Basic application of the Nullstellensatz Background: I have just started learning basic algebraic geometry. My solution to a simple problem involves an application of the Nullstellensatz and I want to know whether this is overkill (or perhaps plainly wrong). 
My question: 
I want to show that $k[X,Y]/(Y-X^2) \simeq k[T].$  
I construct the obvious map $\phi: k[X,Y] \to k[T]$ taking $X\to T$ and $Y \to T^2.$ Since this map is clearly surjective it suffices to show that its kernel is the ideal $(Y-X^2).$
Clearly $(Y-X^2) \subset \text{ker}(\phi).$ To get the reverse inclusion, I need to show that any polynomial $p\in k[X,Y]$ with the property that $p(T,T^2)=0$ is in fact of the form $p(X,Y)=(Y-X^2)q(X,Y)$ for $q\in k[X,Y].$
It's here that I want to use the strong Nullstellensatz: 
The algebraic set $\{(t,t^2)\}$ is exactly the zero set of the radical ideal $(Y-X^2) \subset k[X,Y].$  The Nullstellensatz then tells me that any polynomial which vanishes on this set must in fact be an element of the radical ideal. This is just what I need.
Is my argument wrong, or is there a much easier way to answer this question?
 A: This problem is an instance of the (almost trivial) statement:

Let $R$ be a commutative ring and $a \in R$, then $R[Y]/(Y-a) \cong R$.

Just set $R = k[X]$ and $a = X^2$.
A: In short, no, I do not think the proof is complete. 
It looks like you are trying to show $A(\mathbb{A}^1)\cong A(V(y-x^2))$ and then note $A(\mathbb{A}^1)\cong k[t]$ and $A(V(y-x^2))\cong k[x,y]/(y-x^2)$.
Although the claim that the algebraic set $\{(x,y)\in \mathbb{A}^2: x=t, y=t^2\}$ is the zero set of $y-x^2$ is true, how do you know it is exactly the zero set? To do so, you would need to argue the polynomial $y-x^2$ has exactly one irreducible component and is not in fact a union of varieties (i.e. $\mathbb{A}^1\cong V(y-x^2)$). This is one of the maps to do so, the inverse map is still needed.
Furthermore, you must show $A(V(y-x^2))\cong k[x,y]/(y-x^2)$. This isn't particularly difficult. It suffices to show $y-x^2$ is irreducible in $k[x,y]$ which can be done by Eisenstein's criterion (with the prime ideal $(y)$ in $k[y][x]$).
(Another way to approach this problem, without using geometric methods, is by exhibiting two maps $\pi:k[t]\rightarrow k[x,y]/(y-x^2)$ and $\rho:k[x,y]/(y-x^2)\rightarrow k[t]$ which are inverse to each other. This would exhibit an isomorphism.)
A: Yes, one can avoid the use of  Nullstellensatz. Let $I=(Y-X^2)$, $\alpha:k[X,Y]\to k[X,Y]/I$ be a natural homomorphism. Then we have two obviuous facts:


*

*$\alpha(k[X])=k[X,Y]/I$;

*$k[X]\cap I=\{0\}$.
It follows, that restriction $\alpha'=\alpha|_{k[X]}$ epimorphic and injective, hence $\alpha':k[X]\to k[X,Y]/I$ is an isomorphism.
Verification of statement 1. We will use a bar notation for $\alpha$, i.e. $\alpha(x)=\bar x$. Note, that $Y-X^2\in I$, hence $0=\overline{Y-X^2}=\overline{Y}-{\overline{X}}^2$, i.e. $\overline Y={\overline{X}}^2$. Let $f\in k[X,Y]$. Then 
$$
\overline{f(X,Y)}=
f(\overline X,\overline Y)=
f(\overline X,{\overline X}^2)=
\overline{f(X,X^2)}
=\overline{g(X)},
$$
where $g(X):=f(X,X^2)\in k[X]$, hence $\alpha(k[X])=k[X,Y]/I$.
Verification of statement 2. Let $0\neq f\in k[X]\cap I$. Then $f(X)=(Y-X^2)g(X,Y)$ for some $g\in I$. It follows that $f(X)=f(X,X^2)=0$ - a contradiction.
