Correspondence between prime ideals and irreducible algebraic sets Let $k$ be an algebraic closed field. The Nullstellensatz theorem prove that
$$I(V(J))=\sqrt{J}$$
and we have
$$V(J)\text{ irreducible }\iff I(V(J)) \text{ prime }$$
So if $J$ is prime, $I(V(J))=J$ is prime, hence $V(J)$ is irreducible.

If $V(J)$ is irreducible, do we have that $J$ is prime ?

$V(J)$ is irreducible so $I(V(J))=\sqrt{J}$ is prime, but if I'm not mistaken $\sqrt{J}$ prime does not implies that $J$ is prime as well ($J\subset \sqrt{J}$).

If it is true, does it holds when $k$ is not algebraic closed ?

Thank you for your help.

You're right; $V(J)$ can be irreducible even if $J$ is not prime.  For instance, take $J=(x^2)\subset k[x]$.  Then $J$ is not prime, since $x\cdot x\in J$ but $x\not\in J$.  But $\sqrt{J}=(x)$ is prime, so $V(J)$ is irreducible.
