Compose two ring automorphisms and find another ring automorphism.
Reverse a ring automorphism to find its inverse.
The automorphisms of a ring S form a group under composition.
If S is a ring extension of R, we are often interested in the ring automorphisms of S that fix R. These too form a group. Ring automorphisms are often discussed in the context of integral extensions, because most of the interesting theorems require S/R to be an integral extension. In fact, if R is the subring of S that is fixed by a finite group G of ring automorphisms, then S is integral over R. Consider any u in S, and let c1(u), c2(u), c3(u), etc be the conjugates of u under the various automorphisms in G. Build a polynomial p(x) with these conjugates as roots. The constant term is the product of the conjugates, which is not changed by conjugation. That merely permutes the factors in the product. Hence the product, or the last coefficient of p(x), lies in R. In the same way, the second coefficient is the sum of the conjugates of u, and this too lies in R. The other coefficients are more complicated functions of the conjugates, but they are still symmetric. All coefficients lie in R, and p is monic, and u is a root of p, whence u is integral over R. 

Let S be a ring and let G be a group of automorphisms of S. Let c be an arbitrary member of G. Let W be a multiplicatively closed set in S, such that c(W) lies in W for each c in G. Let c act on the fraction ring S/W by acting on the numerator and denominator. As usual, we need to show this action is well defined.
Assume a1/w1 and a2/w2 are equal, courtesy of u. Write u*(a1w2a2w1) = 0. Apply c, and c(u) becomes the element that proves c(a1/w1) = c(a2/w2). (This works because c is injective, so that c(u) is nonzero, and it lies in W.) The action is well defined.
Represent the sum of two fractions by building a common denominator, then apply c. This produces the same expression as the sum of the image of the two fractions. Thus c respects addition. Similar algebra shows c respects multiplication. We have a ring endomorphism on S/W.
Reverse the action of c on the numerator and denominator to show the endomorphism is onto.
Finally, assume c(a1)/c(w1) = 0, whence some u kills c(a1). Apply c inverse, and a1 becomes a zero divisor, whence a1/w1 was 0 after all. The induced map is a ring automorphism on S/W, and these automorphisms compose according to G, hence a group homomorphism takes G onto H, a group of ring automorphisms on S/W.
Localization is now a handy tool for understanding a ring and its automorphisms. If G acts on S, and fixes R, then it fixes RP for any prime ideal P in R. Let W = RP, and G becomes a group of automorphisms on S/W, fixing the base ring R/W. When R is an integral domain, and P is 0, we have a ring extension of a field. When S is an integral domain, set W to the nonzero elements of S, and G becomes part of the galois group of a field extension.
Let S be a ring and let W be a multiplicatively closed set that is mapped into itself by the automorphisms of G. Thus G becomes a group of automorphisms on S/W.
For any d in W, multiply d by all its conjugates, giving something in W, something nonzero, something fixed by G. There is a fixed ring R.
Let F be the set of fractions whose numerators and denominators are fixed by G. In other words, F is R/(R∩W). This includes 1/1, or v/v for some fixed v in W. Verify that F is a ring. It is a subring of S/W that is fixed by G. In fact a ring homomorphism embeds F into the ring that is fixed by G. We only need prove this map is onto, and the two subrings will coincide. In other words, the fixed fractions of R is the fixed ring of S/W.
Given a fraction n/d, numerator over denominator, multiply top and bottom by all the conjugates of d, ignoring the identity automorphism. This is the same fraction, and now the denominator is fixed by G. Thus it is enough to consider a fraction n/d, fixed by G, where d is fixed by G.
For each automorphism c in G, c maps n/d to an equivalent fraction. Write an equation u*(c(n)dnd) = 0, where u depends on c.
Remember that c maps u into W, and W is multiplicatively closed. Multiply u by all its conjugates, so that u is fixed by G. Thus we can assume each uc is in R.
Fold d into uc. Thus, for each c, some uc satisfies uc*(c(n)n) = 0.
Let v be the product over uc. This is in W, and is fixed by G. Also, v*(c(n)n) = 0 for each c.
Consider the fraction vn/vd, which is equivalent to n/d. The denominator is fixed by G. Apply c to the numerator and get c(n)v, which is equal to nv. That completes the proof. The fractions of the fixed ring produce the same subring as the fixed ring of the fractions.
This theorem is not used elsewhere, so if you are not comfortable with tensor product and base change, you may wish to skip ahead.
Let S/R be a ring extension, with a group G that fixes R. Let T be another R algebra. In other words, a ring homomorphism f maps R into the center of T. Let U = S tensor T, which is an R algebra.
Remember that R is a subring of S, but it might not be a subring of T. Let f(R) become R′ in T. Then review the structure of S×T. Basically, S and T are treated as separate rings, except they share an instance of R. For convenience, the action of R is always passed over to T by vectoring through R′. (You might think of U as an R′ module.) Aside from the shared instance of R → R′, multiplication takes place per component, and terms are added together, like polynomials.
Let c be one of the automorphisms of G, and build a new function d by letting c act on S, and leaving T alone. Note that c fixes R, which is shared between S and T, so there is no contradiction between c moving S about and c fixing T.
Since c respects addition and multiplication, the same holds for d. It is a ring endomorphism on U. Also, d can be reversed by reversing c on the S component of each term, and leaving T alone.
Suppose d(x,y) = 0. This means one of the two components has to become 0. If y is 0 then x,y was already 0. If c(x) becomes 0 then x is 0 (since c is injective). If c(x) is divisible by k in R, such that f(k)*y = 0, so that c(x),y is equivalent to 0, then reverse c to show that x was already divisible by k, and x,y was already 0. If f(k) pulls out of y, such that k*c(x) = 0, then k*x = 0, and x,y was already 0. If d maps a term to 0, then that term was already 0.
Suppose d maps a sum of terms to 0. Collapse the sum of terms as far as possible. If one of the terms is equivalent to 0, throw it out. If something in R can be passed between x1 and y1, so that y1 becomes y2, then this term can be combined with x2,y2. Assume this has already been done, and apply d. We're talking about more than one term, since d is injective on single terms. Somehow c(x1),y1 + c(x2),y2 (and perhaps more terms) collapses to 0. The element k ∈ R that passes between c(x1) and y1, to turn y1 into y2, passes right through c, because c fixes R. The same action passes between x1 and y1, and we should have combined these two terms at the outset. The original sum of terms collapses to one term, which is 0. Therefore d is injective, and c induces a ring automorphism on U. This holds for every c in G.
If R embeds in T, the induced automorphisms remain distinct. Let c1 and c2 map x to two distinct elements z1 and z2. Now x,1 is a nonzero element of U, and d1 and d2 map x,1 to z1,1 and z2,1, respectively. Suppose these values are equal in U. Subtract them to find (z1z2),1 = 0. Since R embeds in T, we can't pass k over to the right to make this term equal to 0. Nor can we pass a unit to the left to kill z1z2, since units don't kill anything. Therefore z1 = z2, which is a contradiction. If R embeds in T, distinct automorphisms in G remain distinct when extended to U.
Don't assume that fixed ring and tensor product commute, as was the case with localization. Clearly R×T = T is fixed by each d, but additional members of U may be fixed. Let's look at an example.
Let R be Z[y], the integer polynomials in y. This will be the ring that embeds in S and in T, i.e. the ring common to both components of the tensor product. Let S = R[x], the integer polynomials in x and in y. Let T = R[z], with yz = 0. The tensor product U = S×T consists of all polynomials in x y and z, but yz drops to 0 wherever it appears.
