# Prove that a regular map is not an isomorphism

Problem: Let $$X$$ be an algebraic curve defined by $$y^2= x^3$$ and $$f(t) =(t^2, t^3)$$ the regular map $$\mathbb{A}^1 \to X$$. Prove that $$f$$ is not an isomorphism. What are the rational functions in $$t$$ that correspond to the functions in the local ring of the point $$(0,0)$$?

My attempt: I tried to construct the inverse as $$P(x) + Q(x)y$$ with $$P(x), Q(x)$$ polynomials, because every regular function on $$X$$ can be written uniquely in this form, but I have trouble constructing the inverse. By writing $$x= t^2$$ and $$y= t^3$$ and composing the two functions I got $$P(t^2) + Q(t^2)t^3$$ and I think I should prove that this last expression is not equal to $$t$$ for no $$P$$ or $$Q$$, but I’m stuck.

For the last question, I wrote the local ring $$O_x= \left\{ \frac{P_1(x) + Q_1(x)y}{P_2(x)} \right\}$$, with $$P_2$$ having a nonzero constant term. How do I express now the rational functions in $$t$$ corresponding to that?

• Hint: $P(t^2)=\cdots+p_2t^4+p_1t^2+p_0$. Do the same thing for $Q$. This should make whether you can get $t$ out of this clear. Your local ring at the end has big issues: why not try writing down the local ring in terms of $x$ and $y$ and then making the substitution? Commented Sep 4, 2020 at 18:46
• I don’t know if I’m catching the hint, but by writing out the polynomials as you suggested, it looks like $t$ cannot be a term of the expression, since there are only higher powers of $t$
– cip
Commented Sep 4, 2020 at 19:33
• For the local ring: do you mean that the expression should be more explicit in $x$ and $y$ or is it wrong all the way?
– cip
Commented Sep 4, 2020 at 19:36
• Yes, $t$ can't appear in that expression, so it can't be equal to $t$. For the local ring, proceed as in your previous question with $x$ and $y$, and then make the substitution to write things in terms of $t$. Commented Sep 4, 2020 at 19:44