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Let $k$ be a field.

Consider the inverse limit

$\varprojlim k[x,y]/(y\cdot x^n)$.

I wonder if there is a nice description of this ring?

Geometrically, we look at the union of the line $y=0$ along with an infinitesimal neighborhood of the line $x=0$. But what happens in the limit? I think we get $k[y][[x]]$? What is the geometric interpretation of this?

Any help will be appreciated.

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Dear anonymous, I'm not convinced that the projective limit is $L=k[y][[x]]$. What would be the image of $1+x+x^2+...\in L $ in $ k[x,y]/(y\cdot x^n)$ ? But it is a nice question:+1 –  Georges Elencwajg Dec 30 '11 at 20:19
    
Thank you for your comment. I now think you are right. I wonder what is the answer. –  anonymous Dec 30 '11 at 20:26
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1 Answer

up vote 6 down vote accepted

Let $R=\varprojlim k[x]/(x^n) = k[[x]]$, the ring of power series in $x$. The natural map $k[x]\rightarrow R$ is injective.

Then your limit is the subring of $R[y]$ with constant terms in $k[x]$.

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Can you please clarify what do you mean by "constant terms in k[x]"? –  anonymous Dec 30 '11 at 20:29
3  
In general, if $R'\subset R$ is a subring of $R$, then there us a subring of $R[y]$ consisting of polynomials with coefficients in $R$ with the additional restriction that the constant term is in $R'$. In this case $R'=k[x]$ and $R=k[[x]]$. –  Thomas Andrews Dec 30 '11 at 20:39
    
Thank you. I still wonder if there is a reasonable geometric interpretation of this result. –  anonymous Dec 30 '11 at 20:57
    
@anonymous: Geometrically, you give yourself a function defined on the $x$-axis, the $y$-axis, and on some neighbourhood of the $y$-axis minus origin. You identify a function with the one obtained by shrinking the neighbourhood of the punctured $y$-axis.The projective limit describes the equivalence classes of these functions. –  Georges Elencwajg Dec 30 '11 at 21:52
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