# Tangent space of a quasiprojective variety

I am reading Shaferevich's Basic Algebraic Geometry. In chapter 2, Section 1, he defined tangent space at a point of a quasiprojective variety in the following way:

A quasiprojective variety is open subset of a closed set in some projective space. A quasiprojective variety is called an affine variety if it is isomorphic to some affine closed set. Let $X$ be a quasiprojective variety and $x\in X$. We define the tangent space to $X$ at $x$, $\Theta_{X,x}:=(\mathcal m_x/\mathcal m_x^2)^*$, the dual vector space, where $\mathcal m_x$ is the maximal ideal of the local ring $\mathcal O_{X,x}$. Let $Y\subseteq X$ is a subvariety i.e., $Y$ is also a quasiprojective variety and $x\in Y$ then my question is:

$\textbf{ Why$\Theta_{Y,x}\subseteq \Theta_{X,x}$}$ ?

I have proved this for $X$ and $Y$ both being affine varieties. I also know that for any quasiprojective variety $X$, $\Theta_{X,x}=\Theta_{U,x}$ where $U$ is an affine neighbourhood of $x$ in $X$. Can this two results be used to prove the above claim? Please help me. Thank you.

• It seems like you have already proved everything. Pick an affine neighbourhood $U$ of $x$ in $X$, then $V:=Y\cap U$ is an affine subvariety of $U$ and also an affine neighbourhood of $x$ in $Y$ so you have $\Theta_{Y,x}=\Theta_{V,x}\subseteq\Theta_{U,x}=\Theta_{X,x}$. – Jesko Hüttenhain Apr 27 '17 at 16:39
• @JeskoHüttenhain Why $V$ is affine? And also why $V$ is closed in U? Please note that when I say I have proved it for affine cases I mean I assumed X is an affine closed set and $Y(\subseteq X)$ is closed in $X$. But in case of quasiprojective varieties $Y(\subseteq X)$ is a subvariety means $Y$ is also quasiprojective or equivalently $Y=Z_1 - Z_2$ where $Z_1$ and $Z_2$ are both closed in $X$. – user436053 Apr 27 '17 at 17:32
• Well, $Y$ is closed in $X$ I assume? Then $V=Y\cap U$ is closed in $U$ by definition of the subspace topology. In particular $V$ is affine because it is a closed subset of the affine variety $U$. – Jesko Hüttenhain Apr 27 '17 at 17:54
• @JeskoHüttenhain If $Y$ is closed in $X$ then I totally agree with you. But if $Y$ is not necessarily closed but $Y$ is only a subvariety (whose definition I gave above) is this not true? – user436053 Apr 27 '17 at 18:25
• Ok so you want $Y$ to be an open subset of a closed subset $Y'$ of $X$. We are only left to show $\Theta_{Y,x}=\Theta_{Y',x}$. But that's easy, just choose an affine open neighbourhood $V\subseteq Y\subseteq Y'$ of $x$, then $\Theta_{Y,x}=\Theta_{V,x}=\Theta_{Y',x}$. – Jesko Hüttenhain Apr 27 '17 at 19:24