For his ground-breaking work in groups, Evariste Galois is sometimes called the father of group theory. But the story doesn't end their. Galois was able to connect field extensions to groups, so that subfields and subgroups correspond. This forms a bridge between group theory and field theory, and theorems in one arena can sometimes be brought to bear in the other.
The connection between groups and fields has since been generalized. If two separate structures can be broken down into a hierarchy of substructures, subfields and subgroups for example, and a map carries substructures to substructures and preserves the partial ordering, that map is an instance of galois theory. Thus Galois created a powerful branch of mathematics that is used to solve everything from quintic polynomials to Fermat's last theorem.
We'll never know what else he could have accomplished, had he lived past the age of 20. Unfortunately he died in a duel, a lover's quarrel. He was gifted with unimaginable mathematical genius, but his judgment and marksmanship were lacking. Many books have been written about him, as a brilliant mathematician and as a troubled young man.
Let F/K be a field extension. The galois group of F over K is the set of automorphisms of F that fix K. Since automorphisms can be composed and reversed, this does indeed form a group.
Let u be algebraic over K, the root of an irreducible polynomial p(x). Let c() be an automorphism taken from the galois group of F over K, hence it fixes K. Apply c to the equation p(u) = 0, and the coefficients of p are fixed, hence c(u) is another root of p(x). There are at most n roots, if p has degree n, so we only have a few choices.
If F is a normal extension, with all the roots of p(x), then there exists an automorphism c, that fixes K, and maps u to any other root v, and maps F onto F. Every root of p(x) is a possible destination for c(u). Using the terminology of group theory, the galois group acts transitively on the roots of p(x).
What if c maps F into F, and fixes K, i.e. a field endomorphism? If F is algebraic, c becomes an automorphism. Suppose there is an element u that is not in the image of c(F), and let u be a root of p(x). Let E be the intermediate field extension generated by K and all the roots of p(x) that are present in F. This includes u, and some or all of its conjugates. Remember that every field homomorphism is injective. Thus c is a monomorphism that carries roots to roots. In other words, c embeds a finite set of roots into itself, and is a permutation. Hence the map is onto. There is some root v such that c(v) = u, and that means c is an automorphism on F. When F/K is algebraic, all endomorphisms are automorphisms. They all belong to the galois group.
If u is transcendental, map u to u2 for a field endomorphism that is not onto.
If K(u) has dimension n over K, and u is a root of p(x), the automorphisms of K(u) carry u to a root of p(x), and nowhere else. There are at most n such roots. Furthermore, the entire automorphism is determined by the image of u. Hence there are at most n automorphisms in the galois group of K(u) over K. Of course n is an upper bound, as shown by adjoining the real cube root of 2 to the rationals. Everything in the extension is real, and the root cannot be mapped onto a complex cube root of 2, so the galois group is trivial, even though the dimension of the extension is 3.
Let E be an intermediate extension between F and K. Every automorphism of F fixing E also fixes K, hence the galois group of F over E is a subgroup of the galois group of F over K.
Let F include a transcendental element u. For every nonzero a in K, the automorphism fixing K and mapping u to a*u becomes an automorphism on K(u), which extends to F; hence the galois group has the multiplicative group of K as subgroup. Replace u with u+a to show the galois group has the additive group of K as subgroup. This is a sidebar; we don't usually consider the galois group of transcendental extensions.
Some rather specialized notation is used to describe the fields fixed by certain automorphisms, and the automorphisms that fix certain fields.
Let G be the galois group of F over K, and let L be an intermediate extension of K within F, and let H be a subgroup of G. The extension H′ is the field fixed by the automorphisms in H. This is called the fixed field of H. The subgroup L′ is the subgroup of automorphisms fixing L.
By definition, the extension F/K is galois if G fixes exactly K. This is equivalent to the condition that some automorphism in G moves u for every u not in K. Using our notation, G′ = K.
