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Let $X = \operatorname{Spec} \mathbf{C}[[x,y]]/(y^{2} - x^{3} - x^{2})$. I would like to describe $X$ set-theoretically. My questions are: Can one explicitly say what the elements in $X$ are? Is it possible to interpret them geometrically? And is $X$ irreducible?

I am not really sure where to get started. Any help would be much appreciated.

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I don't have a problem with it. Maybe you should see this thread on meta. – Ivo Terek Jul 30 '14 at 18:25
up vote 30 down vote accepted

Geometrically you are considering germ of singular node at the origin, that is at the singularity. The node is $ N = \textrm{Spec} ( \mathbf{C}[x,y]/(y^{2} - x^{3} - x^{2})) $ and is irreducible affine curve, because polynomial $y^{2} - x^{3} - x^{2}$ is irreducible in polynomial ring $\mathbf{C}[x,y]$. However your germ $X$ is reducible: this is because polynomial $y^{2} - x^{3} - x^{2}$ become reducible in formal powers series ring : $\; y^{2} - x^{3} - x^{2}=(y-x\sqrt{1+x})(y+x\sqrt{1+x})\in \mathbf{C}[[x,y]]$ where $\sqrt{1+x}= 1+\frac{1}{2}x+...$ can be developed by Newton binomial in $\mathbf{C}[[x,y]]$.

So in scheme sense $X$ contains three points: origin (= singularity )and two generic points of two irreducible components of $N$.

This is very intersting because node remains irreducible in every neighbourhood of origin in affine plane $\textrm{Spec} ( \mathbf{C}[x,y])$ but by going to formal series you obligate curve to split in two components. So intuition should be that going to formal series is strong form of localization.

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+1: this is a really excellent answer: informative, concise, clear. I look forward to more of the same! – Pete L. Clark Dec 5 '10 at 23:46

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