Integral Rings, Ramification / Residue Degree

Ramification / Residue Degree

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 in the previous section. 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 occurred. 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 Q_i be one of the prime ideals lying over P. Mod out by Q_i upstairs and P downstairs, and the embedding of R in S becomes an embedding of the field F = R/P into the field S/Q_i. Let d_i be the dimension of this field extension. This is called the residual degree, or the residue degree, of Q_i.

Note that the residue degree can be computed before or after localization, since the two quotient rings are the same.

Let P*S be the product of Q_i raised to the e_i. Thus e_i is the exponent, yet to be determined. By definition, e_i is the ramification degree of Q_i.

Ramification / Residue Equation

Review the formula for tensoring with a quotient ring. Thus S/(P*S) is the same as S tensor F. This is an F vector space.

Since S is a free R module of rank n, S×F becomes an n dimensional F vector space. Thus S/(P*S) = Fⁿ.

At the same time, S mod Q_i is an F vector space of dimension d_i. What is the connection between d_i and n?

Tensor the ideal Q_i^j with S/Q_i and get Q_i^j mod Q_i^j+1. Since S is a pid, Q_i^j is principal, and looks just like a free S module. Therefore the quotient module looks like S/Q_i, an F vector space of dimension d_i.

Consider the quotient ring S mod Q_i^j+1. This includes the ideal Q_i^j mod Q_i^j+1. We just showed this ideal is an F vector space of dimension d_i. By induction, the quotient ring, which is really S/Q_i^j, is an F vector space of dimension j×d_i. Combine kernel and quotient and find a vector space of dimension (j+1)×d_i. Therefore, S mod Q_i to the e_i becomes an F vector space of dimension e_i×d_i.

Apply the chinese remainder theorem and write S mod P*S as the direct product of all these quotient rings. This becomes the direct product of vector spaces, which adds their dimensions. Therefore n = the sum of e_i×d_i.

In summary, the dimension of the extension is the sum, over the primes lying over P, of the ramification degreee times the residue degree.

Totally Ramified, Unramified

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 previous 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. Thus P is unramified.

A Tower of Extensions

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 Q_i, and push this up to S′. Replace each Q_i*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.

When the Extension is Galois

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. Now G respects ideal multiplication, and G acts transitively on prime ideals. 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.

If the dimension is prime, P is totally ramified or unramified.

A Lumpy Split

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 (refering to a theorem that I haven't posted on my website yet). A pid is integrally closed, so S is integrally closed. If x in E is integral over Z then x is in the fraction field of S, and x is in S. Therefore 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.

Using a computer program, 7 is inert, 31 is flat (-1+2t² times -1+2t+t² times 3+t²), and 5 is lumpy (1+t² times 1+2t-t²). The first factor of 5 has residue degree 5, and the second factor has residue degree 25. 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, and then retains the higher residue degree of 2 when we bring in the cube root of 1 to create a galois extension. 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 original extension, and in the normal closure. 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.