The group G consists of two automorphisms, c and its inverse. Let c replace x with x+y throughout. Clearly c inverse replaces x with xy. Verify that these are automorphisms on S.
Suppose a polynomial is fixed by G, and let xm be the highest power of x. Replace x with x+y and apply the binomial theorem. This introduces a new term myxm1 at degree m1. This cannot be undone by replacing x with x+y in the lower degree terms. The polynomial is not the same. Therefore G fixes precisely R.
Moving to U, consider the element xz. This is not a member of T. However, (x±y)z = xz, and xz is fixed by G.
Let S/R be a ring extension, where S is an integral domain, and R is the subring that is fixed by a group G of ring automorphisms. Thus S is integral over R. Let E be the fraction field of S, and let F be the fraction field of R. Note that S embeds in E, and R embeds in E, and F embeds in E. Thus E/F is a field extension that corresponds to S/R. 

Let G induce a group of automorphisms on E. Since S embeds in E, the automorphisms remain distinct. By the previous theorem, the subring that is fixed by G is precisely the fractions of R, which is F. Thus G is a finite group of field automorphisms of E, fixing F. This makes E/F a galois extension, with galois group G. The automorphisms of S/R are precisely the automorphisms of E/F.
Since a galois extension is separable, apply the primitive element theorem. Select u so that E = F(u), and clear the denominator, so that u is integral over R. Assume we know that u lies in S, as is the case if S is integrally closed in its fraction field E. Let p be the minimum polynomial of u. Make p monic, so that its roots are all the conjugates of u. Since E/F is galois it is a splitting field. All the conjugates of u live in E. Therefore p splits into linear terms in E/F. Since G permutes the roots of p, it fixes the coefficients of p. Thus the coefficients all lie in F. This we already knew, but the coefficients are built using conjugates of u, each conjugate landing somewhere in S, since G maps S onto S, so the coefficients also lie in S. The elements of S that are fixed by G lie in R, hence the coefficients of p lie in R. In other words, p is monic, with coefficients in R. The integral polynomial over R is the irreducible polynomial over F.
The powers of u are linearly independent over F, and over R, and they all live in S, thus S contains a free R module of rank n. We will see below that S is often trapped between two free R modules of rank n, and in many cases S itself is a free R module of rank n. In particular, S is trapped between two free R modules of rank n when R is integrally closed.
Now let v be an arbitrary element of S. Let K be the normal closure of F(v) inside E. Let p(x) be the minimum monic polynomial of v over F. Reason as above to show p has coefficients in R, and becomes the integral polynomial of v.
Each automorphism of G moves v to one of its conjugates, thus permuting the conjugates of v. Conversely, move v to any of its conjugates, and extend this to an automorphism on K. Further extend this to an automorphism on E. Since E/F is galois, this automorphism is part of G. Therefore G acts transitively on the conjugates of v. This holds for any v in S, though the number of conjugates could be less than n. Since the dimension of F(v)/F divides the dimension of E/F, the number of conjugates divides n. It is 1 iff v lies in R.
If c is a ring automorphism, verify that c maps an ideal to an ideal, and c maps the product of two ideals onto the product of their images. It follows that c maps a prime ideal onto a prime ideal.
An ideal is proper, or maximal, or trivial, iff the same is true of its image under c.
Let R be the subring of S fixed by G, hence S/R is an integral extension. Let S be commutative, so that the primes of S lie over the primes of R. Since R is fixed, a prime ideal Q, lying over P, has to move, via c, to another prime ideal lying over P. In other words, G permutes the prime ideals lying over P.
Let S be an integral domain. Let Q1 and Q2 be two different prime ideals lying over P. We want to move Q1 onto Q2.
Let u be an element of Q1 that is not in R. Let v be the product of all the conjugates of u. (This is where we need S to be an integral domain, to take advantage of the previous theorem.) Now v is ± the last coefficient of the monic polynomial P(u), and it lies in R. It also lies in Q1, hence in P. Since Q2 is prime, one of the conjugates of u lies in Q2. Select c so that c(u) lies in Q2.
Let d be the inverse of c. Now u is contained in d(Q2). This can be done for any u in Q1. Therefore Q1 lies in the finite union of all the images of Q2 under G. These images are prime, hence Q1 is contained in the finite union of prime ideals. It follows that Q1 is contained in one of these prime ideals.
For some d, Q1 lies in d(Q2). Since one prime ideal cannot contain another, Q1 = d(Q2). Reverse this, and Q2 = c(Q1). Since Q2 was arbitrary, G maps Q1 to every prime ideal lying over P. In other words, G acts transitively on the fiber of P.
The subgroup of G that stabilizes Q1, times the orbit, gives the order of G. Thus the number of primes lying over P divides G.
Inclusion of R into S induces a continuous map from spec S into spec R. Since every prime P has a prime lying over it, this map is onto. The primes over P become the fiber of P, or the preimage of P under the induced continuous map.
Let G define an equivalence relation on the prime ideals of S, which are the points of spec S. Since G acts transitively, it equates all the primes in the fiber of P.
Since S is integral over R, the map from spec S onto spec R is bicontinuous. Once fibers are compressed down to single points, spec S really is spec R. In other words, spec R is a quotient space of spec S, where G provides the equivalence relation.
An integral extension S is a ring that is integral over R, so it might reasonably be called an integral ring, but the term "integral ring", or "ring of integers", means something more.
Let R be an integral domain in its fraction field F, and let E/F be a finite field extension.
Let S be the integral closure of R in E.
We say S is the integral ring, or the ring of integers, of E/F/R.
Since E contains the fraction field of S, and S is integrally closed in E, S is integrally closed in its own fraction field, thus S is integrally closed as a ring. With S/R defined as above, a subring of S that contains R is called an order. 

Let R be a ufd and let u be algebraic over the fraction field F. u is integral iff its minimum polynomial over F is monic and lies in R[x].
One direction is obvious, so let's move left to right. Being algebraic, u satisfies a polynomial p over F, whose degree is the dimension of the field extension F(u)/F. Make this polynomial monic over F. If the remaining coefficients lie in R then we're done. Otherwise u satisfies some other monic polynomial q(x) with coefficients in R. If p(x) does not divide q(x) over F, u satisfies the remainder polynomial when q is divided by p. This has degree less than p, which is a contradiction. Therefore p divides q over F.
Apply gauss' lemma, and q is the product of two polynomials in R[x], one of them a scaled version of p. The product of the lead coefficients is 1, i.e. the lead coefficient of q. Thus p leads off with a unit, and if we divide through by this unit, p is monic, with coefficients in R, whereupon the minimum polynomial p is monic and lies in R[x]. With R a ufd, the polynomial that proves u is integral is, or can be, the minimum polynomial of u over F.
This theorem can be expanded to additional rings. In fact it applies to any integrally closed integral domain. (Remember that a ufd is such a ring.) We know that u is integral over R by some monic polynomial q. Let p be the minimum polynomial of u over F, thus p divides q over F. Extend F up to E, the splitting field of p. Now E contains all the roots of p. Each root of p is a root of q and is integral over R. Combine these roots to build p(x), and each coefficient of p is integral over R. Yet the coefficients of p lie in F, hence they lie in R. Once again p(x), the minimum polynomial of u, proves u is integral over R.
