I want to prove that if $\mathbb{F}$ is a finite field and $X$ any set, then $R$ a ring defined as : $$ R:= \{f : X \to \mathbb{F}\ \big| \ f \ \text{is a function}\} $$
is Noetherian if and only if $X$ is finite.
I know the condition for Noetherian ring, but how do I work with a set, $X$ has no algebraic structure. I tried to think $X$ as a module, but can't see a suitable ring. Any ideas on how to proceed on this? Am I missing something important, please help. Thanks.
If $x\in X$, then $$ I_x=\{f\in R: f(x)=0\} $$ is an ideal of $R$. If $S\subseteq X$, then $I(S)=\bigcap_{x\in S}I_x$ is an ideal as well.
Prove that if $S\subsetneq T$, then $I(S)\supsetneq I(T)$.
Can you find an infinite descending chain of subsets of $X$, when $X$ is infinite?
For the converse, prove that every ideal of $R$ is of the form $I(S)$ for some $S\subseteq X$.