As shown earlier, adjoining the real cube root of 2 is not galois. The only automorphism is trivial, and that fixes all of F, rather than K.
For the rest of this section, let F/K contain intermediate extensions L and M such that M contains L. Also, let H and J be subgroups of G such that J contains H. Containment need not be proper. Let 1 be the identity element in G, the trivial automorphism on F. The following relationships are not hard to verify.
1′ = F. The trivial automorphism fixes all of F.
F′ = 1. Only the trivial automorphism fixes all of F.
K′ = G. This is really the definition of G.
L′ contains M′. Every automorphism that fixes M also fixes L.
H′ contains J′. Any field fixed by the automorphisms in J is also fixed by the (fewer) automorphisms of H.
L′′ contains L. The field fixed by the automorphisms that fix L must include L.
H′′ contains H. The automorphisms that fix the field fixed by H must include all the automorphisms in H.
L′ = L′′′. The automorphisms that fix L are the automorphisms that fix the field fixed by the automorphisms that fix L.
H′ = H′′′. The field fixed by H is the field fixed by the automorphisms that fix the field fixed by H.
An intermediate extension or subgroup X is closed if X′′ = X.
F is galois over L iff L is closed. Apply the definition of a galois extension; the group of automorphisms that fix L must fix only and precisely L.
Closed extensions and closed subgroups correspond 1 for 1 via the map X → X′. The image X′ is closed, since X′′′ = X′. Apply the bidirectional map again to get X′′ back again, which is X, since X is closed by assumption.
The following assumes the index of H in J is finite, or the dimension of M over L is finite.
The index of M′ in L′ ≤ the dimension of M over L. Let n be the dimension of M over L and proceed by induction on n, with n = 1 implying M = L and M′ = L′. Choose u in M-L, and let p(x) irreducible over L have degree d and root u. The dimension of L(u) over L is d and the dimension of M over L(u) is n/d. If d < n, induction finishes the proof. Otherwise d = n and M = L(u).
Let s be an automorphism in G fixing L, and let M′*s be a left coset of M′ in L′. Any automorphism in this coset maps u to u via something in M′, since M′ fixes M, then maps u to some other root of p(x) via s in L′. Each coset s is associated with a root of p(x).
Suppose two cosets s and t are associated with the same root v. Divide s by t, and the resulting automorphism moves u to v, and back to u. This fixes u, and L(U), which is M. Therefore s/t is something in M′, and s and t are the same coset after all. The map from left cosets of M′ in L′ into the roots of p(x) is injective. Thus the index of M′ in L′ ≤ n.
a similar result holds for subgroups, though the proof is more technical. Suppose the index of H in J is n, yet the dimension of H′ over J′ exceeds n. Let u1 through un+1 be linearly independent elements in H′, where H′ is a vector space over J′. Let t1 through tn represent all the left cosets of H in J. Apply t1 to the n+1 independent elements u1 through un+1. Write this down as a sequence ci = t1(ui), Then write the following homogeneous linear equation.
c1x1 + c2x2 + c3x3 + … + cn+1xn+1 = 0
The variables x1 x2 etc are unknowns taken from the field F.
Repeat the above procedure for t2, t3, and so on. Combine the coefficients into an n by n+1 matrix C, where Ci,j = ti(uj). This matrix represents n equations and n+1 unknowns. Since there are more variables than equations, there is a solution to this system of simultaneous equations.
Let the vector x be a nontrivial solution with the minimum number of nonzero entries. In other words, C*x = 0. Permute the components of x and the columns of C, so that x1 through xr are nonzero. Scale the vector so that x1 = 1.
Let t1 be the automorphism that represents H, hence t1 fixes the independent elements in u. Remember that t1 builds the first row of c. Since x is a solution, the sum of xiui = 0 via the first row of C. If the solution vector is all zeros, except for x1 = 1, then x1u1 = u1 = 0, which is a contradiction. There are at least two nonzero entries in the solution vector x.