If q(x) has root u, where q may or may not be monic, then divide q by p, which is monic, and the remainder also has root u. The remainder has lesser degree than p, which is impossible. Hence the remainder is 0, and p divides q. Every polynomial with root u is divisible by the minimum monic polynomial p. This establishes a principal ideal in the ring of polynomials.
If p = j*k, then either j(u) = 0 or k(u) = 0. Since p has minimum degree, this is impossible. Therefore p is irreducible.
If g is some other irreducible polynomial with root u, then p divides g, as described above, and p = g. The irreducible polynomial associated with u is unique.
Assume R is an integrally closed integral domain. This is the case when R is a ufd or dedekind. If E/F is a finite separable field extension of dimension n, the integral ring S over R is trapped between two free R modules of rank n. This helps us characterize S in a wide range of settings.
R * R * R 
S 
R * R * R 
Let E/F have dimension n, and find a basis for E as an F vector space. Each basis element is algebraic over R, and some multiple becomes integral over R. Scale the basis elements accordingly. The result is still a basis for E over F, but the basis is now integral over R. In other words, we have found a basis for E/F that lies in S. This basis is linearly independent in F, and in R, and spans a free R module of rank n. Linear combinations of integral elements remain integral, so the entire span is integral. Therefore S contains a free R module of rank n.
That was the easy part. Now for the other side, a free R module that contains S.
Let b1 through bn form a basis for E over F, with each bi in S, as described above. Assume E is galois over F. Let y be an arbitrary element of S, and represent y as the sum of aibi. Here ai lies in F, not necessarily in R.
Note that ybj lies in S for each basis element bj. The trace of ybj is the sum of the images of ybj under the n automorphisms of E/F. This lies in F, since F is the fixed field of E/F. Each conjugate of an element in S is another root of the minimal, integral polynomial for that element, and is in S, hence the trace is in S. The trace is integral over R, and in F, and since R is integrally closed the trace is in R. Call the trace of ybj tj, whence each tj lies in R.
The trace of the sum is the sum of the trace, so rewrite tj as the sum of the trace of aibibj. Since ai lies in F it is a constant, and can be pulled out of the trace.
∑ ai * trace(bibj) = tj
There are n such equations, as j runs from 1 to n. Put them together as a system of simultaneous equations, like this.
M * a = t
Here a is a column vector a1 through an, i.e. the coefficients of y using the basis b1 through bn. The right side t is another column vector t1 through tn, where tj is the trace of ybj. The symmetric matrix M holds the trace of bibj. Remember that the trace of anything in S lies in R. Thus the entries of M and t lie in R, and the column vector a lies in F.
Build a new matrix W, whose first row is the basis b1 through bn. This is the trivial automorphism applied to the basis. For the second row of W, take another automorphism of E/F and apply it to the basis. The third row of W is the image of b1 through bn under the third automorphism, and so on. Since E/F is galois there are n automorphisms, and W is a square matrix.
W  = 

Suppose the rows of W are linearly dependent. A linear combination of automorphisms takes the basis to 0, and since F commutes past these automorphisms, the same linear combination of automorphisms maps all of E to 0. However, distinct automorphisms are linearly independent. Therefore W is a nonsingular matrix with a nonzero determinant in E.
Multiply the transpose of W by W, in that order, to get M. Thus det(M) = det(W)2, in an integral domain. With det(M) nonzero, M is nonsingular.
Let d = det(M). This is, by definition, the discriminant of the extension. We'll prove it is basis invariant in the next chapter. Apply cofactors to build the inverse of M, which lies in R over a common denominator d. Multiply by M inverse on the left, and the coefficients a1 through an are elements of R with denominators of d. This holds for every element y in S.
Divide the basis elements by d and find another basis of E/F, that spans a free R module of rank n, that contains every y in S. Therefore S is trapped between two free R modules of rank n.
b1/d b2/d b3/d 
S 
b1 b2 b3 
Note that R and S need not be integrally closed for this to work. Push R up to R′, the integral closure of R in F, and assuming S ⊆ E is integral over R, push S up to S′, the integral closure of S in E. R′ and S′ are integrally closed in their fraction fields. S′ is integral over S, is integral over R and R′. S′ contains R′. Build a basis for E/F inside S′ as above. Assume we know this basis lies in S, or perhaps we have somehow discovered a basis for E/F in S. The free R module spanned by this basis lies in S, and the free R′ module spanned by this basis lies in S′. On the other side, the free R′ module spanned by the basis divided by d contains all of S′, which contains S. S is trapped between a free R′ module of rank n and a free R module of rank n, or if you prefer, S′ is trapped between a free R′ module of rank n and a free R′ module of rank n. If R is integrally closed than we can say S and S′ are trapped between two free R modules of rank n. det(M) continues to be the factor between the two bases of the free modules.
What if E/F is not galois? Let K be the normal closure of E/F. (This is where we need E/F to be separable.) Let T be the integral ring of K/F/R. Note that T contains S.
Now b1 through bn is the first part of a basis for K/F. Let y lie in T, and write it as a linear combination of basis elements. Use the above to find a matrix M and a determinant d, and the coefficients on y lie in R/d. Divide the basis through by d, and now all of T lies in the span of this basis using coefficients from R. Restrict attention to b1 through bn. With coefficients from F, the span is E. Intersect T and E to get S. Therefore S is spanned by b1 through bn alone, without the rest of the basis, and since S lies in T, we only need use coefficients from R. Once again S is contained in a free module of rank n.
If R is noetherian then the same holds for a free R module of finite rank. If R is also integrally closed, the integral ring S is a submodule of a free R module of rank n, and a submodule of a noetherian module is noetherian, thus S is a noetherian R module. An infinite chain of S modules implies a chain of R modules, so S is a noetherian ring.
With S integral over R, a prime ideal P in R lifts to one or more prime ideals in S  however, with S a finitely generated R module, there are finitely many primes over P.
If R is a pid, such as the integers, it is integrally closed and noetherian. In addition, every submodule of a free module is free, thus S is a free R module of rank no larger than n. Since S contains an R module of rank N, its rank can be no smaller. S is a free R module of rank n.
There is another way to prove S is free, though it doesn't come up very often. Let R be noetherian and integrally closed, and locally a pid, and trap S between two free R modules of rank n. Find n generators that span S as an R module, and apply this proof.
Assume R is dedekind. Since R is noetherian, S is noetherian  and S is integrally closed, and integral over R. This makes S dedekind as well. R is locally a pid, so if you can find n generators that span S then S is a free R module.
When R is a pid, it is also dedekind, hence S is a dedekind domain that happens to be a free R module of rank n.
When R is integrally closed, and F is the fraction field of R, we would like to build a correspondence between finite separable extensions of F and integrally closed integral extensions of R. Each field extension admits a ring of integers. Might two different field extensions lead to the same integral ring S? No. Given a finite separable field extension E/F, let S be its integral ring. We showed above that S is trapped between to free R modules of rank n. Represent this with a basis b, and b/d, with S in between. Tensor with F, and both R modules become E, hence S×F = E. In other words, S cannot be the integral ring of E and thee integral ring of some other field extension of F simultaneously.
