Showing an ideal is principal.

Question

Is $(x-1,y-1)$ a principal ideal in $\mathbb C[x,y]/(x^2-y^3)$?

What I've reduced this to is -

$$0 = y^3-x^2 = y^3-1+1-x^2$$

Hence $(y-1)(y²+y+1)=(x-1)(x+1)$

I also know that $x+1$ doesn't belong to the maximal ideal $(x-1,y-1)$. Is there a way to say it is actually a unit?

Answer 1

The first solution is wrong

As mentioned in the comment the ring is isomorphic to $\mathbb{C}[t^3,t^2]$ think about $t^2+t+1$ and $t+1$ which are relatively prime . You have $t^2+t+1-t(t+1)=1$ did you figure out? $x^2=y^3$ WHICH also means $\sqrt[3]{x}=\sqrt{y}$ $(x-1)(x+1)=(y-1)(y^2+y+1)$ $(x-1,y-1)=(\sqrt{y}-1)$ to see this first $(\sqrt{y}-1)(\sqrt{y}+1)=(y-1)$ and $(\sqrt{y}-1)(y+\sqrt{y}+1)=\sqrt[3]{y^2}-1=x-1$ to see the other inclusion. $x-1-\sqrt{y}(y-1)=\sqrt{y}-1$ Hence the ideal $(x-1,y-1)=(\sqrt{y}-1)$ but the later is not in our ring so the ideal is not principal. It is easier to see that in terms of $t$ because the ideal $(t^3-1,t^2-1)=(t-1)$ but the later is not in $\mathbb{C}[t^3,t^2]$ so it is not principal.

As Arthur pointed out my solution does not make sense because the ideal is not the same in the bigger ring. Solving that in $R=\mathbb{C}[t^3,t^2]$ is easier suppose on the contrary the ideal is principal $(t^3-1,t^2-1)=(f(t^3,t^2))$ then we get $f\mid t^3-1 $ and $f\mid t^2-1$. But $f\in R$ so it is degree is at most $2$. Also we know that $\mathbb{C}$ is algebraicly closed so $t^3-1, t^2-1$ factors completely and $1$ is the only common root between them. Hence the only option for $f$ in $R$ is to be a unit or $1$.