Is mapping that sends an idempotent to an idempotent is necessarily a ring homomorphism?

Question

Question: Let $R, S$ be rings. If $f: R\rightarrow S$ is a mapping such that $f$ carries an idempotent to an idempotent. That is, If $a\in R$ is an idempotent element in $R$ then $f(a)$ is an idempotent element in $S$ then is $f$ is necessarily a ring homomorphism?

I know that, other direction holds! that is, a ring homomorphism carries an idempotent to an idempotent.

To solve a given question either I need to find a counter-example that is, the mapping between two rings that maps an idempotent to an idempotent but is not a ring homomorphism 'or' I need to prove this direction also holds... please help.

Answer 1

You could just take $f:\mathbb Z\to \mathbb Z$ and map every element to $1$.

Or if you want non-idempotents to map to non idempotents too, you could just use the map that interchanges $0$ with $1$ and leave everything else alone.

In some sense, the subset of functions from $\mathbb Z$ to $\mathbb Z$ that aren't ring homomorphisms is "huge" compared to the ones that are, and you shouldn't have a hard time finding an element that isn't a ring homomorphism (or even a group homomorphism.)

Answer 2

Let $Z$ be the ring of integers and let $f: Z\to Z$ be a function defined by $f(x)=x^{2}$ then $f(0)=0$ and $f(1)=1$ . Note that $f$ is not a ring homomorphism and $0 , 1$ are the only idempotent elements of $Z$. Generally suppose that $R$ is a ring such that there are elements $x , y$ such that $xy+yx\neq0$, then the function $f:R\to R$ defined by $f(x)=x^2$ is not a ring homomorphism but sends an idempotent to an idempotent.