The galois group of E/F is cyclic, generated by the automorphism r → rb, where b is primitive mod p. That's E/F, let's try to get a handle on the galois group of E/K. Let c generate the cyclic galois group of F/K, which exists by assumption. This is an automorphism on F, and it can be extended to a field homomorphism from F(r) onto F(rb), since r and rb are conjugate. This homomorphism is really an automorphism on E/K. Call this automorphism d.
What is the order of d inside the galois group of E/K? Thanks to c, you must invoke d p times before F gets back to start. Thus the order of d is divisible by p. At the same time, the root r must return to its original position. Thus the order of d is divisible by p1. Since p and p1 are coprime, d = p×(p1), which is the order of the group. The galois group is cyclic, generated by d.
What is the norm of r in E/K? Let L = K(r), and take it step by step. The norm of r in L/K is the product of the conjugates, r * r2 * r3 * … rp1, which is 1. Raise this to the p power to get r = 1 in E/K.
Since r = 1, apply Hilbert's 90, and there is a nonzero w in E with w/d(w) = r. Write w = r*d(w). Since d(w) is not equal to w, w does not lie in K. Let u = 1/w, whence 1/u = r*d(1/u), or d(u) = u*r. Since w is not in K, u is not in K either.
Notice that d fixes up, hence up lies in K. In other words, u is the pth root of a, for some nonzero a in K.
The subgroup generated by the automorphism dp has order p1 in the galois group G, and fixes an intermediate extension of dimension p. Yet d, restricted to F, is c, and cp fixes F. So dp fixes a field of dimension p, and it fixes F, hence it fixes precisely F and nothing more. Since dp fixes u, u is in F.
The field K(u) cannot be intermediate between K and F, since the dimension of F/K is prime; therefore K(u) = F. The polynomial xpa is irreducible over K.
Since F is galois, it is normal, and it splits the polynomial xpa. This means F contains all the conjugates of u, which are all distinct. Take the quotient of any two conjugates and raise this to the p power. The result is 1. Thus F already contains a pth root of 1. If F contains r then K contains r (we showed this earlier). So K contains r after all. We didn't need to adjoin r; it was there all the time.
With r in K, r, in the extension F/K, is rp, or 1. Let r = w/c(w), whence w = r*c(w). Let u = 1/w, whence c(u) = u*r. Reason as above to show F = K(u), where u is the root of an irreducible polynomial xpa.
In summary, let K be a field whose characteristic is not p. The extension F/K is cyclic of order p, iff K contains a primitive pth root of 1, and F = K(u), where u is a root of the irreducible polynomial xpa, for some nonzero a in K.
Let q(x) be an irreducible polynomial over K. I use the letter q to avoid confusion with phrases like characteristic p and pth root.
Assume q(x) is separable, so that its splitting field F is a galois extension. Since K usually has characteristic 0, or is finite, this is rarely an issue.
In this section, a root is a solution of q(x), i.e. an x that satisfies the polynomial, while a radical is a mathematical operation wherein the mth radical of v is u if um = v. There are of course m possibilities for u in the complex plane. This terminology avoids using the word root for two different concepts. Words are often overloaded in mathematics, to be sure, but hopefully not in the same sentence.
Any radical of composite order can be split into successive radicals of prime order, so assume only pth radicals are extracted for various primes p. If you want the sixth radical of 17, take the cube radical, then the square radical.
If K has characteristic p, extracting a pth radical in the literal sense either produces another element in K and can be ignored, or it produces a purely inseparable extension of the previous intermediate field, making q(x) inseparable. Therefore I will assume a pth radical of a over K, with characteristic p, implies a procedure to solve the polynomial xpxa. That is now, by definition, a pth radical when K has characteristic p. This abuses the notation a bit, but it makes the theorem work.
Assume all the roots of q(x) are expressible using field operations, radicals, and elements of the base field K. Build a finite extension of K that contains the first root, extending the field by prime radicals as you go. If there are different formulas for the different roots of q(x), extend the field further as you bring in the second root, and the third, and so on until q(x) splits. We want to show this is a solvable extension.
Let z be a member of the field E, which could already be an extension of K, and look for a field extension F/E that contains the pth radical of z. E is galois by induction, so if one radical of z is in E then they all are, and we are done. Otherwise none of the radicals are present, and a field extension is required.
Let z satisfy an irreducible polynomial s(x) with coefficients in K. If z is already in K then s is just xz. Our radical, actually all the radicals, satisfy s(x) with x replaced by yp. I'll just call this s(y). Its degree is p times that of s(x).
Adjoin the p radicals of 1 around the unit circle, if they are not already present. Then adjoin the p radical of z, which is irreducible (whether K has characteristic p or not), and builds one of our cyclic extensions of dimension p. At this point you might think we are done. A galois extension is placed on top of another galois extension, but the composition of galois extensions might not be galois, so there is more work to do.
Adjoin the p radical of every conjugate of z in s(x). There may be several conjugates of z, thus, perhaps, several extensions of dimension p. Each one is a galois extension of dimension p over the previous. Put them all together and find a field F that splits s(y). In fact we have adjoined the roots of s(y) to E, which is already a splitting field by induction, thus F/K is another splitting field, and normal, and galois. The subfields on the path from E to F have dimension p, like climbing a ladder whose rungs are p apart, except perhaps for the radical of 1, with dimension p1. The galois group of F/E, within the larger galois group of F/K, is a p group, along with (perhaps) a cyclic group of order p1. The p group is a unique sylow group by the third sylow theorem, and is normal, with a cyclic factor group. A p group is also solvable. Therefore F/E is solvable, and since E/K is solvable by induction, F/K is solvable.
