Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ which is called the coordinate ring of the unit circle.
I am trying to prove that $y$ is irreducible in $B.$
I have the following information about $B$:
1- $B$ is an integral domain.
2- $B = \mathbb R[x] \oplus \mathbb R[x]y$ and multiplication by $y$ is a map of $B$ as an $\mathbb R[x]$-module. And $y$ is integral over $\mathbb R[x]$ and so there is a ring inclusion $B \rightarrow M_2(\mathbb R[x]),$ called a representation of $B$ described below:
$$\phi:B \rightarrow M_{2}(\mathbb R[x]), \phi (a_{1} + a_{2}y) = \begin{pmatrix} a_{1} & a_{2}(1-x^2) \\ a_{2} & a_{1} \end{pmatrix} $$
3- Also, I know that $B^* = \mathbb R^*$
I am trying to use Eisenstein criterion for showing that $y$ is irreducible.
Here is my try:
We need to find a prime $p$ such that $p$ divides $0$ because we have only 1 coefficient which is the leading coefficient and every other coefficients are $0$ and any prime can divide zero trivially. Also any prime does not divide 1. But now I do not know why $p^2$ does not divide $a_0$ which is zero in our case ...I need $p^2$ to divide $a_0$ to fullfill the conditions for Eisenstein Criterion. Could anyone explain this to me please?
Also, is there any use for the fact that $y$ is integral in this proof?
Define $N:B\to\mathbb R[x]$ by $N(a_1+b_1y)=\det\phi(a_1+b_1y)$. Notice that $N$ behaves like a norm!
Let $z_i=a_i+b_iy$, $i=1,2$, such that $y=z_1z_2$. Then $N(y)=N(z_1)N(z_2)\Leftrightarrow x^2-1=N(z_1)N(z_2)$. Now we have the following cases:
(i) $\deg N(z_1)=0\Leftrightarrow z_1\in\mathbb R^*$,
(ii) $\deg N(z_1)=2$ $\Leftrightarrow$ $\deg N(z_2)=0$ $\Leftrightarrow$ $z_2\in\mathbb R^*$,
(iii) $\deg N(z_1)=\deg N(z_2)=1$. If $N(z_1)=x-1$, then $a_1^2(x)+b_1^2(x)(x^2-1)=x-1\Rightarrow x-1\mid a_1(x)\Rightarrow\exists a'_1\in\mathbb R[x]$ such that $a_1=(x-1)a'_1$ and plugging this in the foregoing equation we get $a'_1(x)^2(x-1)+b_1^2(x)(x+1)=1$. Looking now at the dominant coefficients of $a'_1$ and $b_1$ we find that one of these is zero (false!) or the sum of their square is zero (false!).