If each xi lies in J′, the first r elements of u are not independent. The linear combination, with coefficients from x, produces 0. So assume x2 is not in J′. Since x2 is not fixed by J, there is some automorphism w in J that moves x2.
Write the matrix equation C*x = 0, then apply w, giving w(C)*w(x) = 0. We will see that C and w(C) are the same equations in disguise.
Apply w to t1(ui), which is the same as applying t1w to ui. Now t1w is in some other coset of H in J. Perhaps t1w is in the same coset as t2. Applying t1w to ui is just like applying t2 to ui. The first row of w(C) looks just like the second row of C. Since w permutes the left cosets of H in J, w(C) looks just like C, with its rows permuted. The solutions to the system of equations represented by w(C) are exactly the same solutions that satisfy C.
This means x satisfies w(C), and more important, w(x) satisfies C. Subtract, and x-w(x) satisfies C. This difference yields x1 = 0, while x2 is nonzero. This contradicts the selection of x as having the fewest nonzero entries. Therefore the dimension of H′ over J′ ≤ n, which was the index of H in J.
As you go from intermediate fields to groups of automorphisms, the index can be no larger than the dimension, assuming the dimension is finite; and as you go from subgroups back to fixed fields, the dimension can be no larger than the index, assuming the index is finite.
Let H have finite index in J, and let H be closed. Switch from groups to fields, and H′ over J′ is no larger than the index of H in J. Switch back to groups and the index of H′′ in J′′ is no larger. Yet h′′ = H, and J′′ contains J, so the index cannot be smaller. Everything reverts to equality, hence J′′ = J, and J is closed. Every group having a finite index over a closed group is closed.
By symmetry, the same relationship holds for fields. If L is an intermediate field that is closed, then all finite extensions of L in F are closed.
|If F is a finite galois extension of K, K is closed, and every intermediate extension L over K is finite, and closed. The dimension of F over K is the index of 1 in G, or the order of G. Therefore G is finite, as is every subgroup of G. Since 1 in G is closed, every subgroup of G is closed. All subfields and subgroups are closed, and correspond to each other 1 for 1.||
Partially order the intermediate field extensions by set inclusion, and partially order the subgroups of G by set containment. The two lattices are isomorphic, with X → X′ implementing the correspondence. The least upper bound of two fields is the smallest field that contains them, and the least upper bound of two groups is their intersection. Beyond the structure of the lattice, pairs of comparable fields and groups have the same dimension/index.
If F is not galois over K it is certainly galois over K′′, and from there, the dimension of F exceeds the order of its galois group. Thus F/K (finite) is galois iff the order of G = the dimension of F over K.
Let F/K be galois and let E be an intermediate extension. By correspondence, the dimension of F over E is the index of 1 in E′, which is the size of the group of automorphisms on F that fix E. Therefore F is galois over E, with galois group E′. This does not mean E is galois over K, unless E′ is normal in G, But that's coming up in the next section.
You've already seen some examples of galois extensions. Let F be a finite field of dimension n. The automorphisms and subfields of F have been completely characterized. If c is the frobenius automorphism, i.e. raising everything to the p, then the application of c, again and again, covers all the automorphisms of F, creating a cyclic galois group. The subgroups of this cycle correspond to the divisors of n, and these correspond to the intermediate fields within F. The lattice formed by these fields, or groups, is the familiar lattice of the divisors of n.
If L is a subfield of F, then F is galois over L, and since L is itself a finite field, it is galois over Z/p.
Another galois extension is the cyclotomic extension over Q. Adjoin the nth roots of 1, and the irreducible polynomial ζ(x) has degree φ(n), and splits into all the primitive roots. An automorphism defines, and is defined by, the image of a primitive root, which is always another primitive root. There are φ(n) choices, and the galois group has size φ(n). This agrees with the dimension, hence the extension is galois. I used this fact to keep the norm within Q, rather than floating above in some intermediate field.
