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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I have two varieties;
$V=V(y-x^2)$ and $W=V(y^2-x^3)$

$\phi: V \mapsto W$; $(x,y) \mapsto (y,xy)$
$\phi^*: C[W] \mapsto C[V]$; $[f] \mapsto [f o \phi]$

This is the morphism i'm looking at;
$\gamma: V \mapsto W$ by $(x,y) \mapsto (x^2,xy)$

Can anyone tell me what $\gamma(x,y)$ is? Also, how do you compute $\gamma^*:C[W] \mapsto C[V] $? Apparently it is the same as $\phi^*$

share|cite|improve this question
You must be losing something in translation, because as written, clearly $\gamma = \phi$. – Michael Joyce May 16 '12 at 11:39
thank you, latex language is hard to type correctly ! – Kiv Efehe May 16 '12 at 11:43
What is the relation between the functions $y$ and $x^2$ on $V$? (Hint: As written, you still have $\gamma = \phi$ and so therefore $\gamma^* = \phi*$, even though the way you write down the two ring homomorphisms may be different.) – Michael Joyce May 16 '12 at 11:53
It is important to distinguish between $x$ and $y$ as elements of the polynomial ring $\mathbb{C}[x,y]$ and the elements $\overline{x}$ and $\overline{y}$ as elements of $\mathbb{C}[V] \cong \mathbb{C}[x,y]/(y - x^2)$. Here $\overline{x}$ refers to $x \pmod{(y - x^2)}$ and similarly for $\overline{y}$. Note that it is not true that $y = x^2$ in $\mathbb{C}[x,y]$, but it is true that $\overline{y} = \overline{x}^2$ in $\mathbb{C}[V]$. (cont.) – Michael Joyce May 16 '12 at 12:24
Geometrically, this says that the functions $y$ and $x^2$ are different functions on the entire affine plane, but they restrict to the same function on the parabola $V$ given by the equation $y = x^2$. – Michael Joyce May 16 '12 at 12:27

Your Answer


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

Browse other questions tagged or ask your own question.