Here is the question I want to answer:
Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ ( the coordinate ring of the unit circle).
Show that $B^* = \mathbb R ^*$.
I know that Since $y$ is integral over $\mathbb R[x]$ there is a ring inclusion $B \rightarrow M_2(\mathbb R[x]),$ called a representation of $B.$ And I found this representation of $B$ because I know that $B = \mathbb R[x] \oplus \mathbb R[x]y$ and multiplication by $y$ is a map of $B$ as an $\mathbb R[x]$-module. And here is this representation $$\phi:B \rightarrow M_{2}(\mathbb R[x]), \phi (a_{1} + a_{2}y) = \begin{pmatrix} a_{1} & a_{2}(1-x^2) \\ a_{2} & a_{1} \end{pmatrix} $$
Then I was thinking about using the determinant to solve my questions and I guess I should work in a way similar to $\mathbb Z \sqrt {3}$ but still I am confused about the details.
Any help will be appreciated.
If $a_1+a_2y$ is a unit, then $\phi(a_1+a_2y)$ must be invertible, so $\det\phi(a_1+a_2y)=a_1^2+a_2^2(x^2-1)$ needs to be a unit of $\mathbb{R}[x]$, i.e., an element of $\mathbb{R}^\times$. If $a_1a_2\neq 0$, write \begin{align*} a_1&=a_{1,m}x^m+a_{1,m-1}x^{m-1}+\dots+a_{1,0}\\ a_2&=a_{2,n}x^n+a_{2,n-1}x^{n-1}+\dots+a_{2,0} \end{align*} with $a_{i,j}\in\mathbb{R}$, $a_{1,n}a_{2,m}\neq 0$. Then $a_1^2+a_2^2(x^2-1)$ has top degree (in $x$) piece $$ \begin{cases} a_{1,m}^2x^{2m} & \text{if }m>n+1\\ a_{2,n}^2x^{2n} & \text{if }m<n+1\\ (a_{1,m}^2+a_{2,n}^2)x^{2m} & \text{if }m=n+1 \end{cases} $$ Can you finish it from here?