Ring normalization

Let $A \subset B$ be a ring extension and $I \subset A$ an ideal. Recall that $b \in B$ is called (strongly) integral over $I$ if $b$ satisfies a relation

$$b^n + a_1 b^{n-1} + \ldots + a_n = 0$$    with $$a_i \in I^i\,.$$

The set $C(I,B) = \{b \in B \mid b$    is integral over $I\}$ is called the integral closure of $I$ in $B$, it is a $C(A,B)$-module, where $C(A,B)$, the integral closure of $A$ in $B$, is a ring.

We are interested in the two most interesting cases:

• $A$ reduced, $B = Q(A)$ the total quotient ring of $A$, and $I = A$. In this case $I^i = I$ and $C\bigl(A, Q(A)\bigr) =: \overline{A}$ is the normalization of $A$.

• $A = B$ and $I \subset A$ arbitrary. Then $C(I,A) =: \overline{I}$ is the integral closure of $I$ in $A$.

Let us consider the normalization first.

Lemma 1.1 (Key-lemma)   Let $A$ be a reduced Noetherian ring and $J \subset A$ an ideal containing a non-zerodivisor $u$ of $A$. Then there are natural inclusions of rings

$$A \,\subset\, Hom_A(J,J)\, \cong \,\frac{1}{u} (uJ:J) \,\subset \, \overline{A}\,,$$

and inclusions $Hom_A(J,J) \subset Hom_A(J,A) \cap \overline{A} \subset Hom_A \bigl(J,\sqrt{J}\!\;\bigr)$ of $A$-modules. Here $uJ:J = \bigl\{h \in A \,\vert\, hJ \subset uJ\bigr\}$.

Proof. If $\varphi \in Hom_A(J,A)$, then $\tfrac{\varphi(u)}{u}$ is independent of the non-zerodivisor $u$ and, hence, $\varphi \mapsto \tfrac{\varphi(u)}{u}$ defines an embedding

$$Hom_A(J,A) \,\stackrel{\cong}{\longrightarrow }\, \{h \in Q(A) \mid hJ\subset A\} \hookrightarrow Q(A)\,.$$

The inclusion $A \subset Hom_A(J,J) \cong \tfrac{1}{u} (uJ:J) = \tfrac{1}{u}\{h \in Q(A) \mid hJ \subset uJ\}$ is given by the multiplication with elements of $A$. To see that the image is contained in $\overline{A}$, consider $\varphi \in Hom_A(J,J)$. The characteristic polynomial of $\varphi$ defines (by Cayley-Hamilton) an integral relation for $\varphi$. To see the last inclusion, consider $h \in \overline{A}$ such that $hJ\subset A$, and let $h^n + a_1 h^{n-1} + \dots + a_n = 0$, $a_i \in A$, be an integral relation for $h$. For a given $g \in J$ multiply the relation for $h$ with $g^n$. This shows that $(gh)^n\! \in J$, hence $gh \in \sqrt{J}$.

Since $A$ is normal if and only if the localization $A_P$ is normal for all $P \in SpecA$, we define the non-normal locus of $A$ as

$$NN(A) := \{P \in SpecA \mid A_P$$    is not normal$$\}\,.$$

It is easy to see that $NN(A) = V(C)$ where $C = Ann_A(\overline{A}/A)$ is the conductor of $A$ in $\overline{A}$, in particular, $NN(A)$ is closed in $SpecA$. However, since we cannot yet compute $\overline{A}$, we cannot compute $C$ either.

The following proposition is the basis for the algorithm to compute the normalization as well as for an algorithm to compute an ideal with zero set $NN(A)$. It is basically due to Grauert and Remmert, [12].

Proposition 1.2 (Criterion for normality)   Let $A$ be a reduced Noetherian ring and $J \subset A$ an ideal satisfying

1.

$J$ contains a non-zerodivisor of $A$,

2.

$J = \sqrt{J}$,

3.

$NN(A) \subset V(J)$.

Then $A = \overline{A}$ if and only if $A = Hom_A(J,J)$.

An ideal $J \subset A$ satisfying 1.-3. is called a test ideal for the normalization.

Proof. If $A = \overline{A}$ then $Hom_A(J,J) = A$, by Lemma 1.1. For the converse, notice that 3. implies $J \subset \sqrt{C}$, hence there exists a minimal $d \ge 0$ such that $J^d \in C$, that is, $J^d \overline{A} \subset A$. If $d > 0$, choose $h \in \overline{A}$ and $a\in J^{d-1}$ such that $ha\not\in A$. Since $ah\in \overline{A}$ and $ahJ \subset hJ^d \subset A$ we have $ah \in Hom_A(J,A) \cap \overline{A}$ and hence, using 2., $ah \in Hom_A(J,J)$, by Lemma 1.1. By assumption $Hom_A(J,J) = A$ and, hence, $ah \in A$. This is a contradiction, and we conclude $d = 0$ and $A = \overline{A}$.

Remark 1.3   As the proof shows, condition 2. can be weakened to

2'.

$Hom_A(J,J)= Hom_A(J,A) \cap \overline{A}$.

However, this cannot be used in practice, since we do not know $\overline{A}$, while, on the other hand, we can always pass from $J$ to $\sqrt{J}$ without violating the conditions 1 and 3, and $\sqrt{J}$ is computable.

Corollary 1.4   Let $J \subset A$ be as in Proposition 1.2 and define $I_{\text{\it NN}} := Ann_A\bigl(Hom_A(J,J)\bigr)$. Then $NN(A) = V(I_{\text{\it NN}})$.

Proof. This follows from 1.2, 1.1 and the fact that the operations which define the annihilator are compatible with localization.

Having a test ideal $J$ and a non-zerodivisor $u \in J$ of $A$, we can compute $A$-module generators of $Hom_A(J,J) \cong uJ:J$ and $A$-module generators for $I_{\text{\it NN}} = Ann_A\bigl(Hom_A(J,J)\bigr) \cong \langle u \rangle :(uJ:J)$, since we can compute ideal quotients using Gröbner basis methods, cf. [15].

Let us describe the ring structure of $Hom_A(J,J)$. For this, let $u_0=u$, $u_1, \dots, u_s$ be generators of $uJ : J$ as $A$-module, and let $(\alpha_0^i, \dots, \alpha^i_s)$ be generators of the module of syzygies of $u_0, \dots, u_s$. Since $Hom_A(J,J)$ is a ring, we have $u_i \cdot u_j = {\sum}_{\ell=0}^s \!\;\beta^{ij}_\ell u_\ell$, $\,1 \le i, j \le s$ for certain $\beta^{ij}_\ell \in A$, the quadratic relations between the $u_i$. Define $Ker\subset A[t_1, \dots, t_s]$ as the ideal generated by the linear and quadratic relations,

$$\alpha^i_0 + \alpha^i_1 t_1 + \dots + \alpha^i_s t_s\,,\quad t_i t_j - (\beta^{ij}_0 + \beta^{ij}_1 t_1 + \dots + \beta^{ij}_s t_s)\,.$$

We get an isomorphism $Hom_A(J,J) \cong uJ : J \cong A[t_1, \dots, t_s]/Ker$ of $A$-algebras, by sending $t_i$ to $u_i$. This presentation is needed to continue the normalization algorithm.

To compute $I_{\text{\it NN}}$, we only need the $A$-module structure of $uJ : J$.