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Next: Test ideals Up: Integral closure of rings Previous: Integral closure of rings

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

$\displaystyle b^n + a_1 b^{n-1} + \ldots + a_n = 0$    with $\displaystyle 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: 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

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

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

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

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

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

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
$ J$ contains a non-zerodivisor of $ A$,
$ J = \sqrt{J}$,
$ 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
$ 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,

$\displaystyle \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

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$.

next up previous
Next: Test ideals Up: Integral closure of rings Previous: Integral closure of rings
Christoph Lossen