Continue this process for each radical, and for all the roots of q(x). The splitting field for q is contained in a solvable group. It is also a normal extension, hence galois. Its galois group is the quotient of a solvable group, and is solvable.
Now for the converse. Let the splitting field F of q(x) have a solvable galois group G. Build a chain of normal subgroups, and a tower of normal extensions, that takes you from K all the way up to F. If every element in F can be derived using field operators and radicals, then the roots can be so derived, and q is indeed solvable.
Let the chain up subgroups of G, and the tower of intermediate extensions of F, have height n. Assume, by induction, that every element in a shorter tower is expressible from its base, using field operators and radicals.
Let H be the maximal subgroup in the composition series of G. Now H is normal in G, with quotient group Z/p. Let H fix E, hence F/E has galois group H and E/K has galois group Z/p.
H is solvable, of length less than n, hence everything in F can be expressed using field operators, radicals, and elements of E. If elements of E can be expressed using radicals and elements of K, then we are done.
The extension E/K is galois, with galois group Z/p, hence it is a cyclic extension. Any element u in E, but not in K, will do the trick. Adjoining u cannot produce an intermediate extension, so K(u) = E. Refer to the previous two theorems, when K does or does not have characteristic p. In either case u can be expressed as the pth radical of z, for some z in K. That brings in all of E, and then F. All the roots of q(x) are expressible by radicals and field operations, and that completes the proof.
Let's put this all together. Let q(x) be a separable polynomial, e.g. any polynomial over the rationals. Now q can be "solved", i.e. its roots expressed using field operators and radicals, iff its splitting field is a solvable extension, iff its splitting field has a solvable galois group.
Galois and others presented compelling reasons why some quintics might be unsolvable, but Abel provided the first concrete example.
Consider the polynomial q(x) = x5  10x + 2 over the rationals. This is irreducible by eisenstein's criterion. We will show that it's splitting field F, and its galois group G, are not solvable.
Since any root can be moved to any other root, the galois group is transitive. The group, acting on the roots of q(x), has one orbit, and G is divisible by the size of this orbit. This means G is divisible by 5.
Apply macay's theorem, and G contains the subgroup Z/5. So G acts on 5 elements, the roots of q(x), and G includes a 5 cycle that shifts these elements around. (If the 5 cycle left these roots alone then it would have to live in a group of automorphisms of order 24, and that is impossible.)
Use a little calculus in the xy plane to show the polynomial attains a maximum at 2¼ and a minimum at 2¼. These are all the real roots of q′(x), and there are no other local maxima or minima.
The graph crosses the x axis once to the left of 2¼, once between 2¼ and 2¼, and once to the right of 2¼. Therefore q(x) has 3 real roots and two complex roots.
Complex conjugation is an automorphism that swaps the two complex roots, and leaves the three real roots alone. This is a transposition. When a group includes a pcycle and a transposition, that group is equal to Sp. Therefore our galois group is S5, which is not solvable. It is not possible to describe the roots of q(x) using field operators and radicals.
In contrast, x5  2 is indeed solvable. Adjoin the fifth roots of 1 first, giving a cyclotomic extension with galois group Z/4. Then adjoin the fifth root of 2, giving a second galois extension with galois group Z/5. The composite extension is two sequential cyclic extensions, and is solvable. That makes the underlying polynomial solvable. Not surprising really, since one of its roots is the fifth root of 2, and that leaves a quartic behind, which is solvable.
We can describe the galois group of the normal closure of the unsolvable quintic extension. If the first root is r then divide x5  10x + 2 by xr and find an irreducible quartic. That is the next step in building S5, a group of order 120. Let s be a root of this quartic, which is also a root of the original quintic. Divide by xs and find an irreducible cubic. Let t be a root of this cubic, divide by xt, and let u be a root of the resulting irreducible quadratic. The "other" root is v. The 5 roots are now r s t u v. The basis for the extension is the powers of r from 1 to r4, cross the powers of s up to s3, cross the powers of t up to t2, cross 1 and u. This basis has dimension 120.
Recall the process of building an automorphism on a normal extension. Map the first root r to any other root of the quintic. Thus r maps to r s t u or v. Once this is done, map s to any root of the resulting irreducible quartic, but that turns out to be a root of the original quintic. Thus s maps to any of the 5 roots other than the image of r. Map t to any of the remaining 3 roots, and map u to either of the remaining 2 roots. Each automorphism determines, and is determined by, a permutation on the 5 roots r s t u and v. That's 120 automorphisms on a galois extension of dimension 120. All is well.
If an element j lies in the original quintic extension, what is the norm of j? j is represented by powers of r, and the image of r in the complex plane is r s t u or v. This determines the image of j. This math is performed here.
For every odd prime p, there is almost certainly an irreducible polynomial of degree p, that crosses the x axis p2 times, and yields a galois group of Sp.
You might be tempted to develop a formula, or procedure, for polynomials over a finite field, since every extension, and every polynomial, is solvable. However, there is already a factoring algorithm, which separates any polynomial into its irreducibles. Given any one of these irreducible factors, x can be set to r, an appropriate root of 1, generating the extension of that degree; thus expressing x using radicals.