0
$\begingroup$

Let $M$ be a free module with rank $2$ over the PID $R$ having basis $B=\{b_1,b_2\}$. I have a few questions regarding the submodule $Rm$, where $m$ is some element of our module $M$. Suppose $m$ is of the form $xb_1+xb_2$ for $x\in R$. How can we show that there is some submodule $T$ such that $M=Rm\oplus T$ if and only if $x$ has an inverse in $R$? C an we also show that if $m=xb_1+yb_2$, then there is $T$ s.t. $M=Rm\oplus T \iff$ the ideal generated by $x$ and $y$ gives the whole ring $R$?

I've given this some thought and have ideas to use a decomposition as done here, but I'm not really sure how to approach this. I'm also not sure how $x$ being invertible is relevant. Thanks in advance.

$\endgroup$
0
$\begingroup$

This is easiest to see working with $2\times2$-matrices over $R$: Let me show that if $m=xb_1 + yb_2$ then there exists $T\subseteq M$ such that $M = Rm \oplus T$ if and only if $Rx+Ry = R$.

First assume that $Rx+Ry=R$. Then there are $w,z\in R$ such that $wx - zy = 1$. Now define $t := zb_1 + wb_2\in M$ and $A := \begin{pmatrix}x & y\\ z & w\end{pmatrix}\in {\rm Mat}_{2\times 2}(R)$ is invertible (its inverse being $\begin{pmatrix}w & -y\\ -z & x\end{pmatrix}$). Thus, we have $$ \begin{pmatrix} m\\ t\end{pmatrix} = \begin{pmatrix} x & y\\ z & w\end{pmatrix}\cdot \begin{pmatrix} b_1\\ b_2\end{pmatrix}. $$ Setting $T:= Rt$, it is clear that $M = Rm\oplus T$. For example linear independence of $m$, $t$: Let $\lambda,\mu\in R$ such that $0 = \lambda m + \mu t = \begin{pmatrix} \lambda & \mu\end{pmatrix}\cdot \begin{pmatrix} m\\ t\end{pmatrix}$. Then $$ 0 = \begin{pmatrix} \lambda & \mu\end{pmatrix}\cdot \begin{pmatrix} m\\ t\end{pmatrix} = \left(\begin{pmatrix} \lambda & \mu\end{pmatrix}\cdot \begin{pmatrix} x & y\\ z & w\end{pmatrix}\right)\cdot \begin{pmatrix} b_1\\ b_2\end{pmatrix}. $$ Since $b_1, b_2$ is a basis, we have $\begin{pmatrix} \lambda & \mu\end{pmatrix}\cdot \begin{pmatrix} x & y\\ z & w\end{pmatrix} = 0$ and hence $$ 0 = \begin{pmatrix} \lambda & \mu\end{pmatrix}\cdot \begin{pmatrix} x & y\\ z & w\end{pmatrix} \cdot \begin{pmatrix} w & -y\\ -z & x\end{pmatrix} = \begin{pmatrix} \lambda & \mu\end{pmatrix}\cdot \begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix} = \begin{pmatrix} \lambda & \mu\end{pmatrix}, $$ i. e. $\lambda = \mu = 0$.

Conversely, if there exists $T\subseteq M$ such that $M = Rm\oplus T$ then $T$ has to be free of rank $1$, generated by, say $t=zb_1 + wb_2\in M$. Since $m, t$ is also a basis of $M$, there exist $a,b,c,d\in R$ such that $b_1 = am + bt$ and $b_2 = cm +dt$. Hence $$ \begin{pmatrix}m\\t\end{pmatrix} = \begin{pmatrix}x & y\\ z & w\end{pmatrix}\cdot \begin{pmatrix} b_1 \\ b_2\end{pmatrix} = \begin{pmatrix} x & y\\ z & w\end{pmatrix}\cdot \begin{pmatrix} a & b\\ c & d\end{pmatrix} \cdot \begin{pmatrix} m\\ t\end{pmatrix}. $$ Since $m,t$ are linearly independent, we conclude $\begin{pmatrix} x & y\\ z&w\end{pmatrix}\cdot \begin{pmatrix} a & b\\ c & d\end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0&1\end{pmatrix}$. In particular, $ax + cy = 1\in R^\times$, i. e. $Rx + Ry = R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.