# More on Integrality

Lemma 1.

Let $A\subseteq B$ be an integral extension. If $\mathfrak b \subseteq B$ is an ideal and $\mathfrak a = \mathfrak b \cap A$, the resulting injection $A/\mathfrak a \hookrightarrow B/\mathfrak b$ is an integral extension.

Proof

Any element of $B/\mathfrak b$ can be written as $x + \mathfrak b$, $x\in B$. Then x satisfies a monic polynomial relation:

$x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0, \quad a_0, \ldots, a_{n-1} \in A$.

Taking the relation modulo $\mathfrak b$ gives a monic polynomial relation for $x + \mathfrak b$ with coefficients in $A/\mathfrak a$. ♦

Lemma 2.

Let $A\subseteq B$ be an integral extension and suppose C is the integral closure of A in B. If $S \subseteq A$ is a multiplicative subset, then $S^{-1}C$ is the integral closure of $S^{-1}A$ in $S^{-1}B$.

Proof

Suppose $\frac x s \in S^{-1}C$ where $x\in C$ and $s\in S$. Since x is integral over A we have $x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0$ for some $a_i \in A$. Then in $S^{-1}C$ we have the equality

$(\frac x s)^n + \frac{a_{n-1}}s (\frac x s)^{n-1} + \ldots + \frac{a_0}{s^n} = 0, \quad \frac {a_i}{s^{n-i}} \in S^{-1}A$

so $\frac x s$ is integral over $S^{-1}A$.

Conversely, suppose $\frac x s\in S^{-1}B$ is integral over $S^{-1}A$ where $x\in B$ and $s\in S$. Then $\frac x 1\in S^{-1}B$ is also integral over $S^{-1}A$ so

$(\frac x 1)^n + \frac {a_{n-1}}{s_{n-1}} (\frac x 1)^{n-1} + \ldots + \frac{a_1}{s_1} (\frac x 1) + \frac{a_0}{s_0} = 0$ for some $\frac{a_i}{s_i} \in S^{-1}A.$

Multiplying by $s_0 \ldots s_{n-1}$, there exist $t\in S$ and $a_0', \ldots, a_{n-1}' \in A$ such that

$t(x^n + a_{n-1}' x^{n-1} + \ldots + a_1' x + a_0') = 0.$

Multiplying by $t^{n-1}$, we see that $tx \in B$ is integral over A so $tx\in C$ and $\frac x s \in S^{-1}C$. ♦

Corollary 1.

• If $A\subseteq B$ is an integral extension, so is $S^{-1}A \subseteq S^{-1}B$.
• If $A$ is a normal domain, so is $S^{-1}A$.

Proof

For the second statement note that $\mathrm{Frac} A = \mathrm{Frac} S^{-1}A$. ♦

Exercise A

Prove that an integral domain A is normal if and only if $A_{\mathfrak m}$ is normal for each maximal ideal $\mathfrak m\subset A$. Thus normality is a local property.

# Spectra of Integral Extensions

The inclusion map $A\hookrightarrow B$ induces $\mathrm{Spec} B \to \mathrm{Spec} A$. It turns out geometrically, such a map is like a finite-to-one map.

For example, let $A = \mathbb C[X]$ and $B = \mathbb C[X, Y]/(Y^2 - X^3 + X)$. The inclusion $A\subset B$ corresponds to projection of the curve $Y^2 = X^3 - X$ onto the X-axis. The map is generically two-to-one, except at the points X = -1, 0, +1 on the Y-axis. This corresponds to the following morphism $W\to V$.

[ Image edited from GeoGebra plot. ]

Lemma 3.

If $A\subseteq B$ is an integral extension of domains, then A is a field if and only if B is a field.

Proof

(⇒) Let A be a field and $b\in B - \{0\}$. We can find $a_0, \ldots, a_{n-1} \in A$ such that $b^n + a_{n-1} b^{n-1} + \ldots + a_0 = 0$. If we assume n is minimal, then $a_0 \ne 0$ since B is a domain which gives

$b(b^{n-1} + a_{n-1} b^{n-2} + \ldots + a_1)a_0^{-1} + 1 = 0$

so b is a unit in B.

