Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a (real) plane, I know the definition of tangent space for irreducible varieities. (If $P \in V(f_1, \cdots, f_n)$ irreducible, then $T_P V = V(f^{(1)}_{1,P},\cdots, f^{(1)}_{n,P})$.

Suppose $V=V_1\cup V_2$ is an irreducible decomposition. Then by taking the irreducible components, one can define tangent spaces for points not in intersection. But if $P \in V_1 \cap V_2$, what is the definition?

For an easy examples, let $V(xy)=V(x) \cup V(y)$. If we follow the definition in the above paragraph, then tangent spaces for $(0,t) \ (t\neq 0)$ will be $V(x)$, similarly for x-axis gives $V(y)$. What's for origin?

It should be the whole plane as I take a look at some books. But I don't know why. False attempt: $O \in V(x,y)$ so that tangent space is just $V(x,y)=O$, a point. What's the problem?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

To answer your question directly, the definition does not care which point we choose on $V.$ Essentially, we are just computing the Taylor expansions of the defining equations at whatever point we like, and using the degree one part to define the tangent space.

By your definition, the tangent space of $V(xy)$ is spanned by the differential $dF$ of $F(x,y)=xy$ at the origin. We compute $dF = \frac{\partial F}{\partial x}(0,0)x+\frac{\partial F}{\partial y}(0,0)y=y|_{(0,0)}x+x|_{(0,0)}y=0,$ which is not surprising, since $F$ is a form of degree two. Thus, the tangent space of $V(xy)$ in $\Bbb A^2$ has defining equation $0,$ i.e. the whole $\Bbb A^2$ is tangent to $V(xy)$ at the origin.

Alternatively, we can proceed to compute the Zariski tangent space.

The Zariski cotangent space to $V(xy)$ at $(0,0)$ is defined to be $\mathfrak m/\mathfrak m^2,$ where $\mathfrak m$ is the maximal ideal of the local ring $\mathcal O_{V(xy),(0,0)}.$ We have

$$\mathcal O_{V(xy),(0,0)}=\left(\dfrac{k[x,y]}{(xy)}\right)_{(x,y)}=\dfrac{k[x,y]_{(x,y)}}{(xy)k[x,y]_{(x,y)}},$$ so

$$ \mathfrak m= (x,y)\left(\dfrac{k[x,y]}{(xy)}\right)_{(x,y)}=\dfrac{(x,y)k[x,y]_{(x,y)}}{(xy)k[x,y]_{(x,y)}},$$

by the fact that localization commutes with taking quotients. So the cotangent space at the origin is given by $$\mathfrak m/\mathfrak m^2=\dfrac{(x,y)k[x,y]_{(x,y)}}{(xy)k[x,y]_{(x,y)}}/\left(\dfrac{(x,y)k[x,y]_{(x,y)}}{(xy)k[x,y]_{(x,y)}}\right)^2=\dfrac{(x,y)k[x,y]_{(x,y)}}{(xy)k[x,y]_{(x,y)}}/\dfrac{(x,y)^2k[x,y]_{(x,y)}}{(xy)k[x,y]_{(x,y)}}=\dfrac{(x,y)k[x,y]_{(x,y)}}{(x,y)^2k[x,y]_{(x,y)}}$$ where the last equality is by the second isomorphism theorem. Again, using the commutativity of localization and quotients we get

$$\mathfrak m/\mathfrak m^2=\left((x,y)/(x,y)^2\right)_{(x,y)}= kx\oplus ky\cong k^2.$$ Taking the dual gives us also that $T_{(0,0)}V(xy)\cong k^2.$

Note that in the first definition, $T_{(0,0)}V\subseteq\Bbb A^2,$ while in the second the definition is intrinsic, not depending on any particular choice of embedding.

share|improve this answer
    
Really nice answer thanks. I don't need to worry about irreducible things. –  Gobi Dec 10 '12 at 19:15
    
Dear @Gobi, that's correct. You're welcome! –  Andrew Dec 10 '12 at 19:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.