Let E be an intermediate extension of F/K. If every K automorphism of F maps E onto itself then E is stable, or at least, stable within F.
If E is a normal extension of K, then restrict any automorphism on F to a homomorphism from E into F. By criteria 4, the homomorphism keeps E within E, and E is stable. If E is normal it is stable within any extension.
Let G be the galois group of F/K and let H be the subgroup that fixes E. Continuing the notation from above, H = E′. Note that F/K need not be a galois extension, but it has a galois group G nonetheless, with H as subgroup. If E is stable then H is a normal subgroup of G. Apply any automorphism c from G, which keeps the elements of E inside E, then apply something from H, which leaves E alone. Finally apply c inverse, which moves the elements of E back where they started from. The result is an automorphism that fixes E, another automorphism in H, hence H is normal in G.
Conversely, assume H is normal in G and let E be the subfield fixed by H, H′ in our notation. By normality, c*H/c returns u in E to its original position. Suppose c(u) = v for some v outside of E. Some automorphism in H moves v, else v would belong to E. Apply this automorphism and move v to w, then apply c inverse. The composite automorphism does not put u back where it started, and does not fix E. Yet H is normal, so cH/c should fix E. Therefore c keeps E inside E, and E is stable.
In summary, E is stable in F iff the subgroup that fixes E is normal in the galois group of F.
Next assume F is galois over K and E is stable. Every u in E-K is moved by some automorphism of F, and this automorphism restricted to E is a valid automorphism of E, hence E is galois over K. A stable extension in a galois extension is galois.
If E is algebraic, and galois over K, then E is normal, and thus stable within any larger extension, as described in the previous section. For any u in E-K, let p(x) be the irreducible polynomial with p(u) = 0, and let u1 u2 … un be the distinct roots of p that are present in E. Let q(x) be the monic polynomial with precisely these roots, and only one copy of each root. In other words, q = x-u1 times x-u2 times … x-un. The last coefficient of q is the product of the roots, and these roots are permuted by any automorphism c, hence the constant coefficient is fixed by every automorphism of E/K. With E galois over K, the constant coefficient of q is in K. At the other end, the second coefficient is the sum of the roots. This too is symmetric, and fixed by every automorphism of E/K. This is also in K. Every coefficient of q is a symmetric polynomial in the roots, hence all the coefficients of q lie in K. Since q divides p, and p is irreducible, q = p. All the roots of p(x) are present in E, and E is a normal extension. Furthermore, all the roots of p have multiplicity 1, and are distinct, which is the definition of a separable extension. (Separable and inseparable fields will be discussed later.) Thus every algebraic galois extension is normal, separable, and stable within any larger extension.
This reasoning can be reversed; if E is normal it is a splitting field, and some automorphism on E moves u to its conjugate v, thus only K is fixed. Again this assumes u has a conjugate other then u, i.e. the roots of every polynomial are distinct and separable. Thus normal and separable implies galois.
Let F have dimension 2 over K, and let the characteristic of K be odd or 0. F is K adjoin u, where u is the root of some irreducible polynomial of degree 2. F is also K adjoin sqrt(d), where d is the discriminant. After all, d can be combined with the coefficients of p to obtain u. That is the quadratic formula. This can be reversed to derive sqrt(d) from u. Thus K(U) = K(sqrt(d)).
If u is present then find sqrt(d), negate it, and build the conjugate of u, i.e. the other root of p(x). If you have one root then you have them both. Therefore F/K is a normal extension. Since the square root of d is nonzero, the two roots of p(x) are distinct. This holds for any u in F. This makes F separable. Therefore F/K is galois.
If K has characteristic 2 the extension need not be galois. Let K be the polynomials in t, with coefficients mod 2, and let u be the square root of t. Thus u satisfies x2+t = 0. This polynomial is irreducible, but both roots are u. That is, x2+t = (x+u) * (x+u). The extension is normal, but it is not separable. The two roots cannot be separated. If you are building the galois group G, you can only move u to u, hence G is trivial. This is smaller than the dimension of F/K, hence the extension is not galois.
