Let $k$ be a field and $c\in k$ not a square. Let $A$ be a commutative $k$-algebra. Consider $A[\sqrt c]:=A[x|x^{2}=c]$. Prove that the subgroup of invertible elements $G(A):=U(A[\sqrt c])$ is identified to a subgroup of $GL_{2}(A[\sqrt c])$.
My idea is the following: write an element in $G(A)$ as $a+bx$ with $a,b\in A$. Then I associate to it a matrix $\begin{pmatrix} a&bx\\bx&a\end{pmatrix}$. Then, $a+bx$ is invertible if and only if there is $a'+b'x\in A[\sqrt c]$ such that $(a+bx)*(a'+b'x)=1$. It is easy to see that the corrispondent matrix is the inverse matrix of the first one. Note that the determinant of the first matrix is $a^{2}-cb^{2}$. Is this correct? Since $G(A)$ is not necessarily a domain, I don't know how to treat the condition $\det(A)\neq0$ for a matrix to be invertible, since it is not sufficient (it must be and invertible element of the algebra).Then how does the fact that $c$ is not a square in $k$ affect the invertibility of a matrix of that form?
Sorry if the question is not very clear, but it reflects my mood on this exercise.
Thank you, any explication is much appreciated!
I will now assume that we want an embedding $G(A)\to GL_2(A)$. Actually we can easily get something a little better, namely a ring (or even $A$-algebra) embedding $f: A[\sqrt{c}]\to M_2(A)$, defined by $$ f: a + bx \mapsto \begin{pmatrix} a & bc \\ b & a \end{pmatrix}.$$
Then $f$ naturally identifies $A$ with the subalgebra of scalar matrices in $M_2(A)$, and you can check that $y=f(x)=\begin{pmatrix} 0 & c \\ 1 & 0 \end{pmatrix}$ commutes with $A$ and satisfies $y^2=c$.
Once you have a ring embedding, it clearly restricts to a group embedding on invertible elements.