(⇐) Let B be a field and $a\in A - \{0\}$. Then $b = a^{-1}$ exists in B. This is integral over A so we have $b^n + a_{n-1} b^{n-1} + \ldots + a_0 = 0$ for some $a_i \in A$. Multiplying throughout by $a^n$ gives us

$1 + a_{n-1}a + \ldots + a_1 a^{n-1} + a_0 a^n = 0 \implies a(\overbrace{a_{n-1} + \ldots + a_1 a^{n-2} + a_0 a^{n-1}}^{\in A}) = -1$

so a is a unit in A. ♦

Exercise B

Find a counter-example when B is not an integral domain.

Corollary 2.

Let $A\subseteq B$ be an integral extension of rings, $\mathfrak q \in \mathrm{Spec} B$ and $\mathfrak p = \mathfrak q \cap A$. Then $\mathfrak p$ is maximal if and only if $\mathfrak q$ is maximal.

Proof

We get an inclusion of domains $A/\mathfrak p \hookrightarrow B/\mathfrak q$. By lemma 3, $A/\mathfrak p$ is a field if and only if $B/\mathfrak q$ is a field. ♦

Definition.

Let $f:A \to B$ be a homomorphism of any rings, which inducees

$f^* : \mathrm{Spec} B \to \mathrm{Spec} A, \quad \mathfrak q \mapsto \mathfrak p = f^{-1}(\mathfrak q)$.

We say that $\mathfrak q$ pulls back to $\mathfrak p$.

# Consequences

Proposition 1.

If $A\subseteq B$ is integral, then $\mathrm{Spec}B \to \mathrm{Spec} A$ is surjective.

Proof

Let $\mathfrak p \subset A$ be prime and $S = A-\mathfrak p$; we get the following diagram

where the rows are injective. Since $S^{-1}B$ is non-trivial it has a maximal ideal $\mathfrak n$, which pulls back to a prime ideal $\mathfrak q\subset B$. By corollary 2, $\mathfrak n$ pulls back to a maximal ideal of $A_{\mathfrak p}$ which must be $\mathfrak p A_{\mathfrak p}$; this pulls back to $\mathfrak p \subset A$. Hence $\mathfrak q\subset B$ pulls back to $\mathfrak p$. ♦

Proposition 2.

If $A\subseteq B$ is integral, then $\mathrm{Spec B}\to \mathrm{Spec} A$ is a closed map, i.e. it takes closed subsets to closed subsets.

Proof

Let $V(\mathfrak b) \subseteq \mathrm{Spec} B$ be a closed subset, for an ideal $\mathfrak b\subseteq B$. Let $\mathfrak a = \mathfrak b \cap A$ so we get an injection $A/\mathfrak a \hookrightarrow B/\mathfrak b$. By proposition 1, we get a surjective map

$V(\mathfrak b) \cong \mathrm{Spec} (B/\mathfrak b) \longrightarrow \mathrm{Spec} (A/\mathfrak a) \cong V(\mathfrak a)$

so $V(\mathfrak b)$ maps surjectively onto the closed subset $V(\mathfrak a)\subseteq \mathrm{Spec} A$. ♦

Note

It follows that if $f:A\to B$ is a ring homomorphism such that $f(A)\subseteq B$ is integral, then $f^* : \mathrm{Spec} B \to \mathrm{Spec} A$ is a closed map, since $f(A) \cong A/\mathrm{ker} f$.

# Going Up Theorem

Proposition 3 (Going Up).

Let $A\subset B$ be an integral extension. Suppose $\mathfrak p_0 \subsetneq \mathfrak p_1$ are prime ideals of A and $\mathfrak q_0$ is a prime ideal of B which pulls back to $\mathfrak p_0$.

Then there exists a prime ideal $\mathfrak q_1$ of B containing $\mathfrak q_0$ which pulls back to $\mathfrak p_1$.

Note that we must have $\mathfrak q_0 \subsetneq \mathfrak q_1$.

