TITLE: Basic application of the Nullstellensatz
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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 $a = X^2$.