MHB Zero Divisor Help: Prove & Example

[corkscrew062]

Came across this problem while reading and I can't understand what it wants. I've tried solving for units to get zero divisors but that doesn't work. It's not the end of the world if I can't figure it out, it's just bothering me because I'm confused. Here's the problem: Let $R$ and $S$ be commutative rings, and let $f:R\longrightarrow S$ be a ring isomorphism. Prove if $x$ is a zero divisor in $R$, then $f(x)$ is a zero divisor in $S$. Give an example to show that it is not necessarily true if $R\not \simeq S$.

Thanks in advance!

 

[Deveno]

What you lack, Sir Lion, is not courage, but a testimonial-so I will give you one.

Rings are terrible beasties-they positively give me fits of apoplexy. That said, one occasionally needs to employ one when one has some -unsavory- business to attend to. So, one must, of needs, descend to the algebraic gutters, and make their acquaintances.

The nicer mannered ones come from the integral school, however, even some of these have produced some rather badly-quotiented offspring, as if dividing by zero wasn't OH SO 20th century. Egads!

You might, for example, consider the modulated by 6 variety, and think of some choice...no prime! idealization it contains which would render the zero-dividing nonsense totally harmless.

(No, seriously-look at $\Bbb Z_6$ which has zero divisors, and consider the ring $\Bbb Z_6/(\overline{2})$).