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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?

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    $\begingroup$ No, this implication is not supported by any theorem. $\endgroup$
    – Crostul
    Apr 13 '16 at 6:13
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    $\begingroup$ This probably has a nonfree f.g. projective module corresponding to the Moebius band, and in that case no. $\endgroup$ Apr 13 '16 at 6:20
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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.

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  • $\begingroup$ In the last paragraph, what is $k$? $\endgroup$ Apr 17 '16 at 16:03
  • $\begingroup$ @BarrySmith $k$ is a field where $-1$ is not a square. $\endgroup$
    – Pedro Tamaroff
    Apr 19 '16 at 2:46
  • $\begingroup$ @PedroTamaroff Is this the answer you talked about few days ago? $\endgroup$
    – user26857
    Aug 31 '17 at 6:56
  • $\begingroup$ @user26857 Yes. $\endgroup$
    – Pedro Tamaroff
    Aug 31 '17 at 7:36
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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.

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  • $\begingroup$ By "in this case", do you mean in the case $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$? Because in this ring $(X,Y) = (1)$ is principal. $\endgroup$ Apr 13 '16 at 6:25
  • $\begingroup$ You're right, I edited. $\endgroup$
    – karmalu
    Apr 13 '16 at 6:35
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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.

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