# Clarification about equivalence irreducibility and prime ideals

It is known the following equivalence: Let $Y \subset \mathbb{A}^{n}$ be an algebraic set. Then $Y$ is irreducible if and only if $I(Y)$ is prime, where:

$I(Y) = \{f \in K[x_{1},x_{2},..,x_{n}]: \textrm{for all p in Y}, \ f(p)=0 \}$.

So as an example, the author considers $J = \langle (xz-y^2,x^3-yz) \rangle \subset k[x,y,z]$. Then let $Y=V(J)$ (the locus set).

So if we want to find whether $Y$ is irreducible we must study $I(Y)=I(V(J))$ and show this ideal is prime (or show it is not).

The author then shows that $J$ is not a prime ideal. But why is this? don't we need to show that $I(V(J))$ is a prime ideal? why we must check $J$? or do we always have $J=I(V(J))$?

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No, we don't always have $I(V(J))=J$ - the Nullstellensatz says that we always have $I(V(J))=\sqrt{J}$, where $\sqrt{J}$ is the radical of the ideal $J$, but there are many ideals that are not equal to their radical. For example, the radical of the non-prime ideal $J=(x^2)$ in $k[x]$ is the prime ideal $(x)$, so that $V(J)$ is irreducible even though $J$ is not prime. So your complaint is valid; it sounds like the book's argument for why $V(J)$ is not irreducible is in error. Could you specify the author/title?
Thanks for your comment. These are notes I borrowed from a friend, not really a book. If it is not too much to ask, how would you show that $V( x^{2}-y,z-1)$ is irreducible? So the first step is to find I(V(x^{2}-y,z-1)) right? how do we proceed here? – user6495 Feb 22 '11 at 7:37
@user6495: Let $J=(x^2-y,z-1)\subset k[x,y,z]$. Then $k[x,y,z]/J\cong k[x,y]/(x^2-y)\cong k[x]$ is an integral domain, hence $J$ is prime, hence $J=\sqrt{J}$, hence $I(V(J))=J$ is prime, hence $V(J)$ is irreducible. – Zev Chonoles Feb 22 '11 at 7:50
oh I see, that's the part I was missing $J=sqrt(J) because$J\$ is prime. Thank you again! – user6495 Feb 22 '11 at 7:57