I am reading about integral bases from Frazer Jarvis' "Algebraic Number Theory", but my question is really about elementary linear algebra. In page 49, author claims the following:
I don't think $M'=CM$ is true. I think it should be $M'=MC$. Who's correct, the book or me?
Of course, the conclusion of the theorem follows regardless. But this has been driving me crazy. I know it is a simple linear algebra, but I just want to confirm my doubts :)
Edit: As Gerry points out, the screenshot does not capture what the matrix $M$ is. It is defined earlier in the book (in exactly the same way as $M'$ defined, except without the primes): $$ M = \begin{pmatrix} \sigma_1(\omega_1) & \sigma_1(\omega_2) & \cdots & \sigma_1(\omega_n) \\ \sigma_2(\omega_1) & \sigma_2(\omega_2) & \cdots & \sigma_2(\omega_n) \\ \vdots &\vdots &\ddots &\vdots \\ \sigma_n(\omega_1) & \sigma_n(\omega_2) & \cdots &\sigma_n(\omega_n) \end{pmatrix} $$
Alright, just to get this out of the unanswered question list, I am posting an answer.
Yes, there is a typo in the book. It should be $M'=MC$. Indeed, $M=(m_{ij})$ where $m_{ij}=\sigma_{i}(\omega_j)$. Then we can compute $(k, i)$-th coefficient of $MC$: $$ (MC)_{ki} = \sum_{j=1}^{n} m_{kj} c_{ji} = \sum_{j=1}^{n} \sigma_{k}(\omega_j)c_{ji} = c_{1i}\sigma_{k}(\omega_1)+c_{2i}\sigma_{k}(\omega_2)+\cdots +c_{ni}\sigma_{k}(\omega_n) = \sigma_{k}(\omega'_{i})=M'_{ki} $$ Thus, $MC=M'$.