Map from $\mathbb{A}^1 \rightarrow \mathbb{A}^2$

Question

Let the map $\varphi_n:\mathbb{A}^1 \rightarrow \mathbb{A}^2$ be defined by $t\rightarrow(t^2,t^n)$.

-Show that if n is even, the image of $\varphi_n$ is isomorphic to $\mathbb{A}^1$ and $\varphi_n$ is 2:1 away from 0.

-Show that if n is odd, $\varphi_n$ is bijective, and give a rational inverse of it.

For the even case: I believe that I've shown that $\varphi_n$ is exactly 2:1, and I believe that I've shown that the standard curve given by $y=x^{\frac n2}$ is the image of $\varphi_n$. How do I go about showing that this is isomorphic to $\mathbb{A}^1$?

For the odd case: I'm not really sure where to start here, I used the same process that I did for the even case, and I think the function $t=\frac{y}{x^m}$ where $(x,y)=(t^2,t^n)=(t^2,t^{2m+1})$ for some m. How to show that this is bijective? Thanks for any help you guys can offer!

Answer 1

For the even case, you are showing that the image is isomorphic to $\mathbb{A}^1$. It is not necessary (and not true) that $\varphi_n$ is the isomorphism. In fact, your answer almost contains the map from $\mathbb{A}^1$ to $\varphi_n(\mathbb{A}^1)$ and the map in the other direction. (You will need to check that they are mutual inverses of course).

In the odd case, I think you want to show that it is bijective onto its image ($\varphi_n$ is not surjective onto $\mathbb A^2$ because $|y|$ is determined by $|x|$). Pick a point $(x,y)$ in the image of $\varphi_n$. Using what you said in your question, you need to show that $\frac{y}{x^m} = \frac{t^{2m+1}}{(t^2)^m}$ gives you an inverse to the map $\varphi_n$.

Answer 2

0) First of all you probably want to assume that the base field is algebraically closed: if the base field is $\mathbb R$ and $n=2$ the image of $\phi_2$ is not even algebraic! So let us suppose that $k=\bar k$.

1) $\textbf n=2k$ even
It is clear that the image of $\phi_n$ is the curve $A_n\subset \mathbb A^2$ with equation $y=x^k$, obviously isomorphic to $\mathbb A^1$, since it is the graph of $\mathbb A^1 \to \mathbb A^1: z\mapsto z^k $.
That the map is $2:1$ away from zero is also easy: given a point $(x,x^k)\in A_n$ choose a square root $\xi\in k$ of $x$ and check that the preimage of your point is the set $\{\xi, -\xi\}$ .

2) $\textbf n=2k+1$ odd
The morphism $\phi_n$ is a bijection onto the curve with equation $y^2=x^n$.
The inverse bijection restricts to an isomorphism $\psi_n: C\setminus \{(0,0)\}\to \mathbb A^1\setminus \{0\}:(x,y)\mapsto \frac {y}{x^k}$, which proves that $\phi_n$ is a birational morphism.
However, not only can $\psi_n$ not be regularly extended to the whole of $C_n$, but $C_n$ is not isomorphic to $\mathbb A^1$ under any isomorphism at all because $C_n$ is singular (with $(0,0)$ as a singularity), while $\mathbb A^1$ is regular.