A K automorphism of E is extendable if it can be extended up to a K automorphism of F. This can always be done when F is normal. We just showed algebraic and galois implies normal, so if F/K is algebraic and galois, every intermediate extension E has automorphisms that are extendable.
Let E be stable in the algebraic galois extension F/K. Thus E′ is a normal subgroup H of G, and E is galois. Its galois group is Q, the quotient group G/H. Map G onto Q by restricting the automorphisms of F to E. Since E is stable, this is well defined, and H is the kernel, mapping to the trivial automorphism on E. Since the automorphisms of E are extendable, the map is onto. Therefore the galois group of E over K is Q, or G/H.
In summary, let E be an intermediate extension in the algebraic galois extension F/K, and let E′ = H. Now E is stable iff E is normal, iff E is galois over K, iff H is a normal subgroup of G; and in that case the galois group of E over K is G/H.
As an application of this principle, adjoin y, the nth root of 1, to the rationals. This is an extension of dimension φ(n), with φ(n) primitive roots, and φ(n) automorphisms mapping y to any of these primitive roots. The extension is galois, and the galois group is multiplication mod n, which is abelian. All subgroups are normal. Every subgroup defines a field extension of Q that is galois. These are not all cyclotomic. Set n = 5, and since n is prime, there is no lesser cyclotomic extension. However, the galois group has a subgroup defining an intermediate field extension L. Since 42 = 1 mod 5, the automorphism of order 2 is y → y4. Let s = y+y4, thus s is fixed by this automorphism, and lives in L. However, s is not fixed by y → y2, which maps s, on the positive x axis, to a negative number. Thus s does not lie in Q, and s generates L.
Note that s is twice the cosine of 72°. The inverse of s is the golden ratio, which is sometimes denoted φ (not to be confused with euler's φ function that counts coprime integers less than n). Since φ and s are inverses, adjoin φ to Q and get the same extension L. In other words, the golden ratio creates a subfield of the cyclotomic extension for n = 5.
There is a cyclic galois group of size d for any d, over Q or the integers mod p. Let F be the finite field of order pd. The frobenius automorphism c raises everything to the p, and c generates the galois group of size d.
For the rationals, select p coprime to d, and p odd, and let p have order k mod d, thus pk = 1 mod d. Adjoin y, such that y is a pk root of 1. Multiplication mod pk is cyclic, of order pk-1×(p-1). Let b generate this cycle, thus b is a base for discrete logs. The automorphism y → yb generates all the others, and is the start of the cyclicc galois group. Let c = y → ybd. Now c generates a normal subgroup H of index d, and the field fixed by c has dimension d, and galois group Z/d, generated by y → yb.
For a nonabelian galois group, adjoin the cube root of 1, which is galois of dimension 2, then adjoin the cube root of 2, which brings in all three cube roots of 2, and is a splitting field, hence galois over the previous cyclotomic extension. Let s be the real cube root of 2, and let w be the primitive cube root of 1. Thus the cube roots of 2 are s, sw, and sw2. One automorphism maps s to sw, and leaves the cyclotomic alone. This shifts the cube roots of 2 around in a cycle of length 3. Another automorphism is complex conjugation, which maps w to w2. This is an automorphism on the cyclotomic, but it extends to the higher splitting field, swapping sw and sw2, and leaving s alone. The galois group acts on the cube roots of 3 by a 3 cycle and a 2 cycle. This generates S3, and since G has size 6 or less, G = S3, and is a nonabelian galois group over Q.
The composition of galois extensions need not be galois. Since Q has characteristic 0, every field extension of the rationals is separable, (I'll prove this later), hence galois and normal are synonymous. Recall that the composition of normal extensions need not be normal, hence the composition of galois extensions need not be galois.