What about other integrally closed rings over R? Does each come from a field extension? Assume R is noetherian, so that each ring of integers is a finitely generated R module. Turn this around and let S be a finitely generated R module, thus S is integral over R. Further assume S is separable, and integrally closed in its fraction field, which I will call E. If x in E is integral over R it is integral over S, and already included in S, Thus S is the integral ring of E/F/R. E/F is separable, but is it finite? Each x in S is a linear combination of finitely many generators, using coefficients in R. Tensor with F to produce linear combinations of these generators, now in E, using coefficients from F. These generators span a possibly smaller finite field extension K. Yet K includes the fraction field of S, as shown here. That completes the correspondence.
If R is dedekind, and all nonzero ideals in R have finite index, the same holds in S. Such is the case when R = Z. Remember that if the index of each prime ideal in R is finite, then the index of each ideal is finite.
Let u lie in a nonzero ideal H in S. View S as an R module, with u*R a submodule of H a submodule of S.
Let b1 through bn be a basis for E/F that is contained in S. Divide each basis element by u, and some multiple of each quotient lies in S. In other words, vibi/u lies in S. Thus vibi lies in H, and together they span a submodule of H.
Remember that some d can be chosen, so that the basis, divided by d, spans all of S. Divide each bi by d, and multiply each vi by d.
A free R module, spanned by v1b1 through vnbn, lives inside a free R module spanned by b1 through bn. The quotient module is isomorphic to the direct product of R/v1 through R/vn. All ideals in R have finite index, so the quotient module is finite.
The ring S, and the ideal H, map to submodules of this quotient module, therefore the cosets of H in S, or the index of H in S, is finite.
Again, R is dedekind. Assume there are finitely many primes in R with index < c for any positive integer c. The same is true for all the ideals of R.
Let H be an ideal in R with index < c. Write H as a product of prime ideals, and note that each prime ideal has index at least 2. There are at most log2(c) primes in the factorization. Each prime in the factorization has index < c, and there are finitely many of these to choose from. We are choosing at most log2(c) items from a finite set, allowing for duplicates, and there are finitely many ways to do this. Therefore the number of ideals in R with index < c is finite.
We would like to propagate this result up to S. As shown above, we only need prove the result for prime ideals, whence it applies to all ideals in S. This assumes S is dedekind, but that follows from R being dedekind, as discussed earlier.
Let Q be a prime over P, such that the index of Q is less than c. Since R/P embeds in S/Q, the index of P is less than c. There are finitely many choices for P, and each such P has finitely many primes Q lying over it, as demonstrated earlier. Therefore the primes of S with index < c is finite, and the same holds for the ideals of S.
Let R be a dvr with one maximal ideal P. S has finitely many prime ideals over P, which makes S a pid. We already know the valuation on F can be extended to E, but this is only an existence proof  not very helpful. In this case we can be more precise.
Let P be the prime in R that seeds the valuation, and let Q be a prime in S over P. Localize about Q to build a valuation ring inside E. In fact SQ is a dvr. Now Q, the prime ideal or its generator, seeds a valuation that is consistent with P. Write v(Q) = v(P)/n for some integer n. (This n need not agree with the dimension of E/F.) The extended valuation on E is discrete, like that of F, but it has a finer granularity, like drawing the half, quarter, and eighth inch lines on a ruler.
Let R be dedekind and let S be the integral ring of E/F/R as usual. The splitting problem attempts to characterize the prime ideals of S that lie over each prime ideal of R. This is, in general, a difficult problem, though it has been solved in a few special cases, e.g. the quadratic and cyclotomic number fields.
Remember that R and S are dedekind, so prime, which usually means prime and nonzero, is equivalent to maximal.
Since S/R is integral, prime ideals in S lie over prime ideals in R, and all the prime ideals of R lift to prime ideals in S. Furthermore, S is a finitely generated R module, hence there are finitely many primes of S over any given prime ideal in R.
The ideals in S contract to ideals in R. Since contraction and product do not commute, this map is not very interesting, except for its prime ideal correspondence. However, extension and product do commute, so the map from the ideals of R into the ideals of S forms a monoid homomorphism.
Let's generalize this to fractional ideals. Let H be a fractional ideal in F, and let H generate an S module inside E. This is the linear combinations of H with coefficients in S. Remember that d*H lies in R for some d in R. Clearly d lies in S. Multiply d by H*S and get something in R*S, thus the extension of a fractional ideal is a fractional ideal. Furthermore, this definition is consistent with the extension of an ideal.
Verify that product and extension commute for fractional ideals. The proof is the same as the one for ideals.
Since R and S are dedekind, fractional ideals in both R and S are invertible. If H is an ideal in R, the extension of the inverse of H has to be the inverse of the extension of H. This because extension and product commute. Therefore the monoid homomorphism on ideals can be extended to a group homomorphism on fractional ideals.
Let P be a prime ideal in R and extend it to H in S. If Q is a prime ideal in S that divides H then Q contains H, and Q contracts to a prime ideal containing P, namely P. The factorization of H consists of primes lying over P.
The splitting problem can now be stated: find the primes over P, and the prime factorization of the extension of P into S.
If a fractional ideal is principal, say c*R, its extension is c*S. Principal fractional ideals in R map to principal fractional ideals in S. Divide by these subgroups and find a well defined homomorphism from the class group of R into the class group of S.
Localize about P, and find RP a subring of SP. Of course I'm abusing the notation a bit, since P is not a prime ideal in S. Understand that SP means the fraction ring of S by RP.
Both SP and RP are dedekind, and SP is integral over RP. The fraction fields have not changed, and dedekind implies integrally closed, hence SP is the integral ring of E/F/RP.
Only P survives localization downstairs, and the primes that persist upstairs are those lying over P. Primes correspond under localization, so each prime Q lying over P corresponds to a prime QP in SP lying over the prime PP in RP. There are finitely many of these, hence SP is a pid.
Let H be the extension of P in S. Apply the denominators from RP, and produce the same ideal as the localization of P, extended into SP. The result is all linear combinations of elements of P, with coefficients from S, and denominators from RP. Therefore extension and localization commute.
Write H as a unique product of primes lying over P  for example, H = Q1Q24Q3. Now product and localization commute, so the same equation holds in the ring SP.
The splitting problem does not change for P as we move to SP/RP, except we have the convenience of working with a pid upstairs and a dvr downstairs.
Let S be the integral ring over R as usual. Consider the extension of a prime ideal P into S, generated by P*S. We would like to factor P*S into a product of prime ideals lying over P. This is the splitting problem, as described above. In this section we will place some constraints on the factors of P*S.
As you recall, the splitting problem does not change after localization, so assume this has already been done. Thus R is a dvr and S is a pid.
To contain is to divide, hence the primes lying over P are precisely the factors that build the extension of P. We only need find the exponents.
Let Qi be one of the prime ideals lying over P. Mod out by Qi upstairs and P downstairs, and the embedding of R in S becomes an embedding of the field K = R/P into the field S/Qi. Let di be the dimension of this field extension. This is called the residual degree, or the residue degree, of Qi.
Note that the residue degree can be computed before or after localization, before or after we turned R into a dvr, since quotient rings are preserved.
Let P*S be the product, over the relevant primes Qi, of Qi raised to the ei. Thus ei is the exponent, yet to be determined. By definition, ei is the ramification degree of Qi.
Review the formula for tensoring with a quotient ring. S/(P*S) is the same as S tensor R/P, or S tensor K. This is a K vector space.
