The problem statement is as follows: Let $R$ be a commutative ring. Let $S$ be an ideal of $R$ and $T$ be a Prime Ideal of $S$. Prove that $T$ is an ideal of Ring $R$.
What I have tried:
We know that $T$ is a prime ideal of $S$. So we can assert 1.) $T$ is itself a ring, and 2.) $T$ is a subset of $S$; hence $T$ is a subset of $R$
Now all we need to show is that for any elements $r \in R$, and $t\in T$, their product $$rt\in T$$
What I have done for that is:
For any element $s\in S$, $$rs\in S$$
Since $T$ is an ideal of $S$, for any $t\in T$, $$(rs)t\in T$$ Or, $$r(st)\in T$$ But $$st \in T$$ hence $st=m$ for some $m\in T$.
Therefore, $$rm\in T$$ Hence $T$ is an ideal of $R$. Now I know this proof is incorrect, because it does not make use of the fact that $T$ is a Prime Ideal. Could you point mistake in my proof, and also give the correct proof? Thanks a lot.