$A=\frac{\mathbb{C}[X,Y]}{(X^2+Y^2-1)}$ 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.
 A: The answer is essentially here, but I will detail it to get this off the unanswered questions list.
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.
