Integral Rings, The Splitting Problem

The Splitting Problem

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 an R module H in F 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.

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 expanded; 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. A subgroup of the fractional ideals of R maps into a subgroup of the fractional ideals of S. This induces a well defined homomorphism from the classgroup of R into the classgroup of S.

Localization

Localize about P, and find R_P a subring of S_P. Of course I'm abusing the notation a bit, since P is not a prime ideal in S. Understand that S_P means the fraction ring of S by R-P.

Both S_P and R_P are dedekind, and S_P is integral over R_P. 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 Q_P in S_P lying over the prime P_P in R_P. There are finitely many of these, hence S_P is a pid.

Let H be the extension of P in S. Apply the denominators from R-P, and produce the same ideal as the localization of P, extended into S_P. In other words, 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 S_P.

The splitting problem does not change for P as we move to S_P/R_P, except we have the convenience of working with a pid upstairs and a dvr downstairs.