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Resolutions of singularities.

Yoshifumi Tsuchimoto

\fbox{05. supplement}

PROPOSITION 05.1   For any ring $ A$ , the map $ \mathbb{P}^n(A)\to \operatorname{Spec}A$ is proper.

PROOF.. See % latex2html id marker 902
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COROLLARY 05.2   For any ring $ A$ and for any $ \mathbb{N}$ -graded ring $ S=\oplus_n S_n$ which is generated by a finite subset of $ S_1$ over the ring $ S_0\cong A$ , the map $ \operatorname{Proj}(S)\to \operatorname{Spec}A$ is proper.


Let $ I$ be an ideal of a ring $ A$ generated by elements $ f_0,\dots f_s$ . Assume that each of the elements $ f_0,f_1,\dots, f_s$ is not a zero divisor in $ A$ . Then:

  1. $ \oplus_{n\in \mathbb{N}} I^n \subset \oplus_{n \in \mathbb{N}} A$ is isomorphic to $ A[ I\cdot T] \subset A[T]$ for an indeterminate $ T$ .
  2. $\displaystyle \operatorname{Proj}(\oplus A[I\dot T] )=
\operatorname{Proj}(\oplus A[f_0 T,f_1 T,\dots, f_s T] )=
\cup_{i=0}^s D_+(f_i T)

  3. $ D_+(f_i T) \cong \operatorname{Spec}A[f_0/f_i,f_1/f_i,\dots, f_s/f_i]$ .

DEFINITION 05.4   Let $ A$ be a commutative ring. Let $ \mathfrak{p}$ be its prime ideal. Then we define the localization of $ A$ with respect to $ \mathfrak{p}$ by

$\displaystyle A_\mathfrak{p}=A[ (A\setminus \mathfrak{p})^{-1}]

DEFINITION 05.5   A commutative ring $ A$ is said to be a local ring if it has only one maximal ideal.

EXAMPLE 05.6   We give examples of local rings here.

LEMMA 05.7  
  1. Let $ A$ be a local ring. Then the maximal ideal of $ A$ coincides with $ A\setminus A^\times$ .
  2. A commutative ring $ A$ is a local ring if and only if the set $ A\setminus A^\times$ of non-units of $ A$ forms an ideal of $ A$ .

PROOF.. (1) Assume $ A$ is a local ring with the maximal ideal $ \mathfrak{m}$ . Then for any element $ f \in A\setminus A^\times$ , an ideal $ I=f A +\mathfrak{m}$ is an ideal of $ A$ . By Zorn's lemma, we know that $ I$ is contained in a maximal ideal of $ A$ . From the assumption, the maximal ideal should be $ \mathfrak{m}$ . Therefore, we have

$\displaystyle f A \subset \mathfrak{m}

which shows that

$\displaystyle A\setminus A^\times \subset \mathfrak{m}.

The converse inclusion being obvious (why?), we have

$\displaystyle A\setminus A^\times =\mathfrak{m}.

(2) The ``only if'' part is an easy corollary of (1). The ``if'' part is also easy.

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COROLLARY 05.8   Let $ A$ be a commutative ring. Let $ \mathfrak{p}$ its prime ideal. Then $ A_\mathfrak{p}$ is a local ring with the only maximal ideal $ \mathfrak{p}A_\mathfrak{p}$ .

DEFINITION 05.9   Let $ A,B$ be local rings with maximal ideals $ \mathfrak{m}_A, \mathfrak{m}_B$ respectively. A local homomorphism $ \varphi:A \to B$ is a homomorphism which preserves maximal ideals. That means, a homomorphism $ \varphi$ is said to be loc al if

$\displaystyle \varphi^{-1}(\mathfrak{m}_B) =\mathfrak{m}_A

EXAMPLE 05.10 (of NOT being a local homomorphism)  

$\displaystyle \mathbb{Z}_{(p)}\to \mathbb{Q}

is not a local homomorphism.

LEMMA 05.11 (Zorn's lemma)   Let $ \mathcal S$ be a partially ordered set. Assume that every totally ordered subset of $ \mathcal S$ has an upper bound in $ \mathcal S$ . Then $ \mathcal S$ has at least one maximal element.

PROPOSITION 05.12   Let $ A$ be a commutative ring. let $ I$ be an ideal of $ A$ such that % latex2html id marker 1085
$ A\neq I$ . Then there exists a maximal ideal $ \mathfrak{m}$ of $ A$ which contains $ I$ .

THEOREM 05.13 (Nakayama's lemma, or NAK)   Let $ A$ be a commutative ring. Let $ M$ be an $ A$ -module. We assume that $ M$ is finitely generated (as a module) over $ A$ . That means, there exists a finite set of elements $ \{m_i\}_{i=1}^t$ such that

$\displaystyle M=\sum_{i=1}^t A m_i

holds. If an ideal $ I$ of $ A$ satisfies

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$\displaystyle I M=M \quad ($that is, $\displaystyle M/I M=0),

then there exists an element $ c\in I$ such that

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$\displaystyle c m= m \qquad (\forall m \in M)

holds. If furthermore $ I$ is contained in $ \cap_{\mathfrak{m}\in \operatorname{Spm}(A)} \mathfrak{m}$ (the Jacobson radical of $ A$ ), then we have $ M=0$ .

PROOF.. Since $ I M=M$ , there exists elements $ b_{i l}\in I$ such that

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$\displaystyle a_i= \sum_{l=1}^t b_{i l} a_l \qquad (1\leq i \leq t )

holds. In a matrix notation, this may be rewritten as

$\displaystyle v=B v

with $ v=^t (m_1,\dots,m_n)$ , $ B=(b_{i j})\in M_t (I)$ . Using the unit matrix $ 1_t\in M_t(A)$ one may also write :

$\displaystyle (1_t -B) v=0.

Now let $ R$ be the adjugate matrix of $ 1_t -B$ . In other words, it is a matrix which satisfies

$\displaystyle R(1_t-B)=(1_t-B)R=(\det(1_t-B)) 1_t.

Then we have

$\displaystyle \det(1_t -B)\cdot v = R (1_t -B) v=0.

On the other hand, since $ 1_t-B=1_t$ modulo $ I$ , we have $ \det(1_t-B)=1-c$ for some $ c\in I$ . This $ c$ clearly satisfies

$\displaystyle v=c v.

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We need a criterion for regularity. Instead of developing the vast theory of regular rings, we site here the following theorem:

THEOREM 05.14 (Matsumura``commutative ring theory'' Theorem14.2)   Let $ (R,\mathfrak{m})$ be a regular local ring. then the following two statements are equivalent:
  1. The images in $ \mathfrak{m}/\mathfrak{m}^2$ of $ x_1,\dots, x_i$ are linearly independent over $ R/\mathfrak{m}$ .
  2. $ R/(x_1,\dots ,x_i$ si an $ (n-i)$ -dimensional regular local ring.

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