# Set of zero divisors and the set of units of the ring of functions from $X$ to $R$

Hope this isn't a duplicate.

I was trying to answer the following question: Let $X$ be a non-empty set and $R$ be a ring. Then define $F(X,R)$ to be the ring of functions from $X$ to $R$. Then what is the set of zero divisors and the set of units of $F(X,R)$?

My attempt:

Let $A$ and $B$ be the set of units and set of zero divisors of $R$, respectively.

Set of units of $F(X,R)$ : $\{f\in F(X,R) : f(x)\in A\;\forall x\in X\}$.

Set of zero divisors of $F(X,R)$ : $\{f\in F(X,R) : f(x)\in B\;\forall x\in X\}$.

Your answer for the units of $F(X,R)$ is correct. But your answer for the zero divisors isn't.

By definition, $f\in F(X,R)$ is a zero divisor iff there exists some function $g\in F(X,R)$, $g\neq0$ (as a function!) such that $fg$ is the identically zero function, i.e. $f(x)g(x)=0$ for all $x\in X$. Note that the condition that $g$ is not the zero function does NOT preclude it from having some zero values — as long as it has at least one non-zero value, it's a non-zero function. For example, if we pick some specific $a\in X$ and $r\in R$, $r\neq0$, then the function defined as $$g(a)=r, \quad g(x)=0 \text{ for all } x\neq a$$ is a non-zero function, isn't it? So your answer to the zero divisors question needs to be corrected, because the $\forall$ quantifier there is wrong.

• The definition I follow for zero divisors is the following : an element a of a ring R is a zero divisor if there exists $b \in R\ \{0\}$ such that ab=0 or ba=0. Hence I include 0 to be a zero divisor and hence following this convention, I think my answer is complete, right?
– user422112
Aug 19, 2018 at 4:48
• @ThatIs: The issue is not whether your answer is complete or not. The issue is that your answer (to the second question) is wrong. The definition you stated here is correct, of course. But you need to find zero divisors in the ring of functions $F(X,R)$, so you need to understand what makes a function a zero divisor, i.e. what this definition says when the ring is $F(X,R)$. Aug 19, 2018 at 5:21
• I think I got your point. So it should be the collection of all functions that map at least one point in X to a zero divisor in R?
– user422112
Aug 19, 2018 at 5:54
• @ThatIs: Yes, exactly! Aug 19, 2018 at 13:03

Your answer is correct $f$ is a divisor of zero if and only if there exists a function $g$ such that $f(x)g(x)=0$ this is equivalent that the image of $f$ is contained in the set of zero divisors of $R$.

A similar argument shows that $f$ is a unit ($f(x)g(x)=1$) if and only if the image of $f$ is contained in the set of units of $R$.