Integral Extensions

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The Cayley–Hamilton Theorem

We begin by stating the generalized Cayley–Hamilton theorem.

Theorem 1 Fix a ring $A$ and an ideal $\mathfrak{a}\subseteq A$, and let $M$ be an $A$-module generated by $n$ elements $e_1,\ldots,e_n$. Then for any $u\in\End_\rMod{A}(M)$, if $u(M)\subseteq \mathfrak{a}M$, there exists a monic polynomial

\[p(\x)=\x^n+p_1\x^{n-1}+\cdots+p_n,\qquad p_k\in \mathfrak{a}^k\]

such that $p(u)=0$.

Proof

See §Determinants, ⁋Proposition 9 (Cayley–Hamilton); note that $M$ need not be free.

Then the following holds.

Proposition 2 Let $A$ be a ring, let $\mathfrak{a}\subseteq A[\x]$ be an ideal, put $B=A[\x]/\mathfrak{a}$, and write $b=\x+\mathfrak{a}\in B$. Then the following hold.

$B$ is generated by at most $n$ elements as an $A$-module if and only if $\mathfrak{a}$ contains a monic polynomial of degree at most $n$. In this case, $B$ is generated by $1,b,\ldots,b^{n-1}$.
$B$ is a free $A$-module if and only if $\mathfrak{a}$ is generated by a monic polynomial. In this case, $1,b,\ldots,b^{n-1}$ form a basis of $B$.

Proof

One direction is obvious. Conversely, suppose $B$ is generated by $n$ elements as an $A$-module. Consider the $A$-module endomorphism $b:B\rightarrow B$ given by multiplication by $b$. Applying Theorem 1 to the ideal $A$, we see that this endomorphism satisfies a monic polynomial $p(x)$ of degree $n$, and substituting the element $b$ must yield $0$. Hence, by the definition of $b$, we have $p(\x)\in \mathfrak{a}$.
First, suppose $\mathfrak{a}$ is generated by a monic polynomial of degree $n$. Then by the preceding result, $B$ is generated by $1,b,\ldots,b^{n-1}$. It remains to show that these elements are linearly independent. If $\sum_{i=0}^{n-1} a_i b^i=0$ in the $A$-module $B$, then $q(\x)=\sum_{i=0}^{n-1}a_i\x^i$ must lie in $\mathfrak{a}$, and degree considerations force $q=0$.
Conversely, suppose $B$ is a free $A$-module of rank $n$. Then $B$ can be generated by $n$ elements, so by the preceding result $\mathfrak{a}$ contains a monic polynomial $p$ of degree $n$, and we can also deduce that $1,b,\ldots,b^{n-1}$ form a basis of $B$. It remains to show that $p$ generates $\mathfrak{a}$. Given any $f\in \mathfrak{a}$, consider the remainder $r$ upon dividing $f$ by $p$. Since both $f$ and $p$ lie in $\mathfrak{a}$, this remainder must also map to $0$ in $B$. But this is the same as evaluating the polynomial $r(\x)$ at $\x=b$, and viewing the result as a linear combination of the basis $1,\ldots,b^{n-1}$ of $B$, we see that all coefficients of $r$ must be $0$.

Integral Extensions

Previously, we observed that an $A$-algebra $E$ consists of the same data as a ring homomorphism $\phi:A\rightarrow E$. More precisely, a ring homomorphism $\phi:A\rightarrow E$ endows the $E$-module $E$ with an $A$-module structure $\phi^\ast E$ via restriction of scalars; because multiplication in $E$ is commutative, this $A$-module structure preserves the multiplicative structure of $E$, so we may regard $E$ as an $A$-algebra.

