$(\bar x , \bar y-1)$ is not a principal ideal of $A=\mathbb R[x,y]/(x^2+y^2-1)$ Let $A=\mathbb R[x,y]/(x^2+y^2-1)$ ; then $A$ is an integral domain . Consider the maximal ideal $P=(\bar x , \bar y-1)$ of $A$ . Then how to show that $P$ is not a principal ideal ?  

We reason by contradiction, considering a generator $g$ of the principal ideal $(\overline{x},\overline{y} - 1)$.
Let $B = \mathbb{R}[t, \frac{1}{1 + t^2}]$, that is, the localization $S^{-1} \mathbb{R}[t]$ where $S \subset \mathbb{R}[t]$ is the multiplicative set generated by $1 + t^2$. The map $(x, y) \mapsto 
(\frac{1 - t^2}{1 + t^2}, \frac{2t}{1 + t^2})$ induces a ring homomorphism $\phi: A \rightarrow B$. (This homomorphism can be checked to be injective, but we won't use that fact.) The ring $B$ is Euclidean, hence a UFD and it easily follows that $\phi(g)$ is of the form $\lambda (1 - t)(1 + t^2)^m$ for some $\lambda \in \mathbb{R} \setminus \{0\}$ and some $m \in \mathbb{Z}$. Let $C$ be the ring of functions from $\mathbb{R}$ to $\mathbb{C}$ of the form $\theta \mapsto \sum_{n = -N}^N \lambda_n e^{in\theta}$ with $\lambda_n \in \mathbb{C}$ and $N \ge 0$.Let $Q$ be the fraction field of $C$. It is immediate to see that $P(X) \mapsto (\theta \mapsto P(e^{i\theta}))$ induces a ring isomorphism from $\mathbb{C}[X^{\pm 1}]$ (resp. $\mathbb{C}(X)$) onto $C$ (resp. $Q$). The map $t \mapsto \tan(\frac{\theta}{2}) \Doteq \frac{e^{i\theta} - 1}{i(e^{i\theta} + 1)}$ induces a ring homomorphism $\psi: B \rightarrow Q$ such that $\psi \circ \phi(A) \subset C$ (check what are the images of $\overline{x}$ and $\overline{y}$ in $Q$). Since $\psi \circ \phi(g) = 
2^m\lambda \frac{e^{i m \theta}((1 + i)e^{i\theta} + 1 - i)}{(e^{2i\theta} + 1)^m(e^{i\theta} + 1)} \in C$, 
we have 
$2^m\lambda \frac{X^m ((1 + i)X + 1 - i)}{(X^2 + 1)^m (X + 1)} \in \mathbb{C}[X^{\pm 1}]$. This is impossible because $-1$ is not a root of the numerator of the latter fraction.
