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Consider the ideals $I = \left< x\right> \cap \left< x,y\right>^2 = \left<x^3,x^2y, xy^2\right>$ and $J=\left< x,y\right>^3=\left< x^3, x^2y, xy^2, y^3 \right>$ in $k[x,y]$.

What is the geometric difference between $I$ and $J$?

I know that the zero set for $J$ is a triple point at the origin on $k^2$ while the zero set for $I$ is the $y$-axis together with a double point at the origin on $k^2$.

But aren't they both colength $3$ ideals in $k[x,y]$?

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Isn't $$\left<x\right>\cap\left<x,y\right>^2= \left\langle x\right\rangle\cap\big\langle x^2,xy,y^2\big\rangle= \big<x^2,xy,xy^2\big> =\big<x^2,xy\big>\quad?$$ –  Zev Chonoles Jun 17 '12 at 20:53
    
Yes, that's true... which is equal to $\left<x \right>\cap \left< x,y\right>$. So it's the $y$-axis together with a point at the origin with multiplicity $1$? –  math-visitor Jun 17 '12 at 20:54
    
That's different from $\big<x^3,x^2y,xy^2\big>$, which is what you wrote in the question. EDIT: The ideal $\big<x,y\big>$ contains $\big<x\big>$, so $$\big<x\big>\cap\big<x,y\big>=\big<x,y\big>\neq \big<x^2,xy\big>$$ –  Zev Chonoles Jun 17 '12 at 21:04
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@math: You are confusing intersection and product of ideals. e.g. <x> is a subset of <x,y>, so their intersection is just <x>. –  Hurkyl Jun 17 '12 at 21:06
    
Thanks Zev and Hurkyl. What I wanted was to write out several examples of ideals that represent $0$-dimensional (any number of) points on $k^2$. –  math-visitor Jun 17 '12 at 21:07

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up vote 2 down vote accepted

The ideals are hugely different since $\sqrt I=\left< x\right> $ whereas $\sqrt J=\left< x,y\right> $, so that the schemes $V(I)$ and $V(J)$ have dimensions $1$ and $0$: a big difference even at the gross level of their underlying topological spaces!

As to your actual question : the length of of $k[x,y]/I$ is infinite whereas that of $k[x,y]/J$ is $6$.
I think that this is what people mean by colength and I don't understand what you mean when you say that both ideals have colength $3$.

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I thought if $K$ is an ideal that represents $6$ points (not necessarily distinct) on $k^2$, then we say $K$ has colength $6$. I thought I saw that somewhere. –  math-visitor Jun 17 '12 at 21:12
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Dear math-visitor: no, you can't just count "physical" points.For example in $\mathbb A^1_k=Spec(k[t])$, the ideals $I_n=\langle t^n\rangle $ all define subschemes with support the singleton set consisting of the origin, but $I_n$ has colength $dim_k( k[y]/I_n)=n$. If the points are not "physically" distinct it is dangerous to try to count them geometrically: better indeed to use algebraic tools like colength . –  Georges Elencwajg Jun 17 '12 at 21:21
    
I see, so it was a misinterpretation on my part when I saw this concept awhile back. Thank you Georges. –  math-visitor Jun 17 '12 at 21:23
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You are welcome, math-visitor. –  Georges Elencwajg Jun 17 '12 at 21:25

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