Is the localization at a prime ideal of a non-zero ring non-zero again?

Question

Let $R$ be a non-zero ring. Is it true that for any prime ideal $p$ in $R$, the localization $R_p$ is non-zero again? Stated in another way: Is it true that

$R \neq 0 \iff \forall p \in Spec(R): R_p \neq 0$?

This would then be a much stronger statement than the one known for modules, that

$M \neq 0 \iff \exists p \in Spec(R): M_p \neq 0$.

I am formulating this as a question, although in fact I have two proofs for the statement. However, this seems too good to be true, so I don't trust the proofs I came up with.

Before giving the proofs, let me explain why I am asking this: if $(X,O_X)$ is any ringed space, then we know that $supp(X) = \{x \in X; O_{X,x} \neq 0\}$ is a closed subset of X, since $supp(X)=supp(1)$, the support of the global section $1 \in O_X(X)$, and we know this to be closed. Now proving the above statement for $(X,O_X)$ a scheme amounts to showing that $supp(X)$ is also open: If $x \in supp(X)$, then let $U=Spec(R)$ be an open affine subscheme containing $x$, then $x\simeq p$ for $p \in Spec(R)$ and by the above statement we find $U\subset supp(X)$.

Here are the two proofs:

1. We know from commutative algebra that for a finitely generated $R$-module $M$ and $p\in Spec(R)$, $M_p \neq 0 \iff p \in supp_R(M) \iff p$ contains $Ann_R(M)$, the annihilator of $M$. Hence for $M=R$ we find $Ann_R(R) = (0)$ $\implies$ $R=0$ or $R_p \neq 0$ for all $p \in Spec(R)$.

2. Assume $R \neq 0$, but we have $p \in Spec(R)$ with $R_p = 0$. We want for any $r \in R$ some $t \notin p$ with $rt=0$, in particular we want such $t$ for $r=1$, but then clearly $t=0$, hence $0 \notin p$, contradiction.

The two proofs are in fact of course reformulations of each other. Are these correct? Thank you in advance for taking the time to help!

EDIT: Thank you for your reassurement, I was uncertain. (I write this here because I am not allowed to post comments.)

Answer 1

We have $R_p / pR_p \cong Frac(R/p)$ by universal properties. Since $p\neq R$, we have $pR_p \neq R_p$.

Concretely, the element $1/1 \in R_p$ equals $0/1$ precisely when there is $s\notin p$ such that $0=s(1\cdot 1 - 1\cdot 0) = s$, but $0$ is in every prime ideal.

Answer 2

This is an expanded version of Hoot's comment:

No local ring can be zero, because it has a prime ideal. Indeed, if $\mathfrak p \subseteq R$ is prime, then $R/\mathfrak p$ is a domain, so $R/\mathfrak p \neq 0$. Hence, $R \neq 0$. Thus, the right hand side is always true: for any $\mathfrak p \in \operatorname{Spec} R$, the ring $R_\mathfrak p$ is not zero.

Maybe the statement you're looking for is: $R = 0$ if and only if $\operatorname{Spec} R = \varnothing$. (Exercise: prove this.)

The reason you cannot apply the module case is that $\forall$ does not imply $\exists$ for the empty set.