Moreover, the composition of ring homomorphisms is again a ring homomorphism. For an $A$-algebra $E$, the ring homomorphism

\[A\overset{\phi}{\rightarrow}E\hookrightarrow E[\x]\]

gives $E[\x]$ an $A$-algebra structure. Hence, by §Algebras, ⁋Proposition 8, there is a unique $A$-algebra homomorphism $\phi[\x]:A[\x]\rightarrow E[\x]$, which concretely sends any polynomial

\[p(\x)=a_n\x^n+\cdots+a_0\in A[\x]\]

to the polynomial

\[(\phi[\x](p))(\x)=\phi(a_n)\x^n+\cdots+\phi(a_0)\in E[\x].\]

Definition 3 Let $\phi:A\rightarrow E$ be a ring homomorphism.

If $\phi$ is injective, we call $E$ an extension of $A$.
An element $x\in E$ is integral over $\phi$ if there exists a monic polynomial $p\in A[\x]$ such that $(\phi[\x](p))(x)=0$. If every element of $E$ is integral over $\phi$, we call $\phi$ an integral homomorphism. If an extension $\phi$ is an integral homomorphism, we call $E$ an integral extension of $A$.
The set of all elements of $E$ that are integral over $\phi$ is called the integral closure of $A$ in $E$, or the normalization of $A$ inside $E$. If $A$ is an integral domain, the integral closure of $A$ in $\Frac(A)$ is simply called the normalization of $A$. An integral domain $A$ is a normal domain if its normalization is $A$ itself.
If the $A$-module $\phi^\ast E$ is finitely generated, we call $\phi:A\rightarrow E$ a finite homomorphism.
If $\phi^\ast E$ is finitely generated as an $A$-algebra, we call $\phi:A\rightarrow E$ a finite type homomorphism.

When the structure morphism $\phi:A\rightarrow E$ is clear from context, we sometimes say that $x$ is integral over $A$ rather than over $\phi$. Similarly, in this case we say that $E$ is integral over $A$ rather than that $\phi$ is an integral homomorphism.

Every finite homomorphism is of finite type by definition. Conversely, for a finite homomorphism $\phi:A\rightarrow E$, the map $x\times-:E\rightarrow E$ is an $A$-algebra endomorphism, so applying Theorem 1 shows that $x$ is integral over $\phi$. Thus every finite homomorphism is integral. The following lemma provides the converse.

Lemma 4 A ring homomorphism $\phi$ is finite if and only if it is an integral homomorphism of finite type.

Proof

One direction was shown above. For the converse, we must show that if $E$ is generated by finitely many integral elements as an $A$-algebra, then $E$ is finitely generated as an $A$-module. We proceed by induction on the number of generators. Suppose $E$ is generated as an $A$-algebra by $n$ integral elements $x_1,\ldots,x_n$. Consider the $A$-subalgebra $E'$ of $E$ generated by $x_1,\ldots,x_{n-1}$; by the inductive hypothesis, $E'$ is finitely generated as an $A$-module. Let $\{s_i\}$ generate $E'$ as an $A$-module. Since $x_n$ is integral over $A$, it is also integral over $E'$, and therefore $E$ is finitely generated as an $E'$-module. Let $\{t_j\}$ be such generators; then $\{s_i t_j\}$ finitely generates $E$ as an $A$-module.

It is natural to expect a relationship between $E$ being integral over $A$ and every element of $E$ being integral. To establish this, we first prove the following lemma.

Lemma 5 Let $\phi:A\rightarrow E$ be a ring homomorphism and let $x\in E$. Then $x$ is integral over $A$ if and only if there exist an $E$-module $N$ and an $A$-submodule $M\subseteq N$ such that $M$ is not annihilated by any nonzero element of $E$ and $xM\subseteq M$.

Proof

First, suppose $x$ is integral over $A$. Taking $N=E$, we know by Proposition 2 that $M=A[x]$ is finitely generated. The converse follows exactly as in the proof of Proposition 2: view multiplication by $x$ as an endomorphism of $M$ and apply Theorem 1.

The following theorem asserts a property one would naturally expect, yet it is virtually impossible to prove directly from the definition without Lemma 5.

Theorem 6 For an $A$-algebra $E$, the integral closure of $A$ in $E$ is again an $A$-algebra.

