Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$ is a prime ideal. 
Show that the ideal $I$ of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$ is a prime ideal, where $k$ is a field.

I tried to prove it by contradiction. Suppose $f$ and $g$ are not of the form $(X_1^3-X_3)F+(X_2^2-X_3)G$ but $fg$ is, but I don't see why this shouldn't happen.
Can anyone point out how?
 A: Hint. Consider the homomorphism $k[X_1, X_2, X_3] \to k[T]$ sending $X_1$ to $T^2$, $X_2$ to $T^3$, and $X_3$ to $T^6$. What is its kernel? What is its image?
A: Expanding on the "degree" comment (though there MUST be a more elegant way..): Consider the map:
$K[X_1,X_2,X_3] \rightarrow k[T]$ sending $X_1$ to $T^2$, $X_2$ to $T^3$, and $X_3$ to $T^6$. 
Suppose $f$ is in the kernel. Now, if $X_3$ appears anywhere in $f$, we can replace it with $X_1^3 - (X_1^3 - X_3)$. In particular, we should be able to write $f = (X_1^3 - X_3)q + f_1$, where $f_1$ does not have any terms with $X_3$ in them. Similarly, we can write $f_1 = (X_1^3 - X_2^2)q' + f_2$, where in $f_2$, $X_2$ appears at most to degree 1. 
Now, if $f$ was in the kernel then $f_2$ must also be in the kernel. If $f_2 = 0$, then we are done, since we have shown that $f$ is a linear combination of $X_1^3 - X_3$ and $X_1^3 - X_2^2$. If $f_2$ is not zero, pick out the nonzero term $\alpha X_1^a X_2^b$ maximizing $2a + 3b$. Since $b$ is either $0$ or $1$, this term is unique, and this implies that the image of $f_2$ under our map will contain a term that looks like $\alpha T^{2a + 3b}$, but this contradicts our assumption that $f_2$ is in the kernel.