Remember that R is a dvr, which is a pid, and S is trapped between two free R modules of rank n. Since S is the submodule of a free module over a pid, S is free of rank n. Tensor with K and get a free K module of the same rank. In other words, S×K is an n dimensional K vector space. Thus S/(P*S) = Kn.
There's another characterization of S/(P*S), based on the chinese remainder theorem. S mod P*S is the direct product of s mod Qi to the ei, for all the primes Qi over P. So we want to characterize S mod Qi to the ei, which I'll call S mod Qe for short.
Remember that S/Q is a K vector space of dimension d, where d is the residue degree of Q.
Tensor the ideal Qj with the quotient ring S/Q, using the tensor quotient formula, and get Qj mod Qj+1. Since S is a pid, Qj is principal, and looks just like a free S module. As an S module, Qj is isomorphic to S. Therefore the quotient module looks like S/Q, a K vector space of dimension d.
Consider the quotient ring S mod Qj+1. This includes the ideal Qj mod Qj+1. We just showed this ideal is a K vector space of dimension d. By induction, the quotient ring S/Qj is a K vector space of dimension j×d. Combine kernel and quotient and find a vector space of dimension (j+1)×d. Therefore, S mod Qe becomes a K vector space of dimension e×d.
Apply the chinese remainder theorem and write S mod P*S as a direct product of quotient rings. This becomes the direct product of vector spaces, which adds their dimensions. Therefore n = the sum of ei×di.
In summary, the dimension of the field extension is the sum, over the primes lying over P, of the ramification degree times the residue degree. This is the degree equation.
Recall that R and S have been localized to produce a dvr and a pid respectively, yet sometimes we localize again, this time about a particular prime Q lying over P. Both rings are still dedekind, and both rings still have the same fraction fields E and F. In fact the dvr downstairs doesn't change at all,since Q intersect R is P, and we have already localized about P. The ring upstairs, I'll call it S′, becomes a dvr, since Q → Q′ is the only surviving prime. Localization and product commute, hence P*S′ = Q′e, where e is the ramification degree of Q. The other primes that were part of the factorization of P*S are not contained in Q, hence they become the entire ring after localization. The residue field is also preserved, as we localize around a maximal ideal. With d and e the same, and the other primes lost, d×e no longer equals n. The degree equation has been derailed. That is because S′ is no longer integral over R. It is not trapped between two R modules and is not itself a free R module of rank n. When a module is not free its tensor with another module is far from obvious. In this case the result is a K vector space of dimension d×e.
To illustrate, start with the gaussian integers over Z, localize about 5, then localize about 2+i, one of the two primes over 5. Within S′, 2+i generates 5 as usual, but 5 also generates 2+i, as you multiply by 1/(2i). Both act as generators for the one and only prime Q′, having residue field 5 and residue degree 1 and ramification degree 1, just as Q did in S, but now Q′1 is all of P.
Localization about Q facilitates a particular extension of the valuation. Since most primes are unramified, P still induces a valuation of 1, and there are no new valuations. This is illustrated by the example above, where 5 and 2+i both have a valuation of 1.
If the ramification degree is n, there is one prime lying over P, with the same residue field as the base ring. This follows from the degree equation. We say P is totally ramified. This is further subdivided into two categories: tame, when the characteristic of the residue field does not divide n, and wild, when the characteristic of the residue field does divide n.
If the ramification degree of each prime is 1, P is unramified. The sum of the residue degrees equals n.
If there are n primes lying over P, each ramification degree and each residue degree is 1, and P is unramified.
Let's look at an example, the gaussian integers. Is this indeed a ring of integers? Adjoin i to the rationals to find a 2 dimensional field extension E. S is the ring spanned by 1 and i. Since S is a pid it is integrally closed. Thus S is the ring of integers of E/F/R.
We already know the primes over primes. 2 is totally ramified, with the ideal {1+i} squared equal to {2}, and all other primes unramified. Primes that are 1 mod 4 split, e.g. 2+i * 2i = 5, while primes that are 3 mod 4 are inert. The splitting problem has been solved for the gaussian integers, and for the eisenstein integers as well.
Consider a tower of ring extensions, with S′ over S over R, and Q′ over Q over P.
Extend P into S via P*S, then extend this into S′ via (P*S)*S′. This is the same as P*S′. In other words, the extension of the extension is the composite extension. Similarly, the primes in S′ lying over P are precisely the primes of S′ that lie over the primes of S, that lie over P.
Fix a prime Q over P, and a prime Q′ over Q. The residue fields form a tower of field extensions, and dimensions multiply as usual. Therefore the residue degree of Q′ over P is the residue degree of Q′ over Q times the residue degree of Q over P.
A similar result holds for ramification degrees. Since product and extension commute, write P as a product of various powers of Qi, and push this up to S′. Assuming S′ is dedekind, replace each Qi*S′ with its factorization in S′ to find the factorization of P in S′. For a given Q′ over P, its ramification degree is the product of the ramification degree of Q′ over Q times the ramification degree of Q over P.
For a fixed prime P, the composition of unramified extensions is unramified, and the composition of totally ramified extensions is totally ramified.
Let E/F be galois, with galois group G. Let R be integrally closed, and let S be the integral closure of R in E. Suppose x is fixed by G, and x is in S, but not in R. x is in F, and integral over R, and since R is integrally closed, that pulls x back into R. R is the fixed ring of G.
G respects ideal multiplication, and G acts transitively on prime ideals. (This was shown in an earlier section.) If one prime is raised to a certain power in the factorization of P*S, than all primes lying over P appear with the same exponent. The ramification degree is constant across the primes lying over P.
If an automorphism carries one prime ideal onto another, mod out by both prime ideals and find an isomorphism between the residue fields. These residue fields establish the residue degree. Again, G acts transitively on prime ideals, Hence the residue degree is constant across the primes over P.
The degree equation has been simplified. The dimension of the extension is the ramification degree times the residue degree times the number of primes lying over P. This is the uniform degree equation.
If the dimension is prime, P is totally ramified or unramified.
Examples are fun, so here we go.
Let t be the cube root of 2, and adjoin t to Z and Q. The latter gives a field extension E of dimension 3. The former gives a ring S, which is a pid, which I will prove later. A pid is integrally closed, so S is integrally closed. S is the ring of integers within the number field E.
Suppose a conjugate of t lies in E. Divide t by its conjugate and get the cube root of 1. This generates an extension of dimension 2, inside an extension of dimension 3, which is impossible. Therefore our extension E/Q is not a normal extension. The residue degree need not be constant across all primes lying over p, and in many cases it is not.
If an element x in S is a + bt + ct2, then the norm of x, or the norm of the ideal generated by x, or the index of that ideal within S, is the absolute value of a3 + 2b3 + 4c3  6abc.
For a prime p, consider S mod p. If p is inert, i.e. if p remains prime in S, the result is a field of order p3. Every nonzero element is invertible. The inverse of x is the product of the two nontrivial conjugates of x, divided by the norm, so the norm is always nonzero mod p except for x = 0. If x is nonzero p31 times then p is inert. On the other hand, if p separates into three distinct primes then apply the chinese remainder theorem to find three fields in parallel, each of order p1. The norm is invertible iff each field presents a nonzero element, and this happens (p1)3 times. The last ramified case is the lumpy case, a field of order p and a field of order p2. The norm is nonzero (p1) × (p21) times. Use this computer program to investigate various primes. 7 is inert, 31 is flat (1+2t2 times 1+2t+t2 times 3+t2), and 5 is lumpy (1+t2 times 1+2tt2). The first factor of 5 has residue degree 5, and the second factor has residue degree 25. Verify with this script.
