Integral Extensions, Quotient Rings

Quotient Rings

Let S be an integral extension of R. Let H be an ideal in S with G = H∩R. Verify that G is an ideal in R. We will show S/H is integral over R/G.

Notice that / is heavily overloaded. Sometimes it means a ring or field extension, sometimes it means a quotient ring, and sometimes it means a fraction ring. Sorry for the confusion. I guess you just have to know what / means by context.

Now where were we? The rings S and R contain the ideals H and G respectively, and the quotient ring S/H is an integral extension of the quotient ring R/G. Verify the following steps.

  1. If x and y are in the same coset of G in R, they are in the same coset of H in S.

  2. If x and y are in different cosets of G in R, they are in different cosets of H in S. Turn this around to prove it; assume x-y = z, where z is in H. Yet z must also be an element of R, so z lies in G, and x and y are in the same coset of G in R.

  3. Mapping x to x carries cosets of G into cosets of H, and the map is an embedding.

  4. One maps to one, and R/G is a subring of S/H.

  5. Select any u in S and let p be the monic polynomial that proves u is itegral. In other words, p(u) = 0. Apply the ring homomorphism S/H to u and to the coefficients of p, and the image of u in S/H is integral over R/G.

Fraction Rings

Let S/R be an integral extension as above, and let T be a multiplicatively closed set in R. Show that the fraction ring R/T embeds in the ring S/T. There are no new denominators, so distinct classes in R/T are not going to merge in S/T.

Since both rings contain T/T, or 1, S/T is a ring extension of R/T. We will show S/T is an integral extension of R/T.

If the numerator of a fraction is x, write p(x) = 0, where p is monic, with coefficients from R. A ring homomorphism carries S into S/T. Here x is mapped to x/1, or xw/w if T does not contain 1. In any case, we can apply the homomorphism to x, and to the coefficients of p(x). Thus x/1 ∈ S/T is integral over R/T. Multiply by 1/j ∈ R/T, and any x/j is integral over R/T. Thus S/T is an integral extension of R/T.

Field over Field

If S/R is an integral extension, and S is an integral domain, then S is a field iff R is a field.

Assume R is a field. Given x in S-R, write p(x) = 0, where p is monic. Suppose p is the product of two smaller polynomials. When evaluated at x, the result is 0, and since S is an integral domain, x is a root of one of the two smaller polynomials. Therefore we can assume p is irreducible. This represents a field extension of R, hence x is invertible.

Conversely assume S is a field. Let x be a nonzero element of R and let y be its inverse in S. Write p(y) = 0 where p is monic. Multiply through by xn-1, and y lies in R after all.

Prime over Prime

Let S/R be an integral extension, let H be a proper ideal of S, and let G be H∩R. We showed above that S/H is integral over R/G.

Suppose H is a prime ideal. This means xy in H implies x is in H or y is in H. This certainly holds when we restrict to R, so G is prime in R. The quotient rings S/H and R/G are integral domains, and by the above, one is a field iff the other is a field. thus, if H is maximal in S, H∩R is maximal in R.

Now assume G is prime in R. We want to lift this to a prime ideal H lying over G.

Let T be the set R-G, which is multiplicatively closed. We showed above that S/T is an integral extension of R/T.

Select a maximal ideal in S/T and let H be the numerators of this ideal. Thus H is a prime ideal in S, and H/T gives the same maximal ideal back again. See ideal correspondence for more details.

Let C = H∩R. If a fraction in R/T and another fraction in H/T represent the same class, establish a common denominator d, and the first numerator is still in R, and the second is still in H. To equate these fractions, write d(a-b) = 0. Since 0 is in H, a prime ideal, and since d is nonzero, a = b. The numerator is in both R and H, hence in C. Conversely, every fraction in C/T is in H/T and R/T. Intersection and localization commute.

Recall that C/T has to be prime, and since H/T is maximal, C/T is also maximal. However, G/T is the largest ideal in R/T. This because RG is a local ring. So G/T is the maximal ideal, and G/T = C/T. Thuse C = G, and the prime ideal H intersects R in G. We have found a prime ideal H lying over G.

Using our earlier result, H is maximal iff G is maximal. There are primes over primes, and maximals over maximals.

Intersection with R implements a map from prime ideals in S onto prime ideals in R. For a fixed element x in R, a prime G in R misses x iff its preimage H in S misses x. Base open sets pull back to base open sets, and contraction to R implements a continuous function from spec S onto spec R.