I have been trying to prove that the residue class of $x$ is irreducible in $\mathbb{R}[x,y]/(x^2+y^2-1)$.
I have tried by assuming say $X=\bar{x}$ is reducible. So $X=p(X,Y)q(X,Y)$ where $Y= y +(x^2+y^2-1)$. But calculations are getting very difficult.
How do I prove this ? Is there any other way to do this than to assume $X=p(X,Y)q(X,Y)$ and proceed ?
Please help. Thanks in advance.
Use the norm: writing $\Bbb R[x,y]/(x^2+y^2-1)$ as the free $\Bbb R[x]$-module on the basis $\{1,y\}$, we can write the matrix representation of the element $p(x)+yq(x)$ as $$\begin{pmatrix} p(x) & (1-x^2)q(x) \\ q(x) & p(x) \end{pmatrix}$$ and the determinant of this map is $p(x)^2-(1-x^2)q(x)^2$. The map $\nu$ which sends $p(x)+yq(x)$ to $p(x)^2-(1-x^2)q(x)^2$ is a monoid homomorphism from the elements of $\Bbb R[x,y]/(x^2+y^2-1)$ under multiplication to the elements of $\Bbb R[x]$ under multiplication, so if $x=ab$ then $\nu(x)=\nu(a)\nu(b)$. Computing, we can see that $\nu(x)=x^2$, so if $x=ab$ with $a,b$ nonunits, then we need to be able to factor $x^2$ as a product of two non-units in the image of $\nu$.
As $\Bbb R[x]$ is a UFD, we'd have to find elements with norm $cx$ for some $c\in\Bbb R^*$, which is equivalent to solving $cx=p(x)^2-(1-x^2)q(x)^2$. But this is impossible: if $c>0$, plugging in $x=-1$ we get that $-c=p(-1)^2$, while if $c<0$, plugging in $x=1$ we get that $c=p(1)^2$.
