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Let $X$ be a nonsingular variety over an algebraically closed field and $x$ be a closed point on $X$.

Then one defines the tangent space in $x$ as the $k-$ vector space

$T_x(X):= \operatorname{Hom}_k(\mathfrak m / \mathfrak m^2, k)$,

where $\mathfrak m$ is the maximal ideal of the point. Now consider the $k-$vector space $\mathfrak m / \mathfrak m^2$: you can also see it as just an $\mathcal O_{X,x}$ - module.

Now the question: under the above hypotheses: is $\mathfrak m / \mathfrak m^2$ already as such $\mathcal O_{X,x}$ - module free?

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1 Answer 1

up vote 3 down vote accepted

No, if $\dim(X)\gt 0$ the $\mathcal O_{X,x}$- module $m/m^2$ is definitely not free, because as a $k$-vector space $m/m^2$ is finite dimensional, whereas $\mathcal O_{X,x}$ is infinite dimensional.
By the way, the non-singularity hypothesis is irrelevant in the above.
(That hypothesis will however guarantee that $\dim(X)=\dim_k(m/m^2)$, which is interesting, but not what you asked about).

Another argument
Fix a non-zero element $x\in m$. It will kill all of the module $m/m^2$. In other words that module is completely torsion, whereas a non-zero free module over the domain $\mathcal O_{X,x}$ is torsion-free.

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@user26857. Could you please abstain from making such nitpicking changes to my posts when they have no mathematical content: I'll handle Tex myself, thank you. I think you have more useful contributions to bring to this site. –  Georges Elencwajg Mar 9 at 20:19
    
Torsionless is one thing, and torsion-free is another thing. Unfortunately you mixed them twice, so these edits are far from a nitpick. –  user26857 Mar 9 at 21:40
    
@user26857: maybe, but then I think it would be more courteous to inform me of the fact and leave to me the the decision to make the change. Anyway, I'm asking you to do that in the future. –  Georges Elencwajg Mar 9 at 21:48

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