Ring of functions from nonempty set into a field has no non-zero nilpotent elements

Question

Let $R$ be a ring of functions from a nonempty set $X$ into a field $F$. Show that $R$ has no non-zero nilpotent elements.

So i've been pouring over this question for awhile now and I'm really having trouble using the fact that the codomain is a field, which I am sure is quite crucial. I was trying to do something like, well every thing in X maps to a unit and maybe units can't be nilpotent or something like that, but I was coming up short on anything solid. Anyone got any advice?

Answer 1

You could actually even say more:

The ring of functions from a (nonempty) set $X$ into a ring $S$ has no nonzero nilpotent elements iff $S$ has no nonzero nilpotent elements.

Perhaps stating it this way makes it clear that straightforward verification using the operations will get you the solution.

So $S$ could really be something much different from a field, like $\mathbb R\times \mathbb H$ or something like that.

Starting with $S=F$ perhaps camouflages this by being overspecific. But it is a good idea to think about such generalizations when faced with a question like this.

Answer 2

Hint:

Start with, let $f \in R$ be nilpotent and use the fact "zero is the only nilpotent element in an integral domain" to conclude $f=0$