Set of zero divisors is an ideal iff the ring is local

Question

Let $R$ be a commutative ring with unity. Show that $Z(R)$, the set of all zero divisors of $R$, is an ideal if and only if $R$ is a local ring.

I have no idea for proving this.

Thanks in advance!

Answer 1

The if is false.

Counter-example:

Let $K$ be a field. consider the ring $R=\bigl(K[X,Y]/(XY)\bigr)_{(X,Y)}$. This ring is local by construction, has dimension $1$, and its zero divisors is the union of its two minimal prime ideals, generated by (the images of) $X$ and $Y$ respectively.

Added (thanks to an idea of @rschwieb):

The only if part is false too:

Consider a non-local domain $D$ and a non-zero torsion-free $D$-module $M$. The set $R=D\times M$, endowed with the ring structure defined by $$\begin{cases}(d,m)+(d',m')=(d+d',m+m')\\ (d,m)(d',m')=(dd',dm'+d'm) \end{cases}$$ is a counter-example.

For ease of notation, we identify the ideal $\{0\}\times M$ with $M$. It's straightforward to check that $M^2=\{0\}$ and $Z(R)=M$. However, as $M$ is nilpotent, the spectra of $R$ and of $D$ are in bijection, hence $R$ is non-local.