# Artin - domains don't seem to match

1. Oct 21, 2014

### Whovian

This isn't a homework problem, but rather a bit of confusion regarding something in the textbook we're using; if this isn't the right place, feel free to move it.

From Artin's Algebra pages 422/423 (slightly paraphrased):

Let $Q=\begin{bmatrix}1&\\3&1\end{bmatrix}$, $A=\begin{bmatrix}2&-1\\1&2\end{bmatrix}$, $P=\begin{bmatrix}1&1\\1&2\end{bmatrix}$, $A'=Q^{-1}AP=\begin{bmatrix}1&\\&5\end{bmatrix}$ (note: blank spaces are to be interpreted as zeroes.)

Let $M$ be the integer lattice with its standard basis ${\bf C}=\left(e_1,e_2\right)$, and let $L$ be the lattice with basis ${\bf B}=\left(v_1,v_2\right)=\left(\left(2,1\right)^t,\left(-1,2\right)^t\right)$. Its coordinate vectors are the columns of $A$. We interpret $P$ as the matrix of a change of basis in $L$, and $Q$ as the matrix of change of basis in $M$.

My question is, if $A$ is interpreted as a map $M\rightarrow L$, wouldn't we have to interpret $P$ as a map $M\rightarrow M$ and $Q$ as a map $L\rightarrow L$ to get $A'$, which is just $A$ with different bases for $M$ and $L$, to be $M\rightarrow L$? Therefore $P$ would be interpreted as a change of basis in $M$ instead of $L$ and $Q$ in $L$ instead of $M$? In fact, if we had set up the problem so $A$ was 2x3, $Q$ 2x2, and $P$ 3x3, $M$ would be $\mathbb{R}^3$ and $L$ would be $\subseteq\mathbb{R}^2$, so $P$ could not be interpreted as a change of basis in $L=\mathbb{R}^2$ as it's 3x3.

Is there something incredibly obvious I'm missing?

2. Oct 22, 2014

### Whovian

Ah, derp. There was something obvious. It turns out M is to be interpreted as the codomain of the free module homomorphism corresponding to A.