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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 torsionless.

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