Actually, lots of primes are lumpy; that's the typical case. I don't have a proof, but there is a pattern that suggests a solution to the splitting problem. Skate past 2 and 3, as these are ramified. If p is 2 mod 3, it becomes lumpy, the first prime u over p having residue degree 1 and the second prime v having residue degree 2. Follow up with a second, quadratic extension, the cube root of 1, producing a galois extension of degree 6 over Q. Within the composite extension, the uniform degree equation applies: residue degree times ramification degree times number of primes = 6. A prime over v cannot have residue degree 2, because 4 does not divide 6. Instead, v splits into two primes, each having residue degree 1 over v, and 2 over p, while u is inert in the second extension so that it two has residue degree 2 over p. The action of complex conjugation swaps the two primes over v, and leaves the prime over u alone. An automorphism fixes some primes and moves others about. This is possible only because the galois group is nonabelian.
If p is 1 mod 3, ask whether 2 is a cube mod p. If it is then the residue degree is 1 in the first extension and also in the normal closure. p extends to 3 primes, and then to 6. If 2 is not a cube then p remains inert in the original extension, and splits into two primes having residue degree 3 in the normal closure. Notice that p is never inert in the normal closure.
Z adjoin the cube root of 3 splits similarly, lumpy when p = 2 mod 3, and asking whether 3 is a cube mod p when p = 1 mod 3.
Splitting gets more chaotic as the galois group grows in complexity. Consider Z adjoin the unsolvable quintic. The galois group is S5, of order 120. A brute force program finds the residue degrees over the first few primes as follows.
2: ramified
3: inert
5:ramified
7:ramified
11: 4 1
13: 3 2
17: 2 1 1 1
19: 2 1 1 1
23: 3 2
29: 2 1 1 1
31: inert
37: 4 1
41: 3 2
43: 3 1 1
47: 3 2
53: inert
59: 3 1 1
Notice that the splitting agrees with the factorization of x5  10x + 2 over the field Z/p. With this in mind, let's rehash the cube root of 2. When p = 2 mod 3 every number has exactly one cube root. Let a be the cube root of 2. Thus x32 is divisible by xa. If the remaining quadratic splits then x32 = (xa)3. Expand, and 3a = 0, which is impossible for p > 3. Thus the uneven factorization of a linear and a quadratic, which corresponds to the lumpy split. When p = 1 mod 3 then the three distinct roots of 1 are present. x32 has no factors and is irreducible if 2 is not a cube, and if 2 is a cube then the polynomial splits. This agrees with the rules we developed earlier, and it suggests a connection between the factorization of the irreducible polynomial mod p and the splitting problem over p. If the galois group G of an integral ring is abelian, then any automorphism c of G moves the primes over P around in parallel cycles of the same length. G acts transitively on a finite set, so this is a consequence of the equal orbits principle. If one prime moves under c, they all do. If a prime moves in a cycle of length 3, all primes spin around in cycles of length 3. This need not be the case when G is nonabelian. An example was givin in the previous section, when complex conjugation swapped two primes and left the third alone.
Let R be integrally closed in its fraction field F, and let E/F be a separable finite field extension. Let S be the ring of integers of R in E. If E/F is galois the uniform degree equation applies. The number of primes over p divides the order of the galois group.
If E/F is not galois its normal closure K has a finite galois group H, and G, the automorphisms of E fixing F, is a finite subgroup of H. If c is an automorphism in G, c maps integral elements to and from integral elements, hence c maps S onto S. G becomes a group of ring automorphisms of S that fixes at least R. Since S contains a basis for E, and G is determined by its action on this basis, two different automorphisms of G cannot be the same on the basis, or on S. The same group G applies to S. Conversely, distinct automorphisms of S become distinct automorphisms of E. The group G is common to both E and S.
Let S be the integral ring of E, and let T be the integral ring of K. If H is the galois group of K, H is also the group of automorphisms of T, and H fixes precisely R. It acts transitively on the prime ideals of T lying over P. As mentioned earlier, the number of prime ideals divides H. Each prime ideal in S, lying over P, lifts to one or more prime ideals in T lying over P. Furthermore, distinct prime ideals in S always lift to distinct prime ideals in T. By acting on the primes of T, H acts transitively on the primes of S lying over P. Multiply the stabilizing subgroup of H that fixes one of these primes in S, times the orbit, and get the order of H. Therefore the number of prime ideals lying over P divides H. Remember that H is the group of K/F, not the group of E/F. This was illustrated by our lumpy split above, wherein the number of primes always divides 6. Drop back to E, and G may not act transitively on these primes any more. Similarly, the number of primes need not divide G. However, the number of primes over P is still finite.
Assume S/R is taken from a purely inseparable extension E/F. There is exactly one prime Q over P, consisting of all elements x in S with xj lying in P. Here j is a sufficiently high power of the characteristic of F. If xj lies in P it lies in Q for any prime Q over P, and by primality, x lies in Q. These values of x lie in each prime Q. Conversely, if x lies in a given prime Q then xj lies in F (because E is purely inseparable over F), x is integral over R (since x lies in S), xj is integral over R, and since R is integrally closed, xj lies in R. With xj in R and in Q, xj lies in P.
The group of automorphisms of E/F is trivial, and by correspondence, the group of automorphisms of S/R is trivial. I suppose it is fair to say the group acts transitively on the prime over P, and the number of primes (1) divides the order of the group (1).
If E/F is any finite extension, write it as a separable extension on top of an inseparable extension of F. Pass through these two extensions, and the number of primes lying over P is finite. If the separable portion is galois, G acts transitively on the primes over P, and the number of primes over P divides G.
Within a dedekind domain, every ideal can be generated by one or two generators. When the dedekind domain is the integral ring over a pid, we can be a bit more specific.
Let p be a prime element, with P the associated prime ideal in a pid R. Let S be the integral ring in the field extension E/F. Let Q be a prime ideal in S lying over P. Note that Q contains the element p, and this will be one of our two generators.
Localize about P, whence QP becomes one of finitely many prime ideals, and SP is a pid. Let w generate QP. There is no harm in clearing the denominator, so that w lies in S. This will become the second generator. If w generates p, then we don't need p after all, and Q is principal.
Since Q is the pullback of QP by prime correspondence, the elements of Q are the numerators of the fractions of QP, which are the numerators for any given denominator, such as 1. This is the contraction of the extension of the principal ideal w*S. All of Q is the contraction of the extension. For a in Q, write a/1 = x/b for some x in w*S. Q consists of w*S and any quotients thereof by elements of R not divisible by p.
Let w generate g*v, where g is in S and v is in RP. Since g*v lies in Q, and v does not, g lies in Q. If w does not generate g directly, there is more work to do.
Since R is dedekind and P is maximal, choose u in R such that u*v = 1 mod p. That is, uv = 1+px. Multiply u by vg, and then subtract pxg. The result is g, hence w and p span g, and w and p generate Q.
As an example, consider S = Z adjoin v, where v is the square root of 5. This is not a ufd, hence not a pid. I'll prove later that it is integrally closed, hence dedekind.
