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Could someone please explain to me why $X$ and $Y$ are generic points of $\mathbb{R}[X, Y]/(XY)$?

And why is the ideal generated by irreducible polynomial is a generic point in $\mathbb{R}[X, Y]$?

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All this follows immediately from the definitions of irreducible element, prime ideal, spectrum, and generic point. Which part of the definitions do you have trouble with? –  Zhen Lin Dec 12 '11 at 11:02
    
X is is generic point of ℝ[X, Y]/(XY) it means V(X)= Spec(ℝ[X, Y]/(XY))? –  Name Dec 12 '11 at 11:17
    
help me please, I don't see the solution –  Name Dec 12 '11 at 11:59

1 Answer 1

up vote 5 down vote accepted

A scheme $S$ has a generic point if and only if its underlying topological space $|S|$ is irreducible, in which case there is a unique point $\eta\in S$ such that $\overline { \lbrace \eta \rbrace}=|S|$.

If $S=Spec(A)$ is an affine scheme, irreducibility amounts to the condition that $Nil(A)$, the nilradical, be prime or equivalently that the reduction $A_{red}=A/Nil(A)$ be a domain.
In your case $A=\mathbb R[X, Y]/(XY)=\mathbb R[x,y] \;$ is already reduced but is not a domain , so that $Spec(A)$ has no generic point.
End of story? No!

If a scheme $S$ is not irreducible, $|S|$ has a decomposition into irreducible components $S=\bigcup S_i$ , each $S_i$ having a dense point $\eta_i$. Those $\eta_i$ are called maximal points or even (by "abuse of language") generic points of $S$.
In the affine case, they correspond to the minimal ideals ${\mathfrak p_i}\subset A$.
In your case you have two maximal points $\eta_x, \eta_y$ corresponding to the only two minimal ideals $(x),(y)$ of $k[x,y]$.
They are the generic (=dense) points of the lines $V(x)$ and $V(y)$, which are the irreducible components of your scheme $S=Spec(k[x,y])$

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So (X) is generic point for V(x) and by abuse of language, we say (X) is generic point for ℝ[X, Y]/(XY)?. –  Name Dec 12 '11 at 12:59
    
Dear @ name: yes, this is correct. To be absolutely precise it is better to write "$(x)$ is the generic point for $V(x)$ " , with a small letter $x$, so as to carefully distinguish between the polynomial $X\in k[X,Y]$ and its class $x$ in the quotient$k[x,y]=k[X,Y]/(X.Y)$ –  Georges Elencwajg Dec 12 '11 at 13:17
    
Thank so much for your help Georges Elencwajg –  Name Dec 12 '11 at 13:21

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