Let E be galois over K and let F be galois over E, with F/K a finite extension. Let G be the galois group of F/K. The size of G agrees with the dimension of F/K iff F is galois. Let H be the galois group of F/E. In this case |H| is the dimension of F/E. If F is galois then the galois group of E/K is G/H, as described in the previous section. Every automorphism of E extends to at least one automorphism on F. Conversely, assume every automorphism of E extends to an automorphism on F. Each automorphism of E defines a unique coset of H in G. The number of cosets is the number of automorphisms of E, is the dimension of E/K. Put this all together and the size of G is the dimension of F/K, hence F is galois. Thus F is galois iff all the automorphisms of E extend to F.
A special case occurs when F/E splits irreducible polynomials that happen to have coefficients in K. These roots can move about, independently of the automorphisms of E. In fact the automorphisms of F/E commute with the automorphisms of E/K. F/K is galois, and the galois group is the direct product of the two smaller galois groups.
One can build a galois extension, somewhere, for any finite group. But first, a lemma.
Let F/K be a field extension with galois group G. The extension need not be galois. Let H be any finite subgroup of G. Let E be the intermediate extension fixed by H, H′ in our notation. By construction, E is fixed by H, and F/E is galois.
Let J be the subgroup of G that fixes E, thus J contains H. In other words, there may be more automorphisms in G that fix E, besides those in H. As you move from automorphism groups to fixed fields, or from fields back to groups, the dimension / index can only decrease. H is finite, and the dimension of 1′ over H′ can be no larger. Thus F/E is finite, a finite galois extension. Its dimension equals the size of its galois group, namely |J|. So J contains H, but is no larger than H, hence J = H. The subgroup H is indeed the galois group for the galois extension F/E.
If H has order n, adjoin n indeterminants to the field K and call the resulting field F. Associate each indeterminant with an element of H. Let H act on these indeterminants by right translation, as though H was acting on itself. Each action is an automorphism on F, and H becomes a finite subgroup of the automorphisms of F over K. Let H fix E, and F/E is galois with galois group H.
Let x be an indeterminant, so that K(X) is the rational functions in x with coefficients in K. This extension is galois iff K is infinite.
An automorphism has to move x somewhere, and that determines the automorphism. Map x to a rational function of x, i.e. a quotient of polynomials f/g, where one or both polynomials are nontrivial, i.e. not in K. Also, the quotient is in lowest terms.
If f/g is algebraic over K, by some polynomial q, then expand q(f/g), clear denominators, and x becomes algebraic over K, which is a contradiction. Thus f/g is transcendental; call it t for convenience.
Let E be the intermediate field extension K(t). The polynomial p(w) = t*g(w) - f(w) is 0 when w = x, thus x is algebraic over E, and K(x) is a finite extension of E.
Let's look ad p(w) again. The coefficient t is a member of the presupposed extension E. It acts like a coefficient, and does not change the degree of g(w). Hence the degree of p is the maximum of the degree of f and the degree of g.
Let's prove p is irreducible. The polynomial p = t*g(w)-f(w) is irreducible over K(t)[w] iff it is irreducible over K[t][w]. This is gauss' lemma. Within the ufd K[t,w], p is linear in t, and only common factors of f and g can be pulled out. Since f and g are relatively prime, a reduced fraction, this is not possible, hence p is irreducible. This shows the dimension of K(x) over E is the maximum of the degrees of f and g.
If the monomorphism x → f/g is to be an automorphism, the dimension of x better be 1, forcing f and g to be linear or constant. These are all the automorphisms of K(x) over K.
Let L be an intermediate extension between K and K(x). Take any nonconstant rational function f/g in L, and build E as before. Now x is algebraic over E, hence K(x) is finite over E, and since L contains E, K(x) is finite over L. Thus K(x) is finite over every intermediate extension, save K.