If 3 is an inert prime then S mod 3 produces a field of order 9, yet 2+v and 2v are zero divisors, so 3 is not inert. In fact the two conjugate primes lying over 3 are generated by 2±v and 3. Mod out by 3 first, giving a quotient ring of order 9, then mod out by 2+v, leaving a quotient ring whose order has to be 3. This is a field, hence Q, generated by 3 and 2+v, is a maximal ideal lying over 3. These are the two generators described above.
But what if we didn't know 2+v would work? Localize about 3, leaving only the two prime ideals over 3. S3 is now a pid, and something generates Q3. A generator for this ideal is 5+v. This has a norm of 30, and thus generates 30. Divide this by 10 to get 3, and subtract 3 from 5+v to get 2+v, thus all of Q.
Let S and R be integral domains, with R the fixed ring of S. Let F be the fraction field of R and let E be the fraction field of S. G is the group of ring automorphisms fixing R, and also the galois group of E/F. Let x be an element of S. All the conjugates of x lie in S, and the norm of x, relative to E/F, is the product of the conjugates of x, perhaps raised to a certain power if F(x) is not all of E.
Let W be a multiplicatively closed set in R. The fractions R/W are fixed by G, and this is precisely the fixed subring of S/W under G. R/W is part of F, and S/W is part of E. In the extreme, W could be all the nonzero elements of R. Now R/W is all of F, and since S contains a primitive element u, S/W contains u and F, and is thus all of E. Intermediate sets W will produce intermediate fraction rings inside F and E.
For any multiplicative set W, G moves x to all its conjugates in S/W, floating above F and floating above R/W. The fraction rings are still E and F, the field extension is still E/F, with the same galois group G, and the norm of x is still the product of its conjugates raised to the same power. In summary, norm and localization commute.
If H is a principal ideal in S, generated by x, let the norm of H be the norm of x, up to associates. If y also generates H then y generates x, and x generates y. x and y are associates, that is, y is some unit u times x. Take norms, and y = x*u. The norms of x and y are associates, thus the norm of H is well defined.
If H and J are principal ideals, the norm of H*J is H * J. This is clear if you step back to generators. This extends to fractional ideals. Thus norm is a group homomorphism from the principal fractional ideals of S into the associate classes of R. In some cases we can generalize this to all ideals; That will be addressed in the next section.
Let R be a pid, and let S be the integral ring of R inside a finite separable extension E/F. Thus S is dedekind. Let the element p generate the prime ideal P in R. Let Q be a prime ideal in S lying over P. Let c be the residue degree, and let d be the ramification degree. What is the norm of the ideal Q?
Recall that norms were defined for principal ideals in the previous section. This is a generalization within the context of an integral ring over a pid.
Localize about P, which produces SP, the integral ring of E/F/RP, and leaving only the primes over P. The result is a pid over a dvr. Norm has been defined for a pid; the norm of QP, (the localization of Q about P), is now the norm of any one of its generators, producing an associate class of RP. Choose, as your preferred associate, a power of p, which conveniently lives in R. Thus the norm of Q is a power of p.
If Q is already principal in S, wherein x generates Q, then x/1 generates QP. Use x/1 to find the same norm in R. Therefore the extended definition of norm is consistent with the earlier definition for principal ideals. But there's a catch. In the previous section we assumed E was galois over F. If this is not the case, then let the norm of x, or x/1, be the product of all the images of x under all the isomorphisms of E into its normal closure. That is the definition of x for any extension E/F, and now it becomes the norm of the ideal generated by x, or by x/1 in SP.
The norm of Q is a power of p, but can we be more specific? Let's look at some examples for inspiration. In the gaussian integers there are two primes lying over 5, generated by 2+i and 2i. Each of these generators, and hence each prime ideal, has a norm of 2+i times 2i or 5. To find the residue degree, consider the complex plane mod 2+i. 1 times 2+i and i times 2+i span a square of area 5. The quotient space has area 5 and contains 5 lattice points. The residue degree is 5, just like the norm. This holds for each prime p = 1 mod 4, with two primes lying over it. In contrast, let p be an inert prime such as 7. The conjugate of 7 is 7, thus the norm is 7×7 or 49. At the same time, 1 times 7 and i times 7 span a square of area 49, giving a quotient field of size 49 and a residue degree of 49. This suggests that norm equals residue degree.
To find the norm, assume the two rings have already been localized about P. Thus R is a dvr and S is a pid. Q is the prime of interest. Let t generate Q. Since R is integrally closed and t is integral over R, the irreducible polynomial of t over F is also the monic polynomial of t over R. (this was shown in an earlier section.) The conjugates of t that are present in E are integral over R and are present in S.
Assume t lies in R. It generates a prime in R containing P, which can only be P, so assume t = p. Q = P*S, and the factorization of P*S is simply Q. By the degree equation, the residue degree of Q is n, where n is the dimension of E/F. At the same time, each isomorphic image of p is p, and there are n automorphisms of E into its normal closure. The norm of t, and of Q, is pn, which is p to the residue degree.
Next assume E/F is galois. An automorphism moves Q to any other prime Q2 over P, and this automorphism carries t to t2 in Q2. t2 generates a maximal ideal, just as t did, hence t2 generates Q2. Each conjugate of t generates the prime that contains it. The conjugates of t are distributed evenly across the primes lying over P.
Let k be the number of conjugates of t, which is also the dimension of F(t) over F. Let l be the dimension of E over F(t). Thus k×l = n. Let m be the number of primes lying over P. The norm of t is the product of all the conjugates, raised to the l.
Localize S about Q, giving SQ. SQ is dedekind, and integrally closed. It is however not integral over R any more. Any ring beyond S, and still within E, will not be integral over R, for S was the integral ring of E/F/R.
We already showed that the residue degree and the ramification degree are the same after localization. QQ still has residue degree c, and QQ raised to the d gives p*SQ. At this point QQ is the only prime over P. The ring extension is a dvr over a dvr. Write QQd = P*SQ, whence td generates the ideal P*SQ. SQ is a dvr, and admits a natural valuation. Set the valuation of p to 1. Since td generates P, the valuation of t is 1/d. There are k/m such generators in Q, and their product has valuation k/(md). The other generators lie outside of Q and are units in SQ, thus their valuation is 0. They do not contribute to the valuation of the norm. Therefore the valuation of t is (kl)/(Md). Since kl = n, the valuation is n/(md), or c, which is the residue degree.
Next assume E is not galois and let E′ be the normal closure of E. Let S′ be the integral ring of E′/F/R. Let V be any prime in S′ lying over Q. Let V have residue degree c2 over Q, and let Q have residue degree c1 over P. Let V have residue degree c3 over P. As shown earlier, c3 = c1×c2. Let u generate V in S′, which is also a pid. Since E′ is galois over E, the norm of u, from S′ down to S, is t to the c2. The norm of u down to F is p to the c3. Remember that the norm of the norm is the norm. The norm of t must be p to the c1, which is the residue degree of Q over P. That completes the proof.
We can say that norm and product commute relative to ideals by definition, and that's ok, but we would like this to be consistent with the traditional definition of norm. Assume x generates a principal ideal that is a product of prime ideals that may or may not be principal. Now we have the norm of x (in the traditional sense), and the norm of the ideal {x} generated by x. The key observation is that everything works locally. If the norm of the ideal {x} did not agree with the norm of x for some prime p, localize about p, with x/1 acting as generator for the product ideal, and in this ring, using elements as generators, norm and product do indeed commute. We do not find inequality at this prime, or any prime; therefore the definitions are consistent.
