So, I am given this excercise:
Given $a,b$ and $A$, the $k$-algebra generated by $a,b$ with the relation $a^2=b^3$. Prove that $A$ is not an UFD.
So, I think it all comes down to seeing $A$ as the quotient $B=\frac{k[X,Y]}{<X^2-Y^3>}$. So, if $\phi: B \to A$ is an isomorphism, such that $\phi(\overline{X})=a$ and $\phi(\overline{Y})=b$.
Then all I got to do is prove that $\overline{X}$ and $\overline{Y}$ are irreducible, hence $a$ and $b$, hence $a^2=b^3$ is not an unique factorization.
Now I have tried proving that if $\overline{X}=\overline{fg}$, then one of those is inverible, but all I could do is a bunch of calculations that seemed endless and I am pretty sure this is not the way.
What I did was this: if $\overline{X}=\overline{fg}$ then $X-fg=p\cdot(X^2-Y^3)$. For some polynomial $p$.
Now writting $f=f_0+f_1+.....+f_h$, such that $f_i$ is the sum of monomials of degree $i$ (same with $g$).
You can see that either $f_0=0$ or $g_0=0$, lets suppose its $f_0$, and without loss of generality you can suppose $g_0=1$, hence $f_1(X,Y)=X$.
Reasoning like this you can find how $f_0,f_1,f_2,g_0,g_1$ are going to be. (Curious fact: up to $f_2$ all $f_i(X,Y)$are of the form $X\cdot r$, with some $r\in k[X,Y]$)
This is all I have, and it cannot possibly be just adding relations for the $f_i$'s and $g_j$'s until I find something useful.
Any help would be appreciated. Thanks in advanced.
The key to the equations $\overline{X}=\overline{fg}$, $X-fg=p\cdot(X^2-Y^3)$ is to choose f,g with $\deg_X(f),\deg_X(g)<2$ and compare the coefficients ($\in k[Y]$) in $k[Y][X]$.