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Let $X=V(xy) \subset \mathbb{C}^2$ be the affine algebraic set. (coordinate axes)

How can it be defined the local ring of $X$ at a point $P=(0,b)$ and $Q=(0,0)$?

My textbook only defines local ring for irreducible varieties. So I don't know the case of not irreducible sets.

I thought for $P$, $P$ is in $V(x)$ so the local ring of $V(x)$ at $P$. Is it right? But I have no ideas for the origin.

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2  
How does your textbook define the local ring for irreducible varieties? If you just change "irreducible variety" to "affine algebraic set" in the definition, does it still work? –  Dustin Cartwright Sep 22 '12 at 14:53
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The local ring really is a "local" property, so yes the local ring of $V(x)$ at $P$ is isomorphic to the local ring of $V(xy)$ at $P$. You might want to try calculating it both ways (see Dustin's comment) and check that you get the same thing. –  Matt Sep 22 '12 at 14:55
    
My book's definition: the local ring of $V$ at a point $P$ is the ring of $f \in k(V)$ where $f$ is regular at $P$. (i.e. has a representation $f=g/h$ with $h(P) \neq 0$). –  Gobi Sep 22 '12 at 15:14
    
So if I want to expand the definition, the field $k(V)$ is at first in trouble, since $k[V]$ is not an int dom. –  Gobi Sep 22 '12 at 15:16
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In general, the local ring at $Q$ in a Zariski-closed subset $X$ of $\mathbb{A}^n$ will be $k[x_1, \ldots, x_n] / I(X)$ localised at the maximal ideal $\mathfrak{m}_Q$ corresponding to $Q$. –  Zhen Lin Sep 22 '12 at 15:51

1 Answer 1

up vote 6 down vote accepted

As Zhen mentions in the comments, the local ring of $V(xy)\subseteq\Bbb C^2$ is the localization $\left(\dfrac{\Bbb C[x,y]}{(xy)}\right)_{(x,y)},$ since $(x,y)$ is the maximal ideal corresponding to the point $(0,0).$

An important property of localization is that it commutes with quotients. In this case, we have $\left(\dfrac{\Bbb C[x,y]}{(xy)}\right)_{(x,y)}=\dfrac{\Bbb C[x,y]_{(x,y)}}{(xy)}.$ Since we know that $\Bbb C[x,y]_{(x,y)}$ is the subring of $\Bbb C(x,y)$ consisting of elements $\dfrac{f(x,y)}{g(x,y)}$ satisfying $g(0,0)\neq 0,$ we can see that $\dfrac{\Bbb C[x,y]_{(x,y)}}{(xy)}$ consists of elements $\dfrac{f(x,y)}{g(x,y)}$ where $g(0,0)\neq 0,$ but also where $f(x,y),g(x,y)$ can have no mixed term monomials $x^\alpha y^\beta$ with $\alpha,\beta\neq 0$ with nonzero coefficient.

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