Proof

Since $\mathfrak q_0 \cap A = \mathfrak p_0$ we have an integral extension of domains $A/\mathfrak p_0 \hookrightarrow B/\mathfrak q_0$. By surjectivity of $\mathrm{Spec} B/\mathfrak q_0 \to \mathrm{Spec} A/\mathfrak p_0$, there is a prime ideal $\mathfrak q_1 /\mathfrak q_0$ of $B/\mathfrak q_0$ which pulls back to $\mathfrak p_1/\mathfrak p_0$, where $\mathfrak q_1$ is a prime ideal of B containing $\mathfrak q_0$. Then $\mathfrak q_1 \in \mathrm{Spec} B$ pulls back to $\mathfrak p_1 \in \mathrm{Spec} A$. ♦

Corollary 3.

Suppose we have a chain of prime ideals $\mathfrak p_0 \subsetneq \ldots \subsetneq \mathfrak p_n$ of A, and a prime ideal $\mathfrak q_0$ of B which pulls back to $\mathfrak p_0$.

Then there is a chain of prime ideals $\mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \ldots \subsetneq \mathfrak q_n$ of B such that each $\mathfrak q_i$ lies over $\mathfrak p_i$.

Proof

Repeatedly apply proposition 3 to extend the bottom chain. ♦

The final piece of the puzzle is the following.

Proposition 4.

If $\mathfrak q_0 \subsetneq \mathfrak q_1$ are prime ideals of B and $\mathfrak p_i = \mathfrak q_i \cap A$ for $i=0,1$, then $\mathfrak p_0 \ne \mathfrak p_1$.

Proof

Suppose $\mathfrak p = \mathfrak p_0 = \mathfrak p_1$; let $S = A-\mathfrak p$. Again we have

Since $\mathfrak q_0 \cap S = \mathfrak q_1 \cap S = \emptyset$, $\mathfrak q_0, \mathfrak q_1$ give us prime ideals $\mathfrak q_0 (S^{-1}B) \subsetneq \mathfrak q_1(S^{-1}B)$ of $S^{-1}B$. They both pull back to $\mathfrak p\in \mathrm{Spec} A$ and hence to $\mathfrak p A_\mathfrak p \in \mathrm{Spec} A_{\mathfrak p}$, the unique maximal ideal. By corollary 2, this means $\mathfrak q_0 (S^{-1}B)$ and $\mathfrak q_1(S^{-1}B)$ are both maximal, hence equal. So $\mathfrak q_0 = \mathfrak q_1$. ♦

As a result we have:

Main Theorem.

Let $A\subseteq B$ be an integral extension of rings. Then $\dim A = \dim B$.

Proof

By Going Up and surjectivity of $\mathrm{Spec}B \to \mathrm{Spec}A$ (proposition 1), any prime chain in A lifts to a prime chain in B. Conversely, any prime chain in B maps to a prime chain in A of the same length by proposition 4. ♦

Example

Since $\mathbb C[X] \subset \mathbb C[X, Y]/(Y^2 - X^3 + X)$ is a finite extension we have

$\dim \mathbb C[X, Y]/(Y^2 - X^3 + X) = \dim \mathbb C[X] = 1$,

since $\mathbb C[X]$ is a PID.

Exercise C

1. Take $A\subset B$ where $A = \mathbb Z[X]$ and $B = \mathbb Z[2i][X, Y]/(Y^2 - X^3 - 1)$. Lift the following prime chains of A to prime chains of B

$0\subset (5) \subset (5, X-2), \quad 0 \subset (X-2) \subset (5, X-2).$

2. Prove that if $A\subseteq B$ is a finite extension with induced $\phi : \mathrm{Spec} B \to \mathrm{Spec } A$, for any $\mathfrak p \in \mathrm{Spec } A$, $\phi^{-1}(\mathfrak p)$ is a finite subset of Spec B. We say that $\phi$ has finite fibres.

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### 4 Responses to Commutative Algebra 40

1. Vanya says:

In the last example at the end “$\mathbb C[X] \subset \mathbb C[X]/(Y^2 - X^3 + X)$ is a finite extension”. shouldn’t it be $\mathbb C[X] \subset \mathbb C[X,Y]...$?

• limsup says:

Oops. Thanks! 😀

2. Vanya says:

The last statement “We say that f^* has finite fibres” should be ” …\phi … \$

• limsup says:

Thanks! Corrected. 🙂