If K′′ contains anything not in K, then let L = K′′. The galois group over L is the same as the galois group over K, but K(X) is galois over L. As shown in the above paragraph, K(x) is a finite extension of L. Hence the galois group is finite, and there are finitely many automorphisms. However, when K is infinite there are an infinite number of automorphisms, x → cx to name just a few. This is a contradiction, hence K(x) is galois over K.
If K is finite, the number of automorphisms of K(x) is finite: x → f/g where f and g are constant or linear. If K(x) is also galois, the dimension of K(x) over K is the order of the galois group, which is finite. Thus x is algebraic, which is a contradiction. Therefore K(x) is galois over K iff K is infinite.
Let K be infinite, whence K(x) is galois over K. Embed this in the larger field K(x,y). Galois extensions are suppose to be normal, but only if the extension is algebraic. In this case an automorphism swaps x and y, and K(x) is not a stable intermediate extension.
This is perhaps a good time to introduce fractional linear transforms. These are the automorphisms described above, where c(x) maps x to the quotient of two linear polynomials. It is called a fractional linear transform because it maps x to a fraction of linear expressions. Since it is an automorphism, the map can always be reversed.
Instead of symbolic polynomials, look at these transforms, in fact all rational functions, as functions from K into K. After all, (x+3) / (x-5) is a function that you can graph in the xy plane, right? But this goes beyond real numbers, and generalizes to any field K. Substitute something from K into x and evaluate the expression; that becomes a function from K into K.
Suppose two expressions implement the same function. The polynomials may be different, but the map from K into K is the same. Clear denominators, and two polynomials implement the same function. Subtract one polynomial from the other, and find a polynomial that implements the zero function. This could happen if K is finite. For example, x7-x is the same as 0 mod 7. However, if K is infinite, each polynomial has finitely many roots, and a nontrivial polynomial cannot become 0 everywhere. Therefore rational functions from K into K and reduced quotients of polynomials correspond 1 for 1. Of course there may be other functions, such as Ex, that are not rational, and cannot be realized as a quotient of two polynomials.
Let s and t be two rational functions. To illustrate, let s = x2, and let t = x+1. Apply s, and double the exponents on every polynomial. Apply t, and replace x with x+1, which turns x2 into (x+1)2. This is the successive application of two field homomorphisms. Thus st maps x to (x+1)2. Now, as maps from K into K, apply s, then t, and get x2+1. This isn't the same, but it's easily fixed. Apply the homomorphisms left to right, and let function composition run right to left. Thus balance is restored. It doesn't matter if you look at polynomials symbolically, or as functions from K into K, composition is associative, and the structure is the same.
Restrict attention to fractional linear transforms, where the numerator and denominator are linear. This is the group of automorphisms of K(x), which is galois over K, and it is also a group of functions from K into K. The two groups are isomorphic.
We have to be a bit careful when talking about functions from K into K. Each such function has at most one pole. If the denominator is x-d, then that function is not defined at d. But it is defined everywhere else.
It is sometimes helpful to add ω, the point at infinity, to K. If K is the reals, or the complex numbers, or even the p-adic numbers, ω is the compactification of the space. When K is the real line, let ω close it up like a giant circle. When K is the complex plane, ω turns it into a sphere, with the origin at the south pole and ω at the north pole. This is just an analogy, but it's a pretty good one.
If g(x) is a polynomial, then augment g by setting g(ω) = ω. This preserves the essence of continuity, if g was continuous to begin with. In other words, g(x) approaches infinity as x approaches infinity. This can be made rigorous with a δ ε proof.
Let g(x) = (ax+b) over (cx+d), where c is nonzero. Since g is not defined at -d/c, let g(-d/c) = ω. This preserves continuity; g approaches infinity as x approaches -d/c.
Suppose g(x) = a/c. Clear denominators, and acx+bc = acx+ad, whence bc = ad. If a = 0 then b = 0, whence g is identically 0, which is not one of our transforms. If d = 0 then b = 0, and g is constant at a/c, not one of our transforms. With all four coefficients nonzero, write a/c = b/d = r, and g is the constant function r, not one of our transforms. Thus g never atains the value a/c. Let g(ω) = a/c. This again preserves continuity; g(x) approaches a/c as x approaches infinity.
