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.

This exercise comes from Shafarevich II.3.6.

Let $\varphi:\mathbb{P}^2 \to \mathbb{P}^2$ be the rational map defined by $\varphi(x_0 : x_1 : x_2) = (x_1x_2:x_0x_2:x_0x_1)$. Consider the point $x = (1:0:0)$ and a curve $C$ that is nonsingular at $x$. The map $\varphi$ restricted to $C$ is regular at $x$ (since the codimension of non-regular points is at least 2), and maps $x$ to some point $\varphi_C(x)$. Prove that $\varphi_{C_1}(x) = \varphi_{C_2}(x)$ if and only if the two curves $C_1, C_2$ touch at $x$ (that is, their tangent spaces are equal at $x$).

I'm having trouble getting a handle on this problem. I want to say something like, "this map maps $x$ to the tangent line at $x$," but I don't know how to view it as such if I don't know anything about the curve.

Another thing I've though about, but haven't been able to make headway on, is blowing up $\mathbb{P}^2$ at $x$, and trying to say something about the preimage of the tangent line to $C_1$.

share|improve this question
    
There is something in the setup of this that I don't understand. If $\varphi$ is defined at the start, and then $\varphi_{C}$ is the restriction to $C$, then how is it possible that $\phi_{C_1}(x)\neq \phi_{C_2}(x)$? To me these should always be equal, because they are both just the restriction of the same map and then applied to the same point. –  Matt Nov 2 '12 at 5:31
    
@Matt: $\varphi$ is not defined at $x$. –  user18119 Nov 2 '12 at 13:29
    
I think the point is that it's not truly a restriction, but an inclusion into the coordinate ring of those curves. –  JeremyKun Nov 2 '12 at 14:20

1 Answer 1

up vote 1 down vote accepted

Let $f(u, v)$ be a defining equation of $C_1$ and suppose for instance that $\frac{\partial f}{\partial u}(0,0)\ne 0$. Let $\mathfrak m_0$ be the maximal ideal of the local ring of $O_{C_1}$ at $(0,0)$. Then $\mathfrak m_0$ is generated by $v$ and we can write $$u=\lambda v (1+\epsilon), \quad \epsilon=vh \in \mathfrak m_0$$ for some constant $\lambda$. At any point $(1:t_1:t_2)\in C_1$ with $t_2\ne 0$, we have $$\varphi(1:t_1:t_2)=(t_1t_2:t_2:t_1)=(t_1:1:t_1/t_2)=(t_1:1:\lambda(1+t_1h(t_1,t_2)).$$ So $$\varphi_{C_1}(1:0:0)=(0:1:\lambda).$$ As $\lambda$ represents the slop of the tangent line to $C_1$ at $(1:0:0)$, this proves the statement you are after.

share|improve this answer
    
\mathbb{P}^2 has three coordinates; are you defining $C_1$ just in the affine slice containing $x$? –  JeremyKun Nov 2 '12 at 14:26
    
@Bean: yes I am working in the affine chart $x_0\ne 0$. –  user18119 Nov 2 '12 at 14:30
    
why is m generated by v? –  JeremyKun Nov 2 '12 at 15:39
    
@Bean: The quotient ring $k[u,v]/(f(u,v), v)$, where $k$ is the base field, is $k$ because in the quotient $v=0$, and $f(u, 0)=0$ which implies that $u=0$ because $f(u,0)=cu+$ (termes of higher degrees in $u$) with $c=\partial f/\partial u (0,0) \ne 0$. Therefore $(f(u,v), v)$ is maximal, and in $O_{C_1}$, this maximal ideal becomes $(v)$. –  user18119 Nov 2 '12 at 21:47

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.