TITLE: Is it true that if $V(\mathfrak a)$ is irreducible, then $\mathfrak a$ is prime ideal?
QUESTION [5 upvotes]: Is it true that if $V(\mathfrak a)$ is irreducible, then $\mathfrak a$ is prime ideal? 

I know that if $V$ is an irreducible affine algebraic set if and only if $I(V)$ is a prime ideal. Can I use this result?

REPLY [6 votes]: Take $\mathfrak a=(X^2,XY)$ in $\mathbb C[X,Y]$. Then $V(\mathfrak a)$ is irreducible iff $I(V(\mathfrak a))$ is a prime ideal. But $I(V(\mathfrak a))=\sqrt{\mathfrak a}=(X)$, hence $V(\mathfrak a)$ is irreducible. But $\mathfrak a$ is not a prime ideal, so the answer to your question is no.