Invertible matrices with coefficients in a quotient ring Let $R$ be a commutative ring and $p\subseteq R$ an ideal. An $n\times n$-matrix $A$ with coefficients in $R$ is invertible iff $\det(A)\in R^\times$. The determinant is a sum of products of the entries of the matrix. Hence the image of the map
$$GL_n(R)\rightarrow GL_n(R/p)$$ is contained in the subgroup of matrices $B\in GL_n(R/p)$ so that $\det(B)$ is in the image of the units $R^\times$ in $R/p$.
Is the image of the described map precisely this subgroup?
 A: Here is another argument due to C. P. Ramanujam, which uses some topology. Consider $R=\mathbb{C}[X,Y,Z,W]$ and $P$ the prime ideal generated by $XY-ZW-1$. Then you can check that both $R$ and $R/P$ has only constants as units and thus if your surjectivity is true, you will also have a surjectivity of $SL_2(R)\to SL_2(R/P)$. In particular the matrix $\left(\begin{matrix} x&z\\ w&y\end{matrix}\right)$ where $x,y,z,w$ denote the obvious elements in $R/P$ can be lifted to $SL_2(R)$. That is, there exists $f,g,h,k\in R$ with $\det\left(\begin{matrix} f&g\\ h&k\end{matrix}\right)=1$ and $f=x$ etc in $R/P$. This gives a continuous map $M_2(\mathbb{C})\to SL_2(\mathbb{C})$, given by
$$\left(\begin{matrix} a&b\\ c&d\end{matrix}\right)\to \left(\begin{matrix} f(a,d,b,c)&g(a,d,b,c)\\ h(a,d,b,c)&k(a,d,b,c)\end{matrix}\right)$$
One easily checks that this is a retraction, which is impossible since $M_2(\mathbb{C})$ is contractible, while $SL_2(\mathbb{C})$ has the homotpy type of a sphere.
A: I think this argument should work.
Let $A+B \in Mat_n(k)$, where $B$ has all of its entries in $p$. I claim that there exists $\beta \in p$ such that $det(A+B)=det(A)+\beta$. This is clear if $n=1$. 
Now suppose the claim is true for some $n$. Write $A+B=(a_{ij}+b_{ij}) \in Mat_{n+1}(k)$. Then $det(A+B)= \sum_{j=1}^{n+1}(a_{ij}+b_{ij})C_{ij}$, where $C_{ij}$ is the $ij$th cofactor of $A+B$. By the induction step, $C_{ij}=\alpha_{ij}+\beta_{ij}$, where $\alpha_{ij}$ is the cofactor of the $ij$th minor of $A$, so 
$$det(A+B)= \sum_{j=1}^{n+1}(a_{ij}+b_{ij})(\alpha_{ij}+\beta_{ij})=\sum_{j=1}^{n+1} a_{ij}\alpha_{ij}+a_{ij}\beta_{ij}+b_{ij}\alpha_{ij}+b_{ij}\beta_{ij}= det(A)+ \beta,$$
where $\beta=\sum_{j=1}^{n+1}a_{ij}\beta_{ij}+b_{ij}\alpha_{ij}+b_{ij}\beta_{ij} \in p$, since $p$ is an ideal. 
A representative for a matrix in $GL_n(k/p)$ is a sum $A+B \in Mat_n(k)$ such that $B$ has all of its entries in $p$ and $det(A+B)+p \in (k/p)^*$. But we're now done. To see this, note that if $det(\overline{A})$ is in the image of $k^*$ (where $\overline{A}=[A+B]$), then $det(A) \in k^*$ by the claim, so $A \in GL_n(k)$ and $\pi(A)= \overline{A}$, where $\pi:GL_n(k) \to GL_n(k/p)$. 
