Prove that $x−1$ and $y−1$ are irreducible in $R$ Let $R$ be the ring of polynomial functions on the circle:
$R=\mathbb{R}[X,Y]/I$ with $I = (X^2 +Y^2 − 1)$ and put $x := X + I$, $y := Y + I$ $∈ R$.
We can define the following Norm map:
$N:R\rightarrow R[x]$ by $N(f + gy):=(f + gy)(f − gy)=f^2 − g^2(1− x^2)$
I have to prove the following statements:


*

*Prove that $x−1$ and $y−1$ are irreducible in $R$
I tried the following using the norm map:
Suppose that $x-1 = ab$ with $a,b \in R$. Then, $x^2-1 = N(x-1) = N(ab) = N(a)N(b)$. Since $N(a)\mid x^2-1$ but $N(a) = x-1$ is impossible, we must have $N(a)=1$ or $N(b)=1$. But the definition of the norm hows that this means $a$ or $b$ is a unit in R. Hence, $x-1$ is irreducible.


*

*Prove that $(x −1)$ and $(y−1)$ are not prime ideals in $R$ and that $R$ is not a unique factorization domain

*Show that $a=(x + y−1)^2=2(x−1)(y−1)$ are two distinct factorizations of $a$ as a product of irreducible elements (and a unit 2).
For the last statement I found the following:
$(x + y - 1)^2=x^2 + y^2 + 2xy - 2x - 2y + 1=2xy - 2x - 2y + 2 + I$
$2(x-1)(y-1)=2xy - 2x - 2y + 2$
Any help would be grateful. Thanks in advance.
 A: This answer amounts to reordering this answer for the sake of the OP.


*

*By repeatedly using the reduction $y^2=1-x^2$ in $R$, one sees that every element can be uniquely written as $a(x) + b(x)y$. Thanks to this we may define $$N : R \to \mathbb{R}[z], \qquad a(x)+b(x)y \mapsto a(z)^2 - b(z)^2(1-z^2),$$ which is multiplicative (check it). 

*With 1. one verifies that invertible elements in $R$ are just the non-zero constants. In fact, if $p(x,y)$ is invertible in $R$ then there exists $q(x,y)$ such that $pq=1$. This implies that $N(p)N(q) = N(pq)=1$ in $\mathbb{R}[z]$, which implies that $N(p)$ (and $N(q)$ too) is a non-zero constant. That is, by writing $p(x,y) = a(x) + b(x)y$,
$$a(z)^2 - b(z)^2(1-z^2) = c \neq 0.$$
If $b(z) \neq 0$, then the leading coefficient on the left hand side would be $a_n^2 + b_m^2$, where $a_n$, $b_m$ are the leading coefficients of $a$ and $b$. However, since $z$ does not appear on the right-hand side, we should have $a_n^2 + b_m^2 = 0$, which is a contradiction. Therefore, $b=0$ and $a(z)^2 = c$ implies that $a(z) = a \in \mathbb{R}$.

*If by contradiction $x-1$ is reducible, then $x-1 = pq$ and hence $$(z-1)^2 = N(x-1) = N(p)N(q).$$
Since neither $p$ (nor $q$) is a unit, $\deg(N(p))>0$ in view of point 2., which leaves as only option $N(p) = z-1$. Write again $p = a(x) + b(x)y$ and $$z-1 = a(z)^2-b(z)^2(1-z^2),$$
which forces $(1-z)\mid a(z)$. Write then $a(z)=d(z)(z-1)$ and plug it into the above equality to find $$d(z)^2(z-1) +b(z)^2(z+1) = 1.$$
The leading coefficient on the left-hand side is $d_n^2+b_m^2$, where $d_n,b_m$ are the leading coefficients of $d(z)$ and $b(z)$. Since $z$ does not appear on the right-hand side, we should have $d_n^2+b_m^2 =0$, which is a contradiction. The same conclusion can be drawn about $1-y$ by switching the roles of $x$ and $y$ in the argument up to here.

*Now, if $\langle x-1\rangle$ was a prime ideal, then $R/\langle x-1\rangle$ would have been an integral domain, but $$R/\langle x-1\rangle \cong \mathbb{R}[X,Y]/\langle X^2+Y^2-1,X-1\rangle \cong \mathbb{R}[X,Y]\langle X-1, Y^2 \rangle \cong \mathbb{R}[Y]/\langle Y^2\rangle,$$ which is not ($y^2 = 0$). Analogously for $\langle y-1\rangle$. In particular, if $R$ was an integral domain, then every irreducible element would have been prime (see Wikipedia) and hence $\langle x-1\rangle$ would have been a prime ideal (see Wikipedia again), which is not the case.


You already solved the last bullet.
