Let $R$ be a commutative ring with unity. $s\in U(S) \iff \det(s) \in U(R)$

Question

Let $R$ be a commutative ring with $1_R$ and let $$ S = \biggl\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \;\Biggm| \; a,b,c \in R \;\biggr\}. $$

If $s = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \in S$, is it true that $$ s \in U(S) \iff \det(s) \in U(R)? $$

My instinct says it isn't but I cant find an example so I can make my mind right.

Answer 1

I assume that by $U(S)$, and $U(R)$ you mean the invertible elements of the ring? This statement is true. Take $s\in U(S)$, there is some $s^{-1}$ such that $ss^{-1}= \text{Id}$. Taking determinants $$ \det(s)\det(s^{-1})=1_R, $$ thus $\det(s)\in U(R)$. Conversely, suppose $\det(s)= ac\in U(R)$, then there is some $d\in R$ such that $acd=1_{R}$. Let $$ t= \begin{pmatrix}cd & -bd\\ 0 & ad\end{pmatrix}. $$ It's easy to see that $$ st= \begin{pmatrix}acd & -abd+abd\\ 0 & cad\end{pmatrix}= \begin{pmatrix}1_R & 0\\ 0 & 1_R\end{pmatrix}. $$

Answer 2

Answer: For any commutative ring $A$ and any $n\times n$ matrix $R\in Mat(n,A)$ with coefficients in $A$ you may define the adjunct matrix $adj(R)$. This matrix has the property that $adj(R)R=Radj(R)=det(R)Id$ where $Id$ is the $n\times n$ identity matrix and $det(R)$ is the "determinant". The determinant $der(R)\in A$ is multiplicative: $det(RR')=det(R)det(R')$. From this it follows that $R$ is invertible (there is a matrix $R^{-1}\in M(n,A)$ with $RR^{-1}=R^{-1}R=Id$) iff $det(R)\in A^*$ is a unit in $A$.

Proof: If $R$ is invertible it follows $RR^{-1}=Id$ hence $det(RR^{-1})=det(R)det(R^{-1})=det(Id)=1$ hence $det(R)\in A^*$. Conversely if $det(R)\in A^*$ it follows $det(R)^{-1}adj(R):=R^{-1}$ is an inverse in $M(n,A)$.

You may check that in your case it follows $adj(s)$ is upper triangular, hence in your case the inverse (if it exists) lives in $S$. There is an explicit formula for the adjunct matrix for $2\times 2$-matrices:

Why the inverse of a matrix involves division by the determinant?