TITLE: Quotient of polynomials, PID but not Euclidean domain?
QUESTION [23 upvotes]: While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that
$$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$
is such a ring. After a good deal of searching, I have not been able to find any other (online) reference to this ring.
Can anyone confirm this result? Is there a reference for it (paper, textbook or website)?

REPLY [6 votes]: Here are some more details for those who, like me, are not so familiar with this material.
$1$) Take a non-zero prime ideal. We wish to show that it is maximal. It contains a prime element (since we are in a UFD, choose an element with a shortest factorisation). Take this to be $ax+by+c$. We will show that the ideal is maximal by showing that the quotient ring by this ideal is a field.
Using the relation $ax+by+c=0$ we can express any element as a polynomial in $x$ with coefficients in $\mathbb{R}$. With the quadratic relation we can make this a linear expression. In searching for an inverse we get two equations in two unknowns, and the determinant of the corresponding matrix is non-zero.

$2$) Non-zero, non-invertible elements exist because otherwise we would be in a field and wouldn't be trying to prove the lemma in the first place.
$3$) Suppose $p$ is not prime. Then there exist $a$ and $b$ such that $p\mid ab$ but $p\nmid a$ and $p\nmid b$.
Write $a=q_a p+r_a$ with $\delta(r_a)<\delta(p)$. Then $r_a$ is invertible, with inverse $s_a$. Similarly, we have $q_b$, $r_b$, $s_b$.
Then $ab=q_a (q_b p+r_b)p +q_b r_a p +r_b r_a$, so $p\mid r_a r_b$. Then $p\mid r_a s_a r_b s_b =1$, a contradiction.

$4$) $A^{\times}=\mathbb{R}^{\times}$ follows because every element of $A$ can be uniquely represented as $p(y)+x q(y)$ for polynomials $p$ and $q$.
$5$) From $1$) we know that each element of $A/pA$ can be expressed as a linear expression in $x$, and we have one relation $x^2(a^2+b^2)+2acx+c^2+1=0$. The roots of this are non-real, so by sending $x$ to one of them we can define an isomorphism to $\mathbb{C}$.
$6$) The projection homomorphism is injective because having an invertible element in $pA$ would show that $p$ is also invertible.
$7$) $\mathbb{R}^{\times}$ and $\mathbb{C}^{\times}$ are not isomorphic because $\mathbb{C}^{\times}$ has order $4$ elements whereas $\mathbb{R}^{\times}$ does not (see linked question).