Proof

Let $x,y\in E$ be integral over $A$. We must show that $x+y$ and $xy$ are integral over $A$. Let $M=A[x]$ and $M'=A[y]$ be submodules of $E$, and let $MM'$ be the subalgebra of $E$ generated by all products $xx'$ with $x\in M$, $x'\in M'$. Since $M$ and $M'$ are each finitely generated, so is $MM'$. Now

\[(xx')MM'=(xM)(x'M)\subseteq MM',\qquad (x+x')MM'\subseteq xMM'+M(x'M')\subseteq MM',\]

so applying Lemma 5 yields the desired result.

Nakayama’s Lemma

We now prove an extremely useful lemma. First, recall that for any ideal $\mathfrak{a}$ of a ring $A$, the nilradical $\sqrt{(0)}$ is given by

\[\sqrt{(0)}=\bigcap_\text{\scriptsize$\mathfrak{p}$ prime} \mathfrak{p}.\]

(§Properties of Localization, ⁋Corollary 8) Similarly, the Jacobson radical of $A$ is defined by

\[J(A)=\bigcap_\text{\scriptsize$\mathfrak{m}$ maximal} \mathfrak{m}.\]

To prove Nakayama’s lemma, we first establish the following auxiliary result.

Lemma 7 Let $M$ be a finitely generated $A$-module and let $\mathfrak{a}$ be an ideal of $A$ such that $\mathfrak{a}M=M$. Then there exists $a\in \mathfrak{a}$ with $(1-a)M=0$.

Proof

From the hypothesis we have $M\subseteq \mathfrak{a}M$, so by Theorem 1 there exists a monic polynomial

\[p(\x)=\x^n+p_1\x^{n-1}+\cdots+p_n,\qquad p_k\in \mathfrak{a}^k\]

such that $p(\id_M)=0$. That is,

\[(1+p_1+\cdots+p_n)M=0,\]

and setting $a=-(p_1+\cdots+p_n)$ gives the desired result.

We can now state Nakayama’s lemma.

Lemma 8 (Nakayama) Let $\mathfrak{a}$ be an ideal of $A$ contained in the Jacobson radical $J(A)$, and let $M$ be a finitely generated $A$-module. Then the following hold.

If $\mathfrak{a}M=M$, then $M=0$.
If the images of $x_1,\ldots,x_n$ in $M/\mathfrak{a}M$ generate $M/\mathfrak{a}M$ as an $A$-module, then $x_1,\ldots,x_n$ generate $M$ as an $A$-module.

Proof

For the first statement, the element $a\in \mathfrak{a}$ supplied by Lemma 7 lies in every maximal ideal by hypothesis. Hence $1-a$ cannot belong to any maximal ideal, so $1-a$ is a unit, and the conclusion follows.

For the second statement, set $N=M/\sum_i Ax_i$. Then one checks that $N/\mathfrak{a}N=0$, and the first statement implies $N=0$.

Localization

We now examine several results related to localization.

Proposition 9 Every unique factorization domain is a normal domain.

Proof

Take $a/b\in \Frac(A)$ with $a$ and $b$ coprime, and suppose $a/b$ lies in the normalization of $A$. Then there exists a monic polynomial such that

\[\left(\frac{a}{b}\right)^n+a_{n-1}\left(\frac{a}{b}\right)^{n-1}+\cdots+a_1\left(\frac{a}{b}\right)+a_0=0.\]

From this we see that

\[\x^n+a_{n-1}b \x^{n-1}+\cdots+a_1b^{n-1}\x+a_0b^n\in A[\x]\]

is a monic polynomial vanishing at $\x=a$. Thus $a^n$ is divisible by $b$, and to avoid a contradiction we must have $b=1$; hence $A$ is a normal domain.

More generally, the following holds.

Proposition 10 Let $A\subseteq B$ be rings and let $p\in A[\x]$ be monic. If there exist monic polynomials $q_1,q_2\in B[\x]$ such that $p=q_1q_2$ in $B[\x]$, then the coefficients of $q_1$ and $q_2$ are integral over $A$.

