Denote $A = \Bbb{R}[x,y] / \langle x^2 + y^2 - 1 \rangle $ (where $\langle \cdot \rangle$ means ideal generated by $\cdot$) and $I= \langle x-1, y \rangle / \langle x^2 + y^2 - 1 \rangle$ . I have to prove that $I$ is not a principal ideal.
In order to do it, I wonder if given a ring $R$ an ideal $J \subseteq R$ the principal ideals of $R/J$ are the image via the quotient projection $\pi : R \rightarrow R/J$ of principal ideals of $R$. If this is true, then supposing that $I$ is a principal ideal, then $\langle x-1, y \rangle $ is principal on $\Bbb{R}[x,y]$.
Here I get stuck since I am not sure of the thing I said about the principal ideals and quotient and I do not see how to follow, so any possible help would be appreciated :)
EDIT
Since in the discussion in comments that @Amateur_Algebraist mentioned is only about hints and it does not show the accurate way for the proof, I am writing the proccess that it should be followed to prove it, but I am getting stuck.
Supposing that $I$ is principal, it exists $p \in A$ such that $\langle p \rangle = I$. Then, $p$ divides both $x-1$ and $y$ , so it exists $a, b \in A$ such that $$ap = x-1, bp = y$$
By definition of $A$, we have that $x^2 + y^2 - 1 = 0$ (on $A$) $\Rightarrow y^2 = 1-x^2 = (1+x)(1-x)$.
Then: $ap (x+1) = x^2 - 1 = -y^2 = -b^2 p^2 \Rightarrow b^2 p^2 + a(x+1) p = 0$.
$p$ is not $0$ because if it would be, then $I= \{0\}$, which is false. Therefore, $b^2 p + a (x+1) = 0$. Here is where I do not know how to follow, it comes to mind to evaluate on $(-1,y)$ so we have that $b^2 (-1,y) \cdot p(-1,y) = 0$ and then one of this polynomials on $\Bbb{R}[y]/\langle x^2 + y^2 - 1\rangle$ would be $0$, but I do not know if this leads to a contradiction.