# Showing a point is generic

How does one show that something is a generic point? I feel I may be missing some subtly. Vakil in his notes asks one to show that the prime ideal $(y-x^2)$ is a generic point of $V(y-x^2)$? That is we need to show that the closure of $(y-x^2)$ is $V(y-x^2)$. But this seems trivial since $V(y-x^2)$ is by definition the smallest closed subset containing $y-x^2$. I am tired, though...

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$V(y-x^2)$ is by definition the set of primes containing $(y-x^2).$ Is it obvious that it is the smallest closed set containing $[(y-x^2)]$?

Well, if a closed set contains $[(y-x^2)],$ then it is of the form $V(S)$ for some $S\subseteq (y-x^2).$ Hence, the closure of $[(y-x^2)],$ which can be written as the intersection of all closed sets containing $[(y-x^2)],$ is $$\bigcap_{S\subseteq (y-x^2)}V(S)=V(\sum_{S\subseteq (y-x^2)}\langle S\rangle)=V((y-x^2))=V(y-x^2),$$

as we expected all along.

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1) If $A$ is a domain and $V=Spec(A)$, the generic point of $V$ is $\eta=[(0)]$, corresponding to the zero ideal.
The rationale is that integral subschemes $W\subset V$ correspond to points of $Spec(A)$ : associate to $W$ the ideal $I(W)$ of elements $f\in A$ vanishing on $W$.
So if $W=V$, then only $f\in A$ vanishing on $W$ is $f=0$ and this gives the generic point $\eta$ of $V$.
In your case, the generic point of $V=V(y-x^2)=Spec(k[x, y]/(y-x^2))=Spec(A)$ is thus $\eta =[ (\bar 0)]$.

2) However in your example, you have a closed immersion $j:V\hookrightarrow \mathbb A^2_k$ corresponding to the quotient morphism $k[x,y]\to A=k[x, y]/(y-x^2)$.
The immersion $j$ sends $\eta \in V$ to $H=j(\eta)\in \mathbb A^2_k$, the point of $\mathbb A^2_k$ corresponding to the prime (but not zero!) ideal $(y-x^2)\subset k[x,y]$.
It is good hygiene to conceptually distinguish between $\eta\in V$, with ideal $(\bar 0)\in A$, and $H\in \mathbb A^2_k$ , with ideal $(y-x^2)\subset$k[x,y].
As always, once you clearly understand the distinction, you are welcome to "abuse the language" and go back and forth between both notations.

3) In the above $H$ is capital "eta", not capital "eitch".

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