I am solving an exercise of Reid's Undergraduate Commutative Algebra, which asks the following. Let $A$ be Noetherian, and $X\subseteq \text{Spec A}$ be Zariski closed. Then $X$ is irreducible (i.e. not the union of two distinct closed subsets) if and only if $I(X)=\cap_{P\in X} P$ is prime. My problem is that I wrote a proof in which I don't see flaws, but I did not use the fact that $A$ was Noetherian. Could it be that Noetherianity is not needed?
Here's a sketch. For the if part, by contraposition: if $X$ is reducible, there are closed $X_1,X_2$ such that $X=X_1\cup X_2$, both being non-nested and distinct. This implies that $I(X_1)$ and $I(X_2)$ are also distinct and non-nested, so there are $f\in I(X_1)\setminus I(X_2)$ and $g\in I(X_2)\setminus I(X_1)$, while obviously $fg\in I(X)$, which therefore is not prime.
For the only if part, again by contraposition. Say $I(X)=I$ is not prime, so there is $fg\in I$ with $f,g\notin I$. The proof then relies on the claim\begin{equation*} X=V(I,f)\cup V(I,g)\end{equation*} Both directions are obvious: if $P\in X$, then $P\supseteq I$ so $fg\in P$ and thus either $f$ or $g$ are in $P$ so that $P\in V(I,f)$ or $P\in V(I,g)$. On the other hand if $P$ is in the RHS, say $P\in V(I,f)$, then obviously $P\supseteq I$, so that $P\in LHS$.
Can anybody see flaws?
You are correct that this result does not require $A$ to be Noetherian. Your proof is missing one detail, though. In the second direction, you also need to check that $V(I,f)$ and $V(I,g)$ are distinct from $X$. This follows from the definition of $I$: since $f\not\in I$, there is some $P$ such that $P\in X$ and $f\not\in P$, and then $P\not\in V(I,f)$, and similarly for $g$.