Radical of Polynomial Quotient Ring I'm working on a problem in Bosch's Algebraic Geometry and Commutative Algebra:

Consider the polynomial ring $K[X,Y]$ over a field $K$ and set $R =$
$K[X,Y]/\left (X-XY^2, Y^3\right )$.  Writing $\overline{X}$,
$\overline{Y}$ for the residue classes of $X$ and $Y$, show that
$\text{rad}(R) = \left (\overline{X}, \overline{Y}\right )$ is the
only prime ideal in $R$ and $R/\text{rad}(R) \simeq K$.

I'm rather lost in showing what the radical of $R$ is.  I was thinking that I could use the fact that $\text{rad}(\mathfrak{a}) = \pi^{-1}\left (\text{rad}\left (R/\mathfrak{a}\right ) \right )$ and instead try to compute $\pi \left (\text{rad}(\mathfrak{a} ) \right ) = \text{rad}\left (R/\mathfrak{a}\right )$, but then I would have to compute the radical of $\left (X-XY^2, Y^3\right )$.  Clearly $Y$ will be an element of the radical, but why must $X$ be?
It's then rather unclear to me why $\left (\overline{X}, \overline{Y}\right )$ is the only prime ideal.  How would I go about showing this?  I suppose the last statement will come from the first isomorphism theorem.
Forgive me if my attempt looks a little scant; I've spent most of my training in differential geometry/Lie groups and so actually have relatively little training in algebra by comparison.
Attempt #1
In response to Viktor Vaughn's comments below, if I set $\mathfrak{a} = \left (X-XY^2, Y^3\right )$ then $V(\mathfrak{a})$ would be to set $Y^3 = 0$ hence $Y=0$, and this forces $X = 0$ for the other equation.  In terms of $\mathbb{A}^2$, would this be the union of the two coordinate axes?  Then we would need $I(V(\mathfrak{a})) = (X,Y)$ since we would need all functions to vanish along both axes.  Is my logic here sound?  I think this forces $\text{rad}\left (R/\mathfrak{a}\right ) = \pi\left (\text{rad}(\mathfrak{a}) \right ) = \left (\overline{X}, \overline{Y}\right )$, but I don't understand why this ideal is prime, and why it's the only prime in $R/\mathfrak{a}$.
 A: Two hints:
(1) What does the vanishing locus $\newcommand{\a}{\mathfrak{a}} \mathbb{V}(\a)$ of the ideal $\a = (X(1-Y^2), Y^3)$ look like as a subset of the plane $\mathbb{A}^2$? Prime ideals of $R$ correspond to irreducible subvarieties of $\mathbb{V}(\a)$.
(2) The radical of $I$ is the intersection of all prime ideals containing $I$. (See here, for instance.) So, given a prime ideal $\newcommand{\p}{\mathfrak{p}} \p \supseteq \a$, can you show that $X \in \p$?

Given a prime ideal $\p \supseteq \a$, then $Y \in \p$ and $X(1-Y^2) \in \p$, so $X \in \p$ or $1 - Y^2 \in \p$. But if $1 - Y^2 \in \p$, then $1 = 1 - Y^2 + Y^2 \in \p$, contradiction. Thus $X \in \p$. Since $\displaystyle \sqrt{\a} = \bigcap_{\substack{\p \supseteq \a}\\ \text{$\p$ prime}} \p$, then $X \in \sqrt{\a}$.
(As an aside, one can actually show that $X \in \a$:
$$
X = X (1-Y^2) \cdot (1+Y^2) + Y^3 \cdot Y \in \a \, .
$$
This shows that $\a = (X, Y^3)$, which makes the radical calculation easy.)
Now
\begin{align*}
\frac{R}{\sqrt{\a}} &= \frac{K[X,Y]/\left(X-XY^2, Y^3\right)}{(X,Y)/\left(X-XY^2, Y^3\right)} \cong \frac{K[X,Y]}{(X,Y)} \cong K
\end{align*}
by the third isomorphism theorem. By the lattice isomorphism theorem, the prime ideals of $R$ are in bijection with the prime ideals of $K[X,Y]$ containing $\a$. Since prime ideals are radical, given a prime ideal $\p \supseteq \a$, then $\p \supseteq \sqrt{\a} = (X,Y)$. But $(X,Y)$ is maximal (the quotient is a field), so $\p = (X,Y)$.
