# Tangent space of quasi-projective varieties

If $X$ is a quasi-projective variety and $X_i,\;\;i=1,\ldots,k\;$ are its irreducible components, then why $$\mathrm{dim}\;T_{X,x}=\mathrm{max}_{i=1,\ldots,k}\;(\mathrm{dim}\;T_{X_i,x})?\qquad \qquad (x\in X)$$

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What definition of $T_{X,x}$ are you using? With at least one reasonable interpretation (i.e. the Zariski tangent space), the statement you give is false.

E.g. if $X = \{(x,y) \in \mathbb A^2 \, | \, xy = 0\}$ (two lines crossing at the origin) then the tangent space of the origin in each component is one-dimensional (since each component is smooth, being a line), but the Zariski tangent to the pair of lines at the origin is two-dimensional.

Let $Z$ be an irreducible component of $X$ passing through $x$. Then $\dim T_{x,Z} \leq \dim T_{x,X}$.
The proof is as follows: The local ring $\mathcal O_{x,Z}$ is a quotient of the local ring $\mathcal O_{x,X}$, by an ideal $I$, say. Then $\mathfrak m_{x,Z} = \mathfrak m_{x,Z}/I,$ and so $\mathfrak m_{x,Z}/\mathfrak m_{x,Z}^2 = \mathfrak m_{x,X}/(I + \mathfrak m_{x,X})^2$ is a quotient of $\mathfrak m_{x,X}/\mathfrak m_{x,X}^2$. Passing to duals, we find that $T_{x,Z}$ embeds as a subspace of $T_{x,X}$, which gives the required bound.
Taking the maximum over all components $Z$ passing through $x$, we find that $$\max_{Z \text{ a comp. of X passing through z}} \dim T_{x,Z} \leq \dim T_{x,Z}.$$ As the example given above shows, this inequality can sometimes be strict.
you are right, then maybe $$\mathrm{dim}\;T_{X,x}\geq\mathrm{max}_{i=1,\ldots,k}\;(\mathrm{dim}\;T_{X_i,x}‌​)?\qquad \qquad (x\in X)$$? – Jacob Fox Jun 13 '11 at 14:48
If $X$ is quasi-projective, my definition of $T_{X,x}$ is $(\mathcal{m}_x/\mathcal{m}^2_x)^*$ where $\mathcal{m}_x$ is the maximal ideal of the stalk in $x$. How can I prove that the disequation hold? – Jacob Fox Jun 13 '11 at 17:03