Resolutions of singularities.

Yoshifumi Tsuchimoto

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

PROOF.. See http://amathew.wordpress.com/2010/10/23/a-projective-morphism-is-proper/. $\qedsymbol$

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.

PROPOSITION 05.3  

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. $$\operatorname{Proj}(\oplus A[I\dot T] )= \operatorname{Proj}(\oplus A[f_0 T,f_1 T,\dots, f_s T] )= \cup_{i=0}^s D_+(f_i T)$$

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

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

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.

• Any field is a local ring.
• For any commutative ring $A$ and for any prime ideal $\mathfrak{p}\in \operatorname{Spec}(A)$ , the localization $A_\mathfrak{p}$ is a local ring with the maximal ideal $\mathfrak{p}A_\mathfrak{p}$ .

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

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

which shows that

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

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

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

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

$\qedsymbol$

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

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

EXAMPLE 05.10 (of NOT being a local homomorphism)  

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

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 $A\neq I$ . 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

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

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

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

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

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

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

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

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

$$v=B v$$

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 :

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

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

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

Then we have

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

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

$$v=c v.$$

$\qedsymbol$

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.