Commutative algebra

Yoshifumi Tsuchimoto

Supplement:

PROPOSITION 07.1   Let $A$ be a ring. Then $\dim A=\sup_{\frak m \in \operatorname{Spm}(A)}A_{\frak m} = \sup_{\mathfrak{p}\in \operatorname{Spec}(A)}A_\mathfrak{p}$ .

DEFINITION 07.2   Let $\mathfrak{p}$ be a prime ideal of a ring $A$ . Then we define its height $\operatorname{ht}\mathfrak{p}$ to e the supremum of the lengths of prime chaines

$$\mathfrak{p}=\mathfrak{p}_0 \supsetne... ...rak{p}_1 \supsetneq \mathfrak{p}_2 \supsetneq \dots \supsetneq \mathfrak{p}_r.$$

We have $\operatorname{ht} \mathfrak{p}= \dim A_\mathfrak{p}$ .

DEFINITION 07.3   For an ideal $I$ of a ring $A$ , we define its height $\operatorname{ht}I$ to be

$$\operatorname{ht}I=\inf \{\operatorname{ht}\mathfrak{p}\vert I \subset \mathfrak{p}\in \operatorname{Spec}(A)\}$$

DEFINITION 07.4   Let $A$ be a ring and $M$ be an $A$ -module. Then a prime ideal $\mathfrak{p}$ of $A$ is called an associated prime ideal of $M$ if It is the annihilator ann(Mx of some $x\in M$ . The associated primes of the $A$ -module $A/I$ are refereedd to as the prime divisorsof $I$ .

THEOREM 07.5   Let $A$ be a Noetherian ring, and $I=(a_1,\dots,a_r)$ an ideal generated by $r$ elements; then if $\mathfrak{p}$ is a minimal prime divisor of $I$ we have $\operatorname{ht}\mathfrak{p}\leq r$ .

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THEOREM 07.6   Regular local ring $(A,\mathfrak{m})$ is UFD.

Step 1.

Induction on $\dim(A)$ .

If $\dim(A)=0$ , Then $A$ is a field. Thus it is UFD.

Assume $\dim(A)>0$ . Take $x \in \mathfrak{m} \setminus \mathfrak{m}^2$ . It suffices to prove:

1. If $A[x^{-1}]$ is UFD, then $A$ is UFD.
2. $A[x^{-1}]$ is UFD.