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Let $V = V(X^2-Y^3,Y^2 - Z^3)$. Let $P= (0,0,0)$ that is a point in $V$. I want to compute $\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2$ where $\mathfrak{m}_P$ is the maximal ideal in the local ring $$ \mathcal{O}_P(V) := k[V]_{(\overline{X},\overline{Y},\overline{Z})}. $$

By Lemma 1.31 in Milne's notes here, it sufices to compute the dimension of $I/I^2$ where $I = (\overline{X},\overline{Y},\overline{Z})$ since $\mathfrak{m}_P = I_{(\overline{X},\overline{Y},\overline{Z})}.$ Now on one hand it is clear to me that $\dim_k I/I^2 = 3$. However suppose we consider the map $$\begin{eqnarray*} \alpha: &k[X,Y,Z]& \to k[T] \\ &(X,Y,Z)& \mapsto (T^9,T^6,T^4). \end{eqnarray*}$$

This induces an isomorphism $\overline{\alpha}$ between $k[V]$ and a subring of $k[T]$ since $$I(V(X^2-Y^3,Y^2 - Z^3))= (X^2-Y^3,Y^2 - Z^3).$$

The image of $I$ under $\overline{\alpha}$ is $(T^4,T^6,T^9) = (T^4)$ while $\overline{\alpha}(I^2) = (T^8)$. This means that

$$\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2 = \dim_k (T^4)/(T^8) = 4.$$

My question is: How can I get two different answers by trying to compute the dimension in two different ways?

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2 Answers 2

up vote 2 down vote accepted

Your variety $V$ is a curve whose only singularity is the origin $O=(0,0,0)$.

Its desingularization (=normalization) is the morphism $$a:\mathbb A^1_k\to V: t \mapsto (x,y,z)=(t^9,t^6, t^4)$$ whose associated comorphism is your $k$-algebra morphism $\alpha: k[V] \to k[T]$.

The crucial point is that $a$ is a bijective morphism but not an isomorphism: the inverse rational map $$a^{-1}:V \to \mathbb A^1_k:(x,y,z )\mapsto t=\frac {x}{z^2}$$ is an isomorphism between the open subsets $V\setminus \{O\}\subset V$ and $\mathbb A^1_k \setminus \{0\}\subset \mathbb A^1_k$, but is not regular at $O$.

Hence the $k$-algebra morphism $ \alpha$ is not an isomorphism (since $a$ is not an isomorphism), it does not induce an isomorphism of $\mathfrak{m}_P/\mathfrak{m}_P^2$ with $(T^4)/(T^8)$ and thus, as already remarked by Zhen, there is no contradiction.

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Thanks for your answer! Although I am curious to know why this results in no contradiction; it seems to me that the computation of $\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2$ is a purely algebraic thing, similarly $\dim_k I/I^2$. Why does the inverse not being regular mean $\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2 \neq \dim_k I/I^2$? – user38268 Mar 18 '13 at 21:09
Dear Benja, it is perfectly true that $\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2=\dim_k I/I^2 = 3$. What I tried to explain is that on the other hand $\dim_k(T^4) /(T^8)=4$, but that there is no contradiction because there is no isomorphism relating the corresponding algebras, and I tried to show geometrically why no such isomorphism should be expected. – Georges Elencwajg Mar 18 '13 at 21:53
Dear Georges, sorry I mistyped my last line, it should read instead: "Why does the inverse not being regular mean $\dim_k I/I^2 \neq \dim_k (T^4)/(T^8)$?". – user38268 Mar 18 '13 at 22:08
Dear Benja, if you have a morphism of algebras, like here $\alpha: k[V] \to k[T]$, which is not an isomorphism you can't really use dimensions of vector spaces on the right like $(T^4)/(T^4)$ to calculate dimensions of vector spaces on the left like $I/I^2$. – Georges Elencwajg Mar 18 '13 at 23:11
I realise the mistake I made: The equality $(T^4) = \overline{\alpha}(I)$ holds only in the subring of $k[T]$ that is isomorphic to the coordinate ring, similarly with $(T^8)$ and $ I^2$. – user38268 Mar 19 '13 at 10:42

Well, one of the answers is obviously wrong: the tangent space can't have dimension greater than the ambient affine space. The homomorphism $\overline{\alpha} : k[V] \to k[t]$ you construct is also not an isomorphism. (What maps to $t$?)

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