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Let $W=V(y^{2}-x^{3}) \subseteq \mathbb{A}^{2}$ and $k$ algebraically closed. Clearly the dimension of the tangent space at the origin is $2$. I want to compute this using the definition the fact that $\operatorname{dim} T_{(0,0)}W=\operatorname{dim}_{k} \mathfrak{m}/\mathfrak{m}^{2}$. Where $\mathfrak{m}$ is the maximal ideal of the local ring $\mathcal{O}_{(0,0),W}$.

OK according to Hartshorne the local ring is isomorphic to the localization of the corresponding coordinate ring localized at the ideal $(x,y)$ right?

So we need to compute the maximal ideal of $k[x,y]/(y^{2}-x^{3})$ localized at $(x,y)$. I know that in general given a ring $A$ then if $\mathfrak{p} \in \operatorname{Spec}(A)$ we have $\mathfrak{p}A_{p}=\{\frac{g}{h}: g,h \in P\}$ is the unique maximal ideal. However I don't see how to simplify things here. How to compute this?

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Well, $W$ is irreducible, so it has a unique generic point, and so every local ring embeds as a subring of the field of rational functions. Since $W$ is an algebraic curve, the field of rational functions is going to be a certain finite extension of the field $k(t)$. (You can think of $t = y / x$ in this case.) Does this help? – Zhen Lin May 23 '12 at 21:19
Be careful with your definition, $\mathfrak{p} A_{\mathfrak{p}} = \{ \frac{g}{h} : g \in \mathfrak{p}, h \notin \mathfrak{p} \}$. The idea is to show that the images of $x$ and $y$ are a basis of $\mathfrak{m} / \mathfrak{m}^2$ as a $k$-vector space. It is clear they span, so you have to show that they are linearly independent. – Michael Joyce May 23 '12 at 21:21
up vote 1 down vote accepted

If $M=(x,y)\subset k[X,Y]/(Y^2-X^3)=k[x,y]\stackrel{\text {def}}{=}A \;$, the maximal ideal you are interested in is $\mathfrak m=MA_M \subset A_M$ the maximal ideal of the local ring $A_M=\mathcal O_{(0,0),W}$.
The $k$-vector space $\mathfrak m/\mathfrak m^2$ is generated by $\bar x$ and $\bar y$ and all you have to check is that they are linearly independent.
A linear dependence relation $q\bar x+r \bar y=0$ ($q,r\in k$) means $qx+ry \in \mathfrak m^2=(x^2,xy,y^2)$.
Lifting to actual polynomials this means $$qX+rY\in (X^2,XY,Y^2)+(Y^2-X^3)=(X^2,XY,Y^2)\subset k[X,Y]$$ and implies $q=r=0$.
So $\bar x$ and $\bar y$ are linearly independent in $\mathfrak m/\mathfrak m^2$ and $dim_k(\mathfrak m/\mathfrak m^2)=2$

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I don't understand why we aren't localizing? isn't by (a result that appears in Hartshorne's book says that the localization at $\mathfrak{m}_{p}$ is isomorphic to the local ring $\mathcal{O}_{p}$) sorry I'm new with this and getting confused. – user31509 May 23 '12 at 21:44
Dear user31509, we are localizing. I have edited the second line in order to emphasize that. It is true that in the calculations the fact that the elements of the localization are fractions does not appear explicitly. – Georges Elencwajg May 23 '12 at 22:09

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