This definition extends to fractional ideals in a natural way. The norm of Q1 is Pc, and so on.
There are even more general ways to define the norm of an ideal, but I'm not going to get into them here.
The decomposition group D is a particular subgroup of the group G of ring automorphisms of S/R. Unfortunately the terminology is somewhat misleading. This has nothing to do with the decomposition of G into a direct product of summands, or the decomposition of G into a chain of normal subgroups and simple factor groups. Well  it's called the decomposition group anyways, so on we go.
Let S be an integral domain with finite automorphism group G and fixed ring R. Thus S is integral over R, and G is the galois group of the corresponding field extension. Given Q lying over P, let D be the stabilizing subgroup of G that maps Q onto itself.
Recall that S/Q is an integral extension of R/P. I will call these rings S′ and R′ respectively. Both are integral domains. If P is maximal then Q is maximal and both are fields.
Each automorphism c in D fixes R, and R′, and becomes an automorphism on S′. Thus D is a group of ring automorphisms of S′ fixing at least R′.
Let KQ be the fraction field of S′, and let KP be the fraction field of R′. (The notation looks like localization but it is not.) Thus KQ/KP is a field extension.
Q  S′  KQ 
P  R′  KP 
Assume R is integrally closed in its fraction field, and S is also integrally closed, or a basis for the corresponding field extension can be found with in S. As shown above, S is now trapped between two free R modules of rank n, where n is the order of G. Assume R is noetherian. S is now the submodule of a noetherian module and is noetherian, with generators g1 through gl. In many cases S is free itself, as when R is a pid, and in that case l = n. Mod out by Q, and S′ is an R module using the same generators. P has no effect, so S′ is an R′ module using the same generators. Adjoint these generators one at a time to KP. Each is integral over R, hence algebraic over KP. The result is a field extension containing S′. It contains KQ, and is contained in KQ, hence it is KQ. Therefore KQ/KP is a finite field extension.
Assume S′ is separable over R′. This means every x in S′ is separable over KP. Consider x/y in KQ. This is included in KP adjoin x adjoin y, which is a separable extension. Therefore KQ/KP is separable.
By the primitive element theorem, let u′ generate KQ over KP. Assume P and Q are maximal, so that S′ = KQ, and u′ belongs to S′. Represent u′ by an element u in S. Let f(x) be the monic irreducible polynomial of u. As shown earlier, the coefficients lie in R; f is the polynomial that proves u is integral over R. Recall that f splits in S, with G acting transitively on the roots, i.e. the conjugates of u.
Reduce f(x) mod Q and get f′(u′) = 0, with f′ in KP[x] and u′ in KQ. f′ splits in KQ/KP, just as f splits in S, and whatever the irreducible polynomial of u′ is, it divides f′, and it splits too. Therefore KQ is the splitting field for a separable polynomial, and KQ is galois over KP.
Assume the primes Qi over P are pairwise coprime in S. That means any two prime ideals combine to span S. This happens when P and each Qi is maximal.
If x+y = 1, with x in Qi and y in Qj, raise this to a high power, and expand via the binomial theorem. Each term lies in a high power of Qi or a high power of Qj. Thus the prime ideals raised to various powers are still pairwise coprime.
Assume S is dedekind, hence some product of the prime ideals over P, to various powers, yields P*S. These prime powers are coprime, as shown above. Apply the chinese remainder theorem, and write S mod P*S as a direct product of quotient rings S mod (Qi to the ei).
Let's get more specific about the value of u. For our particular prime Q, let u = u′ mod Q. For every other Qi, let u map to 0 mod Qi.
Let c be an automorphism on S. Before c is applied, u is in every prime Qi, except for Q. After c is applied, u is in every prime Qi, except for c(Q). The image of u lies outside the image of Q.
Consider an arbitrary automorphism on KQ. It moves u′ to one of its conjugates, say v′. This is the image of v, a conjugate of u, another root of f(x). Since G acts transitively on the roots of f, some automorphism c in G moves u to v.
Since v′ is nonzero in KQ, v lies outside of Q. This means c maps Q onto Q. Therefore c belongs to D.
The automorphism on KQ is completely determined by the image of u′, which is v′. Thus our automorphism on KQ, which was arbitrary, comes from an automorphism c in D. The decomposition group D maps onto the galois group of KQ/KP.
There is no need to specify Q in all of the above. An automorphism on S induces an isomorphism between S/Q1 and S/Q2. Hence KQ/KP does not depend on Q. Similarly, the decomposition groups D(Q1) and D(Q2) are isomorphic, courtesy of an inner automorphism on G, i.e. conjugation by c, where c maps Q1 onto Q2. Such a c exists, because G acts transitively on the prime ideals lying over P.
We saw above that KQ/KP does not depend on Q (up to isomorphism). Thus the residue degree, i.e. the dimension of KQ over KP, is constant across all primes lying over P. Ramification degree is constant as well; this is the uniform degree equation that we saw earlier.
Map D onto the galois group of KQ/KP. Let T be the fixed ring of D, and verify that S is integral over T, is integral over R. Let Q lie over J in T, which lies over P. Now D acts transitively on the primes over J, and D fixes Q, so only Q lies over J.
The residue field KJ is intermediate between KP and KQ. The larger extension is galois, hence KQ is galois over KJ.
Remember that D induces every automorphism of KQ over KP. There is nothing above KP that can remain fixed by D. Therefore KJ = KP.
A group homomorphism takes D onto the galois group of KQ/KP, which is the same as the galois group of KQ/KJ. The size of this galois group is the residue degree. Thus D is the residue degree times the size of its kernel.
How large is D? When looking at the primes over P, G acts transitively, and D is the stabilizing subgroup. Therefore n = D times the number of primes over P.
Let I0 be the kernel of the homomorphism on D. This is the "inertial group", the automorphisms that fix all the cosets of Q in S. Assume R and S are dedekind, so that the uniform degree equation holds. Put this all together and the size of I0 equals the ramification degree.
If there are n primes lying over P, then I0 is trivial. The residue degree is also 1, hence D is trivial. Only the identity map fixes a prime Q over P.
Let the decomposition group D map onto the galois group of KQ/KP as described in the previous section. In many cases KP will be finite, e.g. when R = Z. Thus KQ is a finite field, and the galois group is cyclic, generated by a particular automorphism on KQ, perhaps raising everything to the l, where l is the order of KP.
Let P be unramified, so that the inertial group is trivial. (This is the typical case.) Thus D is isomorphic to the galois group of KQ. D is a cyclic group. Our generating automorphism has a unique preimage α in D.
Let c be an automorphism in G that maps Q onto Q2. Conjugation by c produces an inner isomorphism between D(Q) and D(Q2). Let β be the counterpart of α. The isomorphism tells us β generates D(Q2), just as α generates D(Q). Furthermore, β acts on its isomorphic version of KQ just as α acts on KQ. In particular, consider x a cosrep of Q2 in S, apply c inverse, then α, which raises c inverse of x, as a coset of Q, to the l, then apply c, whereupon x, as a cosrep of Q2, is raised to the l.
Select any prime Q over P, build KQ over KP, find a generator for the cyclic galois group of KQ, e.g. raising to the l, then pull back to α in DQ, and in G. The frobenius conjugacy class is the set of conjugates of α in G. Since G acts transitively, there is one conjugate for each prime over P. These are the automorphisms of S that lead to a designated generator of the galois group of each KQ/KP.