Show that the map $F: \Bbb A^1 \to \Bbb V(y-x^2) \subset \Bbb A^2, t \longmapsto (t,t^2)$ is an isomorphism from the line to a parabola. 
Show that the map $F: \Bbb A^1 \to \Bbb V(y-x^2) \subset \Bbb A^2, t \longmapsto (t,t^2)$ is an isomorphism from the line to a parabola.

I've learned about morphisms between affine varieties and the pullback and this is an example problem where I think I should use the fact that if $V$ and $W$ are affine varieties in $\Bbb A^n$, then $V \cong W \iff \Bbb C[V] \cong \Bbb C[W]$, here $\Bbb C [V]$ and $\Bbb C[W]$ are the coordinate rings of $V$ and $W$.
So if I define the pullback $F^\#: \Bbb C[x,y]/(y-x^2) \to \Bbb C[t]$ such that $x \longmapsto t, y \longmapsto t^2$ I have an surjective algebra homomorphism with kernel $\ker F^\# =\{p \in \Bbb C[x,y]/(y-x^2) \mid F^\#(p)=p(t,t^2)=0 \}$.
I think I should use the isomorphism theorem for algebras here to get something like $$\Bbb C[x,y]/(y-x^2) \Big / \ker F^\# \cong \Bbb C[t]$$ but I don't know what this kernel is and similarly I don't know how to handle this "double quotient". Any hints what to do here?
 A: First, here's what I'd call the "right" way to do this problem. Define
\begin{align*}
G: \mathbb{V}(y-x^2) &\to \mathbb{A}^1\\
(x,y) &\mapsto x \, .
\end{align*}
$G \newcommand{\Ft}{\tilde{F}}$ is a polynomial map, hence is a morphism, and one can easily check that $F$ and $G$ are mutually inverse, using the fact that a point on the parabola is of the form $(x, x^2)$. Thus $F$ (and $G$) is an isomorphism.
If you'd prefer to work with coordinate algebras, we can similarly define the map
\begin{align*}
G^\#: \mathbb{C}[t] &\to \mathbb{C}[x,y]/(y-x^2)\\
t &\mapsto x
\end{align*}
and show that $F^\#$ and $G^\#$ are mutually inverse.
To pursue your desired approach and show that $\ker(F^\#) = 0$: consider the composition
\begin{align*}
\Ft : \mathbb{C}[x,y] \overset{\pi}{\to} \mathbb{C}[x,y]/(y-x^2) \overset{F^\#}{\to} \mathbb{C}[t]
\end{align*}
with the quotient map $\pi$. So $\Ft$ is just the evaluation map $x \mapsto t, y \mapsto t^2$. We now apply the following lemma. (See here for a proof.)
Lemma.
Let $R$ be a unital commutative ring and $R[x]$ be the one-variable polynomial ring over $R$.  For $\alpha \in R$, let
\begin{align*}
\varphi = \text{eval}_\alpha: R[x] &\to R\\
x &\mapsto \alpha
\end{align*}
be the evaluation homomorphism.  Then $\ker(\varphi) = (x-\alpha)$.  Moreover, the induced map $\overline{\varphi}: R[x]/(x-\alpha) \to R$ is an isomorphism.
Considering $\mathbb{C}[x,y]$ as $(\mathbb{C}[x])[y]$, we get that
\begin{align*}
\frac{\mathbb{C}[x,y]}{(y-x^2)} &\overset{\sim}{\to} \mathbb{C}[x]\\
y &\mapsto x^2
\end{align*}
is an isomorphism, which, after changing $x$ to $t$, is exactly the map $F^\#$.
One can also show that the kernel is $(y-x^2)$ more directly, as done in this answer.