Proof

By adjoining roots, we can find a ring $C$ containing $B$ such that in $C[\x]$ both $q_1$ and $q_2$ split into linear factors $\prod (x-\alpha_i)$ and $\prod(x-\beta_j)$. By definition the $\alpha_i$ and $\beta_j$ are all integral over $A$, so the subring $C'$ of $C$ generated by them is an integral $A$-algebra. On the other hand, expanding $p=q_1q_2$ and inspecting its coefficients shows that they lie in $C'$.

Hence the following holds.

Corollary 11 Over a normal domain $A$, every monic irreducible polynomial is prime.

Meanwhile, normalization commutes with localization; the proof is immediate.

Proposition 12 Let $A\subseteq B$ be rings and let $S$ be a multiplicative subset of $A$. Then for the integral closure $A'$ of $A$ in $B$, the localization $S^{-1}A'$ is the integral closure of $S^{-1}A$ in $S^{-1}B$.

Another result related to localization is a slight strengthening of §Properties of Localization, ⁋Proposition 4. A ring $A$ is called semilocal if it has only finitely many maximal ideals. Then the following holds.

Proposition 13 Let $A$ be a semilocal ring and let $M,N$ be finitely presented $A$-modules. If $M_\mathfrak{m}\cong N_\mathfrak{m}$ for every maximal ideal $\mathfrak{m}$, then $M\cong N$.

Proof

Let $\mathfrak{m}_1,\ldots,\mathfrak{m}_n$ be the maximal ideals of $A$, and for each $k$ let $u_k:M_{\mathfrak{m}_k}\rightarrow N_{\mathfrak{m}_k}$ be an isomorphism. Since $A_{\mathfrak{m}}$ is a flat $A$-module and $M$ is finitely presented by assumption, we have an isomorphism

\[\Hom_{A_{\mathfrak{m}_k}}( M_{\mathfrak{m}_k}, N_{\mathfrak{m}_k}) \rightarrow \Hom_A(M,N)_{\mathfrak{m}_k}\]

(§Properties of Localization, ⁋Proposition 5). The image of $u_k$ under this isomorphism is obtained by inverting an element of $A\setminus \mathfrak{m}_k$ in $\Hom_A(M,N)$; thus, if necessary, we may multiply $u_k$ by this denominator to obtain $v_k\in\Hom_A(M,N)$.

Now, since $\mathfrak{m}_k$ is prime, if $\bigcap_{l\neq k} \mathfrak{m}_l\subseteq \mathfrak{m}_k$ then $\mathfrak{m}_l\subseteq \mathfrak{m}_k$ for some $l$, which is impossible. Hence

\[\bigcap_{l\neq k} \mathfrak{m}_l\not\subseteq \mathfrak{m}_k,\]

so there exists $a_k\in\bigcap_{l\neq k}\mathfrak{m}_l$ with $a_k\not\in\mathfrak{m}_k$. Define $v=\sum_{k=1}^n a_kv_k$. Then $v$ is the desired isomorphism; to verify this, it suffices by §Properties of Localization, ⁋Proposition 4 to check the localization at each maximal ideal $\mathfrak{m}_k$.

More generally, we claim that for any local ring $(B,\mathfrak{n})$ and maps $s,t:K\rightarrow L$ between finitely generated $B$-modules, if $s$ is an isomorphism and $t(K)\subseteq \mathfrak{n}L$, then $s+t$ is also an isomorphism. Applying this to the local ring $(A_{\mathfrak{m}_k},\mathfrak{m}_kA_{\mathfrak{m}_k})$ and the maps $s=a_k v_k$, $t=\sum_{l\neq k} a_l v_l$ from $M_{\mathfrak{m}_k}$ to $N_{\mathfrak{m}_k}$ completes the proof.

