Throughout this article, A is a subring of B; we will also call B a ring extension of A.
An element is said to be integral over A if we can find (where ) such that
For example, is integral over since . One intuitively guesses that is not integral over , but this is quite hard to show at this point.
The following are equivalent for .
- is integral over .
- , the A-subalgebra of B generated by b, is finite over A.
- is contained in a ring C, where and C is finite over A.
(1) ⇒ (2). is the set of all polynomials in b with coefficients in A. Since b is integral over A, for some n > 0 we can express as an A-linear combination of . By induction, this holds for all . Hence is generated by as an A-module.
(2) ⇒ (3). Just take .
(3) ⇒ (1). Let generate C as an A-module; we can write each as an A-linear combination of . In matrix form,
where . Write M for the RHS matrix; denoting for the identity matrix, we get where is the column vector . Note that this relation holds in the ring C.
- Now for every square matrix M with entries in any ring C, we can define its adjugate matrix N, which is a matrix of the same size with entries in C such that . All entries of N can be obtained by polynomials in the entries of M with integer coefficients so this holds in any ring.
Applying this to our matrix , we obtain . In particular . Expanding this determinant gives us a monic polynomial relation in b with coefficients in A. ♦
Prove that if are ring extensions such that B is finite over A, C is finite over B, then C is finite over A.
The above gives us:
If are integral over A, so are .
Consider the extensions
Since is integral over A, it is also integral over , so both extensions are finite by the main proposition. Thus by the above exercise is also a finite extension of A.
Since , by the main proposition they are integral over A. ♦
We see that and are integral over since each summand is clearly integral over .
By proposition 1, the set of integral over A forms a ring extension of A contained in B. We call this the integral closure of A in B and denote it by .
We say B is integral over A if .
We say A is integrally closed in B if .
Next, we see that integrality is transitive.
Let be ring extensions, where is integral.
- If is integral over B, then it is integral over A.
Hence if C is integral over B and B is integral over A, then C is integral over A.
Since c is integral over B there are such that
By the main proposition, we have a sequence of ring extensions, each finite over the previous.
so is finite over A; again by the main proposition, c is integral over A. ♦
Recall that for a field extension , a finite extension is algebraic but the converse may not be true; heuristically, this is because we can attach infinitely many algebraic elements to k. This inspires the following.
If B is an A-algebra of finite type, and B is integral over A, then B is finite over A.
Write . In the extensions , each ring is finite over the previous. Hence B is finite over A. ♦
In this section, for an integral domain A, denotes its field of fractions.
We say A is a normal domain if it is integrally closed in K.
The normalization of A is the integral closure of A in K.
Prove that the normalization of A is a normal domain.
In some books, A is called an integrally closed domain, but we consider it potentially confusing here. On the other hand, the term normal is used excessively in mathematics. However, in the context of commutative algebra this should not cause any confusion.
Let where . Then . Now A is not a normal domain because is integral over A but does not lie in A.
One reason for defining normal domains lies in the following.
Let be integral domains, where is normal. For an integral , consider it as an element of and take its minimal polynomial
Find a monic polynomial such that . Note that divides in . Now pick a field extension in which factors as a product of linear factors
Since each is integral over A. Thus the coefficients of are integral over A, being elementary symmetric polynomials in the . Hence the coefficients of lie in . ♦
However, to use the above result, we need a criterion for normal domains.
A UFD is a normal domain.
Let A be a UFD and suppose is integral over A where a and b have no common prime factor. Pick such that
Then . Since a and b have no common prime factor, b is a unit so . ♦
1. Let , which has minimal polynomial over . Since is a UFD, it is a normal domain. Thus b is not integral over which solves our problem at the beginning of the article.
2. Let and . The minimal polynomial of over is , which does not lie in . On the other hand is integral over A because . This happens because A is not a normal domain.
1. Decide if each of the following is a normal domain. Find its normalization.
[ Hint: conjecture and prove a result about , where and A is a UFD in which 2 is invertible. ]
2. Consider where and . Is this an integral extension?
3. Let . Prove that there is a such that .
Please, can you give hints to (a) and (b) of Exercise B?
In part (1) I obtained that all except is normal. The latter one is not normal because is integral over . Is that right?
In part (2), B is not integral over A because Y is not integral over A: its minimal polynomial must be the integral equation of dependence. Is that right?
You got it right for (1), but I think you meant is integral over .
For (2), almost. But remember to use that you need the base ring to be a normal domain (see proposition 4). The way to overcome this is to extend and so that becomes a normal domain.