Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

2 Answers 2

up vote 1 down vote accepted

$V(y-x^2)$ 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.

share|improve this answer

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".

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.