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Let $A$ be the local ring $$A=\left(\frac{\mathbb{C}[x,y]}{(y^2-x^3-x^2)}\right)_{(x,y)}$$ $\mathfrak{m}=(x,y)A$ be the maximal ideal in $A$. I'm asked to calculate the dimension of the $\mathbb{C}$-vector space $\mathfrak{m}/\mathfrak{m}^2$ which is the dimension of the cotangent space at $(0,0)$. I do it as follows.

$$ \begin{aligned} \mathfrak{m}/\mathfrak{m}^2&=\left(\frac{(x,y)}{(y^2-x^3-x^2)}\right)_{(x,y)}/\left(\frac{(x,y)^2}{(y^2-x^3-x^2)}\right)_{(x,y)}\\ &=\left(\frac{(x,y)_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}\right)/\left(\frac{(x,y)^2_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}\right)\\ &=\frac{(x,y)_{(x,y)}}{(x,y)^2_{(x,y)}}=\left(\frac{(x,y)}{(x,y)^2}\right)_{(x,y)}\simeq\mathbb{C}^2 \end{aligned} $$ This seemingly correct argument gives the desired result 2, but I realize that I didn't use any information about $(y^2-x^3-x^2)$, and it will be the same at any other point on $y^2-x^3-x^2$ which is impossible. I think there must be some problem about my argument, but I cannot find it out. Where am I wrong? What is the correct way to do it? Thanks in advance.

Y.Guo
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1 Answers1

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I’ll be really sloppy here; you can clean up after me.

Consider the nonsingular point $P=(-1,0)$, so that you want to localize at $(x+1,y)$. Your generators seem to be $x+1$ and $y$, but since $x$ is a unit in the localization, you could generate the maximal ideal just as well by $(x+1)x^2$ and $y$. In other words, by $y^2$ and $y$, thus by $y$ alone.

Lubin
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