Support of a non-trivial local ring is the whole spectrum?

Question

$\newcommand{\Supp}{\operatorname{Supp}}\newcommand{\Spec}{\operatorname{Spec}}$Suppose $A$ is a local ring. There are a few places where $\Supp(A)=\Spec(A)$ seems necessary, for example $V(x)= V(x)\cap \Spec(A)=V(x)\cap \Supp(A)=\Supp(A/xA)$, also for example in a filtration of a finite module $0=M_0\subset M_1\subset \dots \subset M_n = M$, where $\Supp(M)=\Supp(M_{n-1})\cup \Supp(M_n/M_{n-1})= \dots =\Supp (0)\cup \Supp(M_n/M_{n-1})\cup \dots\cup \Supp(M_1/0) = \Supp(A/P_n) \cup\dots\cup \Supp(A/P_1) = V(P_1)\cup \dots\cup V(P_n).$

Here $V$ refers to the Zariski closure.

But how do I know $\Supp(A)=\Spec(A)$?

What is a non-trivial local ring? – Martin Brandenburg Aug 16 '21 at 10:58 — Martin Brandenburg, Aug 16 '21 at 10:58

score 1 · Accepted Answer · answered Aug 16 '21 at 10:59

1

If $A$ is any commutative ring, then $A_{\mathfrak{p}} \neq 0$ for every prime ideal $\mathfrak{p}$ (in fact, this is a local ring), hence $\mathrm{supp}(A) = \mathrm{Spec}(A)$.

answered Aug 16 '21 at 10:59

Martin Brandenburg

163,620

$A_p\neq 0$ is not intuitively obvious to me since $A$ is not assumed to be integral. I had to check the details using definitions, but turned out ok. Thanks. – Jun Xu Aug 16 '21 at 12:39
$S^{-1} A = 0 \iff 0 \in S$. – Martin Brandenburg Aug 16 '21 at 16:20

Support of a non-trivial local ring is the whole spectrum?

1 Answers1