To prove the claim, first note that $t$ induces the zero map $K\rightarrow L/\mathfrak{n}L$, while $s$ induces an epimorphism $K\rightarrow L/\mathfrak{n}L$; hence $s+t$ also induces an epimorphism $K\rightarrow L/\mathfrak{n}L$. By Lemma 8 (Nakayama), the morphism $s+t:K\rightarrow L$ is itself an epimorphism. Now take the inverse $s^{-1}$ and consider the surjective endomorphism $s^{-1}(s+t):K\rightarrow K$. By Theorem 1, $s^{-1}(s+t)$ is also an isomorphism, and therefore $s+t$ is a monomorphism, as required.

We now examine further the properties of ring homomorphisms from Definition 3, beginning with the following proposition.

Proposition 14 Let $\phi:A\rightarrow E$ and $\rho:A\rightarrow A'$ be ring homomorphisms, and put $E'=A'\otimes_A E$. Then the following hold for $\rho_!\phi:A'\rightarrow E'$.

If $\phi$ is integral, then $\rho_!\phi$ is integral.
If $\phi$ is finite, then $\rho_!\phi$ is finite.

Proof

The proofs are similar, so we prove only the first statement. Assume $\phi$ is integral, and let $x_i$ be integral elements generating $E$. Then each $x_i$ satisfies a monic polynomial $p_i\in A[\x]$. Hence each $1\otimes x_i$ generates $E'$ and satisfies $(\rho[\x])(p_i)$. It follows that $E'$ is integral over $A'$.

Moreover, the following holds.

Proposition 15 Let $\phi:A\rightarrow E$ be a ring homomorphism, and suppose $(a_1,\ldots,a_n)=A$. Then the following hold.

If each $A_{a_i}\rightarrow E_{a_i}$ is integral, then $A\rightarrow E$ is integral.
If each $A_{a_i}\rightarrow E_{a_i}$ is finite, then $A\rightarrow E$ is finite.

Proof

As before, the proofs are similar, so we prove only the first statement. Fix $x\in E$, and let $\mathfrak{A}$ be the ideal of polynomials in $A[\x]$ vanishing at $x$. Then the set $\mathfrak{a}$ of leading coefficients of these polynomials is an ideal of $A$.

By hypothesis, each $A_{a_i}\rightarrow E_{a_i}$ is integral, so the image of $x$ in $E_{a_i}$ is integral over $A_{a_i}$. Writing down its integral equation and clearing denominators, we obtain for each $i$ an integer $n_i$ such that $a_i^{n_i}\in\mathfrak{a}$. Raising both sides of

\[1=\sum_{i=1}^n \alpha_i a_i\]

to a sufficiently high power, we see that these $a_i^{n_i}$ generate the unit ideal, and therefore $1\in\mathfrak{a}$. Hence $x$ is integral by definition.

Proposition 16 Let $\phi:A\rightarrow E$ be a ring homomorphism and let $x\in E$. Then $x$ is integral over $\phi$ if and only if the image of $x$ in $E_\mathfrak{p}$ is integral over $\phi_\mathfrak{p}:A_\mathfrak{p}\rightarrow E_\mathfrak{p}$ for every prime ideal $\mathfrak{p}$ of $A$.

Proof

One direction follows from Proposition 14. For the converse, suppose that for every prime ideal $\mathfrak{p}\subset A$, the image of $x$ in $E_\mathfrak{p}$ is integral over $A_\mathfrak{p}$; we show that $x$ is integral. Let $E'$ be the $A$-subalgebra of $E$ generated by $\phi(A)$ and $x$; it suffices to show that $A\rightarrow E'$ is integral.

By hypothesis, there exist $a_i\in A_\mathfrak{p}$ such that in $E_\mathfrak{p}$ we have

\[x^d+\phi_\mathfrak{p}(a_{d-1})x^{d-1}+\cdots+\phi_\mathfrak{p}(a_0)=0.\]

Clearing denominators in the $\phi_\mathfrak{p}(a_k)$, we may assume that all $a_k$ lie in $A_f$ for some suitable $f\in A\setminus\mathfrak{p}$; then the above equation holds in $E_f$, and it follows that $A_f\rightarrow E_f'$ is a finite homomorphism.

References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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The Cayley–Hamilton Theorem

Integral Extensions

Nakayama’s Lemma

Localization

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