If $S$ contains a zero divisor, then $S^{-1}R$ is the zero ring.

Question

I have $R$ a commutative ring with neutral element and $S$ a multiplicative subset. I also have $S^{-1}R=\{\frac{r}{s}, r\in R,s\in S\}$.

I have to prove that if $S$ contains a zero divisor, then $S^{-1}R$ is the zero ring and I also have to say for which Ideals $I$, $S=R\setminus I$ is closed under multiplication.

Regarding the statement I don't know if the following proof is okay: Let $x\in S$ be a zero divisor. This means that there exists $y\ne 0$ such that $xy=0 \in S$ since $S$ is closed under multiplication. And when $0\in S $ we know that $S^{-1}R$ is the zero ring (we had a proof of that in the lecture).

And could also somebody help me with the ideals, which ones make $S=R\setminus I$ closed under multiplication?

Thanks in advance for the help.

Answer 1

These are exactly the prime ideals. You probably defined a prime ideal as an ideal $P$ where $xy\in P$ implies either $x\in P$ or $y\in P$. It is straightforward that the statement "$x,y\in R\setminus P$ implies $xy\in R\setminus P$" is an equivalent definition.

Answer 2

Recall that $$ S^{-1}R = (S\times R)/{\sim}$$ where $$ (s_1,r_1)\sim(s_2,r_2)\iff \exists s\in S\colon ss_1r_2=ss_2r_1,\tag1$$ which - writing $\frac rs$ for the equivalence class of $(s,r)$ - allows us to define desired ring operations in the usual way as $\frac ab\cdot\frac cd=\frac{ac}{bd} $ and $\frac ab+\frac cd=\frac{ad+bc}{bd}$, using $(1)$ to show that these operations are well-defined.

Now consider the case $R=\Bbb Z\times \Bbb Z$ and $S=\{\,(1,n)\mid n\in\Bbb Z\,\}$. Note that $S$ contains the zero divisor $(1,0)$, but $S^{-1}R$ is isomorphic to $\Bbb Z$ because $(1)$ readily turns into $(s_1,r_1)\sim(s_2,r_2)\iff r_1=r_2$.