Integrality

Throughout this article, A is a subring of B; we will also call Bring extension of A.

Definition.

An element $b\in B$ is said to be integral over A if we can find $a_0, a_1, \ldots, a_{n-1} \in A$ (where $n>0$) such that

$b^n + a_{n-1} b^{n-1} + \ldots + a_1 b + a_0 = 0$ in B.

For example, $b = \sqrt 2 \in \mathbb C$ is integral over $\mathbb Z$ since $b^2 - 2 = 0$. One intuitively guesses that $\frac 1 {\sqrt 2}$ is not integral over $\mathbb Z$, but this is quite hard to show at this point.

Main Proposition.

The following are equivalent for $b\in B$.

1. $b$ is integral over $A$.
2. $A[b]$, the A-subalgebra of B generated by b, is finite over A.
3. $b$ is contained in a ring C, where $A\subseteq C \subseteq B$ and C is finite over A.

Proof

(1) ⇒ (2). $A[b]$ 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 $b^n$ as an A-linear combination of $1, b, \ldots, b^{n-1}$. By induction, this holds for all $b^n, b^{n+1}, \ldots$. Hence $A[b]$ is generated by $\{1, b, \ldots, b^{n-1}\}$ as an A-module.

(2) ⇒ (3). Just take $C = A[b]$.

(3) ⇒ (1). Let $x_1 = 1, x_2, \ldots, x_n$ generate C as an A-module; we can write each $bx_i$ as an A-linear combination of $x_1, \ldots, x_n$. In matrix form,

$b\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$

where $a_{ij} \in A$. Write M for the RHS matrix; denoting $I$ for the $n\times n$ identity matrix, we get $(b\cdot I - M)\cdot \mathbf x = 0$ where $\mathbf x$ is the column vector $(x_1, \ldots, x_n)^t$. 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 $MN = NM = \det(M)I$. 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 $b\cdot I - M$, we obtain $\det(b\cdot I - M)I\cdot \mathbf x = 0$. In particular $0 = \det(b\cdot I - M)x_1 = \det(b\cdot I - M)$. Expanding this determinant gives us a monic polynomial relation in b with coefficients in A. ♦

Easy Exercise

Prove that if $A\subseteq B \subseteq C$ are ring extensions such that B is finite over AC is finite over B, then C is finite over A.

Consequences

The above gives us:

Proposition 1.

If $b_1, b_2 \in B$ are integral over A, so are $b_1 + b_2, b_1 b_2$.

Proof

Consider the extensions

$A \subseteq A[b_1] \subseteq A[b_1, b_2].$

Since $b_2$ is integral over A, it is also integral over $A[b_1]$, so both extensions are finite by the main proposition. Thus by the above exercise $A[b_1, b_2]$ is also a finite extension of A.

Since $b_1 + b_2, b_1 b_2 \in A[b_1, b_2]$, by the main proposition they are integral over A. ♦

Example

We see that $\sqrt 2 + \sqrt[3] 3$ and $\sqrt 2 + \sqrt[3] 3 + \sqrt[5] 5$ are integral over $\mathbb Z$ since each summand is clearly integral over $\mathbb Z$.

Definition.

By proposition 1, the set of $b\in B$ 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 $\mathrm{int. cl.}_A B$.

We say B is integral over A if $\mathrm{int. cl.}_A B = B$.

We say A is integrally closed in B if $\mathrm{int. cl.}_A B = A$.

Next, we see that integrality is transitive.

Proposition 2.

Let $A\subseteq B\subseteq C$ be ring extensions, where $A\subseteq B$ is integral.

• If $c\in C$ 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.

Proof

Since c is integral over B there are $b_0, b_1, \ldots, b_{n-1} \in B$ such that

$c^n + b_{n-1} c^{n-1} + \ldots + b_1 c + b_0 = 0.$

By the main proposition, we have a sequence of ring extensions, each finite over the previous.

$A \subseteq A[b_0] \subseteq A[b_0, b_1] \subseteq \ldots \subseteq A[b_0, \ldots, b_{n-1}] \subseteq A[b_0, \ldots, b_{n-1}, c]$

so $A[b_0, \ldots, b_{n-1}, c]$ is finite over A; again by the main proposition, c is integral over A. ♦

Recall that for a field extension $k\subseteq K$, 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.

Proposition 3.

If B is an A-algebra of finite type, and B is integral over A, then B is finite over A.

Proof

Write $B = A[b_0, \ldots, b_{n-1}]$. In the extensions $A \subseteq A[b_0] \subseteq \ldots \subseteq A[b_0, \ldots, b_{n-1}] = B$, each ring is finite over the previous. Hence B is finite over A. ♦

Normal Domains

In this section, for an integral domain A, $K := \mathrm{Frac} A$ denotes its field of fractions.