With these enhancements in place, each fractional linear transform is a well defined map from K+ω onto K+ω. In fact it is a bijection, and a bicontinuous map whenever K has a supporting topology.
For every distinct a, b, and c in K, a fractional linear transform maps a to 0, b to 1, and c to ω.
g(x) = (b-c)/(b-a) × (x-a)/(x-c)
If a b or c = ω, the following transforms do the trick.
a = ω: (b-c) over (x-c)
b = ω: (x-a) over (x-c)
c = ω: (x-a) over (b-a)
Carry three points onto 0 1 and ω, then on to three other points, and there is a unique transform that carries any three distinct points in K+ω onto any other three distinct points in K+ω.
This makes sense if you think about it. Start with (ax+b)/(cx+d) and clear c from the numerator and denominator, leaving (ax+b)/(x+d). Three coefficients determine the transform.
Let K = C, the complex plane. Each fractional linear transform g maps lines and circles to lines and circles. If you consider a line to be a circle that passes through ω, then g maps circles to circles.
Think of a transform as r times (x+d) / (x+d), which is the constant fuhnction r, plus s / (x+d). Each step in this construction maps circles to circles.
Add a constant and slide the plane over, which maps circles to circles.
Use Demoivre's formula to multiply by a constant. This magnifies the plane by r, and rotates it by θ, and maps circles to circles.
The only tricky case is 1/z. Apply this transform, which replaces θ with -θ and r with 1/r. The former is merely a reflection through the x axis, and preserves shapes. The latter is symmetric about the origin, so rotate the plane through any angle that proves efficacious. If the "circle" is a line, Spin the plane so that the line is horizontal. If it passes through the origin the image is another line passing through the origin, with 0 mapped to ω and ω mapped to 0. If the line has the equation y = d, write it in polar form as r = d/sin(θ). Replace r with 1/r and get r = sin(θ)/d. Let c = 1/2d for notational convenience. Thus r = 2c×sin(θ). Replace sine with y/r, and r2 = 2cy. Replace r2 with x2+y2, subtract 2cy from each side, and fine the equation of a circle centered at 0,c and passing through the origin.
Reverse this algebra to show that a circle passing through the origin, with center at y = c, is mapped to a horizontal line. Or just remember that each fractional linear transform is invertible.
Now consider a circle that does not pass through the origin. Spin the plane so that the circle has its center on the positive x axis. Magnify the plane by m, take the reciprocal of distance, then magnify by m again, which effectively maps r onto 1/r. Magnification takes circles to circles, so scale the plane by any factor that is convenient. Scale the plane so that the circle has center at x and radius 1. Draw a radius, having length 1, and draw the segment from the end of the radius to the origin. Bring in the segment from 0 to x, giving a triangle with angle θ at the origin. Use the law of cosines to solve for r.
x2 + r2 - 2xr×cos(θ) = 1
Replace r with 1/r and multiply through by r2 to get the following. Remember, r cannot be 0; we've already handled the case where the circle passes through the origin.
x2r2 + 1 - 2xr×cos(θ) = r2
(x2-1)r2 - 2xr×cos(θ) = -1
Remember this equation, and start down a new path. If the image is really a circle, its left and right points are at 1/(x+1) and 1/(x-1). Let c = x2-1 for notational convenience. The center is now x/c and the radius is 1/c. Write the equation for this circle, in polar form, using the law of cosines, as we did above.
x2/c2 + r2 - 2xr/c×cos(θ) = 1/c2
Multiply through by c and move the first term to the right.
cr2 - 2xr×cos(θ) = 1/c - x2/c = (1-x2)/c = -1
This is the same equation we arrived at earlier by applying the transform. That completes the proof.