# An exercise from Shafarevich (regular functions agreeing at a pt <=> tangent spaces are equal)

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 $C_1$.

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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
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.
\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
@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