Definition.

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.

Exercise A

Prove that the normalization of A is a normal domain.

Note

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.

Example

Let $A = \mathbb Z[2i] = \{a + 2b\cdot i : a, b\in \mathbb Z\}$ where $i = \sqrt{-1}$. Then $K = \mathbb Q[i]$. Now A is not a normal domain because $i \in K$ is integral over A but does not lie in A.

One reason for defining normal domains lies in the following.

Proposition 4.

Let $A\subseteq B$ be integral domains, where $A$ is normal. For an integral $b\in B$, consider it as an element of $\mathrm{Frac} B$ and take its minimal polynomial

$q(X) = X^n + c_{n-1} X^{n-1} + \ldots + c_0 \in K[X]$.

Then $q(X) \in A[X]$.

Proof

Find a monic polynomial $p(X) \in A[X]$ such that $p(b) = 0$. Note that $q(X)$ divides $p(X)$ in $K[X]$. Now pick a field extension $L\supseteq \mathrm{Frac} B$ in which $q(X)$ factors as a product of linear factors

$q(X) = (X - \alpha_1) \ldots (X - \alpha_n), \ \alpha_i \in L$.

Since $p(\alpha_i) = 0$ each $\alpha_i$ is integral over A. Thus the coefficients of $q(X)$ are integral over A, being elementary symmetric polynomials in the $\alpha_i$. Hence the coefficients of $q(X)$ lie in $\mathrm{int. cl.}_A K = A$. ♦

However, to use the above result, we need a criterion for normal domains.

Proposition 5.

A UFD is a normal domain.

Proof

Let A be a UFD and suppose $\frac a b \in K$ is integral over A where a and b have no common prime factor. Pick $a_0, \ldots, a_{n-1} \in A$ such that

$(\frac a b)^n + a_{n-1} (\frac a b)^{n-1} + \ldots + a_0 = 0\implies a^n + a_{n-1}(a^{n-1}b) + \ldots + a_0(b^n) = 0.$

Then $b|a^n$. Since a and b have no common prime factor, b is a unit so $\frac a b \in A$. ♦

Examples

1. Let $b = \frac 1 {\sqrt 2} \in \mathbb C$, which has minimal polynomial $X^2 - \frac 1 2$ over $\mathbb Q$. Since $\mathbb Z$ is a UFD, it is a normal domain. Thus b is not integral over $\mathbb Z$ which solves our problem at the beginning of the article.

2. Let $A = \mathbb Z[2i]$ and $\alpha = \sqrt i$. The minimal polynomial of $\alpha$ over $\mathrm{Frac} A$ is $X^2 - i = 0$, which does not lie in $A[X]$. On the other hand $\alpha$ is integral over A because $\alpha^4 + 1 = 0$. This happens because A is not a normal domain.

Exercise B

1. Decide if each of the following is a normal domain. Find its normalization.

\begin{aligned}\mathbb Z[X], \quad \mathbb Z[\sqrt 2],\quad, \mathbb Z[\sqrt 5], \quad \mathbb C[X, Y, Z],\quad \mathbb C[X, Y]/(Y^2 - X^3 + X),\\ \mathbb C[X, Y, Z]/(Y^2 - X^3 + X),\quad \mathbb C[X, Y]/(Y^2 - X^3),\quad \mathbb C[X, Y, Z]/(X^2 + Y^2 + Z^2).\end{aligned}

[ Hint: conjecture and prove a result about $A[X]/(X^2 - f)$, where $f\in A$ and A is a UFD in which 2 is invertible. ]

2. Consider $A\subset B$ where $A = \mathbb Z[2i][X]$ and $B = \mathbb Z[2i][X, Y]/(3Y^2 - 2X^3 - 1)$. Is this an integral extension?

3. Let $f\in \mathbb Z[X, Y]$. Prove that there is a $g \in \mathbb Z[X, Y] - \{0\}$ such that $fg \in \mathbb Z[X^{123}, Y^{789}]$.

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4 Responses to Commutative Algebra 39

1. Vanya says:

Please, can you give hints to (a) and (b) of Exercise B?

2. Vanya says:

In part (1) I obtained that all except $C[X,Y]/(Y^2 - X^3)$ is normal. The latter one is not normal because $1/X$ is integral over $C[X]$. 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?

• limsup says:

You got it right for (1), but I think you meant $Y/X$ is integral over $\mathbb C[X]$.

• limsup says:

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 $A$ and $B$ so that $A$ becomes a normal domain.