Integral Rings, An Introduction

Introduction

An integral extension S is already a ring that is integral over R, so what's so special about an integral ring?

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. Now 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, and is "integrally closed" as a ring.

With S/R defined as above, a subring of S that contains R is called an order.

As you might imagine, there are many theorems that apply in this situation, that do not hold for arbitrary ring extensions.

In many cases R is a ufd, pid, or dedekind domain. You may see a theorem, now and then, that does not address E/F directly, but deals with an integral extension of a pid. In practice, such theorems are applied to integral rings. Here is an example.

Integral over a UFD

Let R be a ufd and let u be algebraic over the fraction field F. (If u is integral over R it is certainly algebraic over F.) Being algebraic, u satisfies a polynomial p over F, whose degree is the dimension F(u)/F. We may as well make this polynomial monic. If the remaining coefficients lie in R then u is integral.

Conversely, assume u is integral. It satisfies a monic polynomial q(x), and if p(x) does not divide q(x), 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.

Apply gauss' lemma, and q is the product of two polynomials in R[x], and one is a scaled version of p. The product of the lead coefficients is 1, which is 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, and u is integral courtesy of p(x).

In summary, u is integral iff its minimum polynomial over F is monic and lies in R[x].

Although gauss' lemma fails when R is dedekind, this theorem can be salvaged nonetheless. In fact it applies for any integrally closed integral domain. Extend F to the splitting field of p. Each root of p is integral over R, hence each coefficient of p is integral over R. 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.