What are the elements of $k[X,Y]/(X^2-Y^3)$ like? What are the elements in $k[X,Y]/(X^2-Y^3)$ like, where $k$ is a field?
For example, in $k[X]/(x^2+2x+3)$, all elements are of a degree lower than $2$. But I can't quite figure out the multi-variable case. 
My first guess was that we could treat $Y$ as a constant and ensure that all the elements of $k[X,Y]/(X^2-Y^3)$ had their degree of $X$ as less than $2$. $Y$ obviously could then have any degree. 
But then again, we could do also ensure that all elements had their degree of $Y$ as less than $3$, letting $X$ take any degree. 
Having two representations for the elements of $k[X,Y]/(X^2-Y^3)$ sounds a little spurious to me. 
Any help would be much appreciated.  
 A: There is nothing spurious here, and both of your representations of the elements are perfectly fine.  In general, choosing a basis for the elements of a $k$-algebra will depend on some arbitrary choices.
For example, one standard way of choosing a basis is the technique of Gröbner bases, which depends (in your example) on a choice of order for the monomials of $k[X, Y]$.  There are different choices, which give different bases.  And in fact, the two representations you give can be regarded as coming from two different Gröbner bases of your example.
A: You can think of $k[x,y]/(x^2 - y^3)$ as the subring $k[t^2,t^3]$ inside $k[t]$. 
Let $k[x,y] \mapsto k[t]$ by $x \mapsto t^3$ and $y \mapsto t^2$, and more generally $f(x,y) \mapsto f(t^3,t^2)$.  This substitution map is a $k$-algebra homomorphism and its image is clearly $k[t^2,t^3]$. Check that the kernel is precisely the multiples of $x^2 - y^3$ in $k[x,y]$, so we get the desired $k$-algebra homomorphism. 
Note $k[t^2,t^3]$ is strictly smaller than $k[t]$ since it does not contain $t$. The ring $k[t^2,t^3]$ is $k + t^2k[t]$: all polynomials in $k[t]$ with linear coefficient equal to $0$.  
