Is $\mathbb{Q}[X,Y]/\langle X^2+Y^2-1\rangle$ a PID? I know that $\mathbb{Q}[X,Y]$ is not a PID. Does this imply that the quotient ring $\mathbb{Q}[X,Y]/\langle X^2+Y^2-1 \rangle$ is not a PID?
 A: Generally speaking no, in fact $\mathbb{Q} [X, Y]/(Y) \simeq \mathbb{Q} [X] $ is a PID. 
In this case the quotient isn't a PID. 
A: The following can be found in T.Y. Lam's book Lectures on modules and rings. 
Suppose that $R$ is a ring and $R\subseteq S$ for another commutative ring, and suppose that $P$ is an invertible $R$-submodule of $S$, meaning there is another $R$-submodule $Q$ such that $PQ=\{\sum p_iq_i:p_i\in P,q_i\in Q\}=R$. 
Using the dual basis theorem it is relatively easy to see that $P$ must be finitely generated projective (use an equation $p_1q_1+\cdots+p_nq_n=1$). On the other hand $P$ will be free if and only if $P=uS$ for a unit $u\in S$. This last claim is not too immediate, but one can see this by noting that if $PQ=R$ then $P\otimes_R Q$ is isomorphic to $R$ by the canonical map. If $P\simeq R^n$ then $Q\simeq P^\ast\simeq R^n$, so the tensor product is $R^{n^2}$ so since $R$ is commutative, $n=1$. 
Take now $R=\Bbb Q[X,Y]/(X^2+Y^2-1)=\Bbb Q[x,y]$ (or replace $\Bbb Q$ for any field where $-1$ is not a square. This is a domain, and one can consider its field of fractions $S$. 
The claim is that $\mathfrak p=(1-y,x)$ is an invertible (prime) ideal that is not free, so it cannot be principal. To see that $\mathfrak p$ is invertible, note that $x^2 = 1-y^2=(1-y)(1+y)$ so the element $z=x(1-y)^{-1}$ maps $\mathfrak p$ into $R$, and in fact $$2=(1+y)+(1-y)=(1-y)+zx$$
so $\mathfrak p^{-1} = (1,z)$. This shows that $\mathfrak p$ is invertible, so it is projective and finitely generated.
Consider now the quadratic extension $k(x,y)/k(y)$. The norm of an element $f+xg$ is then $f^2-x^2 g^2=f^2+y^2g^2-g^2$, and this has even degree at least two. If we had $\mathfrak p =(f+xg)$, then we would be able to write $$1-y=(f+xg)\alpha\\ x =  (f+xg)\beta$$ for $\alpha,\beta\in R$. Taking norms and subtracting gives $$(N(\alpha)-N(\beta))N(f+xg) = (1-y)^2
+ x^2 = 2(1-y)$$
which contradicts our observation that $N(f+xg)$ has degree at least $2$. Thus $\mathfrak p$ is not principal. Note, however, that $\mathfrak p^2 =(1-y)$ is principal.
A: No, it doesn't! 
$R=\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is not a PID, but for other reasons. Actually it is not even a UFD for $x$ is irreducible, but not prime.
