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Consider the ideals $I = \left< x\right> \cap \left< x,y\right>^2 = \left$ 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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    Thanks Zev and Hurkyl. What I wanted was to write out several examples of ideals that represent $0$-dimensional (any number of) points on $k^2$.2012-06-17

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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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    You are welcome, math-visitor.2012-06-17