# $A=\frac{\mathbb{C}[X,Y]}{(X^2+Y^2-1)}$ is a PID.

I was given an exercise to show $A=\frac{\mathbb{C}[X,Y]}{(X^2+Y^2-1)}$ is a PID. But I wonder if it is at all true. Note that PID $\implies$ UFD. But we have $$X\cdot X = 1-Y^2 =(1-Y)(1+Y)$$ in $A$ which contradicts UFD.

Is there something wrong in the above factorization? Any suggestions / hints.

Edit: I can prove mechanically it is PID. My main concern was the above factorization. Thanks for the comments.

• See mathoverflow.net/questions/5591/… @ZachTeitler Here the base field is $\mathbb{C}$, which allows you to break $x^2+y^2=(x+iy)(x-iy)$. You will not have this decomposition when the base field is $\mathbb{R}$ or $\mathbb{Q}$. Aug 31, 2017 at 6:11
• This is a good question. I've seen many solutions of some similar questions based on such a quick "argument", but this isn't enough. You have to show that $x,1-y,1+y$ are irreducible and non-associates. But $x$ (for instance) is not irreducible in $A$ since we have $x=\frac{(x+iy+i)(x+iy-i)}{2(x+iy)}$. Aug 31, 2017 at 7:08
• Oh yes, I see. And $1+y = \frac{1}{2}(x-iy)(x+iy+i)^2$. So it's not that $x$ is associated to $1\pm y$, it's that $x$ and $1\pm y$ are reducible. Aug 31, 2017 at 14:46
• The comments seem to solve the question, perhaps any of you should submit that as an answer. Sep 1, 2017 at 4:21
• @YoTengoUnLCD Taking into consideration the great number of upvotes the first comment got I'd say that the question is already answered on Mathoverflow. If this isn't so (which I actually believe) then maybe one of the seven upvoters decide to show us in an answer how that MO thread answers the present question. (I have to admit that I'm not able to figure this out.) Or maybe is the comment itself the answer to the question! Who knows? Sep 18, 2017 at 8:14

As Georges Elencwajg observed in his answer to the corresponding MO post, the linear substitution $u:=x+iy$ and $v:=x-iy$ is invertible and therefore $\Bbb C[u,v]=\Bbb C[x,y]$. Furthermore, $$uv = (x+iy)(x-iy)= x^2+y^2$$ so the ring in question is $\Bbb C[u,v]/(uv-1)$, which is the same as $\Bbb C[u,u^ {-1}]$. Now, let us invert this linear transformation to understand the factorization of $x^2$.
We have $x=u-iy$ and $y=i(v-x)$. Hence, $x=u+v-x$, meaning $$x=\frac{u+v}2=\frac{u+u^{-1}}2=\frac{u^2+1}{2u}=\frac{u-i}{2u}\cdot\frac{u+i}{2u}$$ is not even irreducible and similarly, $$y = i\cdot \left(v-\frac{u+v}2\right)=i\cdot\frac{v-u}2=\frac{u-u^{-1}}{2i} = \frac{u^2-1}{2iu}=\frac{u-1}{2iu}\cdot\frac{u+1}{2iu}$$ isn't either. Now, $$1-y=\frac{2iu-u^2+1}{2iu}=\frac{(u-i)^2}{2ui}$$ and $$1+y = \frac{2iu+u^2-1}{2iu} = \frac{(u+i)^2}{2iu}$$ explains that you have simply redistributed your irreducible factors and it all really makes sense.