how to show that $V( Y-X^2 )$ is irreducible?

show that $V( Y-X^2 )$ is irreducible.

$Y-X^2$ is an irreducible polynomial ($Y-X^2$ cann't be factored into more irreducible components). Can we conclude that $V(Y-X^2)$ is irreducible??

Solution 1:

$<y-x^2>=I(\{(t,t^2):t\in k\})$.Then since $<y-x^2>$ is an ideal of a set, hence $I(V(y-x^2))=<y-x^2>$ , therefore $I(V(y-x^2))$ is prime (as $k[x][y]/<y-x^2>\cong k[x]$) and hence $V(y-x^2)$ is irreducible.

Solution 2:

Easiest way is to show that the corresponding ideal, which is $(Y - X^2)$, is prime. This in turn can be seen easiest by showing that its quotient ring $k[X,Y]/(Y-X^2)$ is an integral domain (or even by directly arguing that $Y - X^2$ is irreducible). Now, what is this quotient ring isomorphic to?

But, to actually answer your question: yes, from the fact that $Y - X^2$ is irreducible, you can conclude that $V(Y - X^2)$ is irreducible.