Integral Extensions, Integrally Closed is a Local Property

Local Property

The term locally is an adverb that describes a property of a ring, rather than the ring itself. For instance, imagine a ring can be green. We say that the ring R is locally green if every localization R_P is green. Furthermore, green is a local property if the following holds. A ring is green iff all its localizations are green. In other words, a ring is green iff it is locally green.

We will see that "integrally closed" is a local property when R is an integral domain. but first we need to show that fractions and integral closure commute.

Integral Closure and Fractions Commute

Let S be a ring extension of R and let C be the integral closure of R in S. Here R and S are commutative, but they could have zero divisors.

Let T be a multiplicatively closed set in R. If T does not contain 1, toss 1 in. This doesn't change the ring S/T, or its subrings; it is merely a convenience. we can represent x by x/1, rather than xw/w for some w in T.

In an earlier section we showed that C/T is integral over R/T. We only need show that C/T includes all the integral elements of R/T.

Let x/y be a fraction in S/T that is integral over R/T. If x/y is 0 then we are done, so assume nothing in T kills x. Write p(x/y) = 0, where p is monic, and has coefficients in R/T. Let d be a common denominator for the coefficients of p, and multiply through by (dy)ⁿ. This builds a monic polynomial, of the same degree, with root dx, and coefficients in R. Thus dx lies in C. Remember that d is in T, and so is y, hence x/y is in C/T.

We can take the integral closure, and then the fraction ring, or vice versa; the result is the same.

S Integral over R is a Local Property

If S is integral over R, then each fraction ring S/T is integral over R/T. Conversely, suppose x ∈ S is an element that is not integral over R. View S as an R module, and find a maximal ideal P in R such that x/1 remains nonzero in S_P. In other words, we are taking fractions with denominators in T, where T = R-P. By assumption, x/1 is integral over R/T, and x is not killed by anything in T. Set y = 1 and apply the previous theorem, whence x is integral over R.

To see if S is integral over R, it is sufficient to look locally, relative to the maximal ideals of R.

Integrally Closed is a Local Property

Let R and S be integral domains, with S an extension of R. This allows S and its various subrings to embed in the fractions of S.

Let T be a multiplicatively closed set in R. If R is integrally closed in S then take the integral closure, which is R, then the fractions by T, and the result is the integral closure of R/T. Thus R/T is integrally closed in S/T, and every localization R_P is integrally closed in S_P.

We often set S to the fraction field of R, whence R integrally closed implies R_P integrally closed for each prime ideal P.

Now for the converse. Let C be the integral closure of R in S. Yet every localization R_P is integrally closed in S_P.

We can jump up to C, then localize, or localize and then take the integral closure. The result is the same. Thus C_P = R_P for each prime ideal P.

Let x be an element of C. Select any prime ideal P, and x embeds in C_P via x/1. (This is where we need S to be an integral domain.) Thus x is in every C_P, and in every R_P. The intersection of the localizations R_P produces R. Thus x is in R, and R is integrally closed.

Being integrally closed with respect to S, or with respect to the fraction field of R, is a local property.

Note that it is sufficient to localize about maximal ideals, rather than all prime ideals.

Extensions of Integrally Closed Rings

Don't assume that an integral extension of an integrlly closed ring remains integrally closed. For example, start with Z, which is a ufd, and integrally closed. Adjoin q, the square root of -3. Clearly Z[q] is an integral extension of Z. Let r = (q+1)/2. Note that r is not in Z[q], yet r is in the fraction field of Z[q]. Also, r³ = -1, hence r is integral over Z[q]. Therefore Z[q] is not integrally closed.

When you study algebraic number theory, you will need to prove various integral extensions of Z are integrally closed. It is often helpful to look locally, since "integrally closed" is a local property. This seems to make things worse, because you have to prove the extension is integrally closed for each and every prime P; but in practice you only need examine a few primes, as determined by the polynomial that defines the integral extension. All this will make more sense when you explore algebraic number theory. For now, let's return to integral extensions.