Field extension trace/norm confusion pertaining to multiplication matrix I've gotten stuck on page 37 on P Samuel's 'Algebraic Theory of Numbers', on an equation that also features at the start of chapter 12 of Ireland and Rosen. The setup is, if we have a field $K$, and $x\in\overline K$, and we form the algebraic extension of fields $K[x]/K$, and we consider the endomorphism $m_x$of $K[x]$ given by multiplication by $x$, then $m_x$ can be written as a matrix of dimension $n$, where $n=\operatorname{deg} f$, where $f$ is the minimal polynomial of $x$ over $K$; equivalently $n=[K[x] : K]$. If we form a basis $(y_i)_{i=1,\ldots,n}$ for $K[x]$ as a vector space over $K$, and write this matrix $M=(a_{ih})$ with respect to this basis, then Samuel claims that $xy_i=\sum_h a_{ih}y_h$. I don't see why this is. For example, taking the "standard" basis $1,x,\ldots,x^{d-1}$ of the extension $\mathbf{Q}[x]$ for $x$ satisfying the minimal polynomial $f(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_0$, we have the following multiplication matrix in this basis:
$$M=\begin{bmatrix}0&0&\cdots&0&-a_0\\1&0&\cdots&0&-a_1\\0&1&\cdots&0&\vdots\\\vdots&0&&\vdots&\vdots\\\vdots&\vdots&&\vdots&\vdots\\0&0&\cdots&1&-a_{n-1}\end{bmatrix}$$
This is because $x\cdot x^k=x^{k+1}$ for $k<n-1$, and $x\cdot x^{n-1}=x^n=-a_{n-1}x^{n-1}+\cdots+-a_0$. Well, then if the basis $y_i=x^{i-1}$ for $i=1,\ldots,n$, then by the formula we should have $xy_1=\sum_h a_{1h}y_h=-a_0y_n=a_0x^{n-1}$. But in this setup $xy_1=x\cdot x=x^2$. It seems to me that the formula should be $xy_i=\sum_h a_{hi}y_h$, but since Ireland & Rosen write the same thing as Samuel I must be confused.
 A: If I understand what you're writing, then your mistake is in thinking that the $y_i$s are supposed to be the basis elements. They're not -- they are the coefficients of the basis elements when you write the element you want to multiply with $x$ as a $K$-linear combination of basis elements.
In particular, you can't have $y_n=x^{n-1}$, because the $y_i$s live in $K$, not in $K[x]$.
If you want to find $x$ times $y$ where $y=x^{n-1}$, you start by writing
$$ y = 0\cdot 1 + 0 \cdot x + \cdots + 0 \cdot x^{n-2} + 1\cdot x^{n-1} $$
so all the $y_i$s are $0$ except $y_n=1$. You then get
$$ (xy)_i = \sum_h a_{ih}y_h = -a_iy_n = -a_i\cdot 1 = -a_i $$
which is just what you want:
$$ x^n = -a_0 - a_1x - a_2 x^2 - \cdots - a_{n-1}x^{n-1} $$
since $x$ is a root of the minimal polynomial.
A: Owen, I agree with you. Samuel states clearly that $(y_i)$ is supposed to be a basis of $K[x]$ over $K$, so I think he wrote $a_{ih}$ where he should have written $a_{hi}$, as you say.
I checked to see if perhaps Samuel used unusual conventions for indices in a matrix. His book Géométrie projective, as far as I can tell, doesn't contain any indexed matrices. Bourbaki uses the usual order for indices.
However, intriguingly, the  book Commutative Algebra, co-written with Oscar Zariski, has the same order of indices in a similar formula on pages 86-88 of Volume 1. But in that book, the order is correct, as I will now explain.  
Zariski-Samuel fixes a basis $\omega_1, \ldots, \omega_n$ of $K/k$, and then chooses the coefficients $a_{ij}$ so that
$$x\omega_i = \sum_j  a_{ij} \omega_j.$$
The authors then point out that if $A$ is the matrix $[a_{ij}]$ and $\Omega$ is the column matrix $$\begin{bmatrix} \omega_1 \\ \vdots \\ \omega_n \end{bmatrix}$$
then we have $x\Omega = A\Omega$.
Thus in Commutative Algebra, the same order of indices is used for the summation formula as in Algebraic Number Theory. However, in Commutative Algebra the claim is never made that the matrix $A$ represents the matrix of multiplication by $x$, as it is in the later book.
Therefore I would venture the following guess. Samuel used the order of indices he was used to, and forgot to change it now that he was defining $A$ as the matrix of multiplication by $x$.
Edit: Ireland and Rosen have a similar formula
$$\alpha \alpha_i = \sum_j a_{ij} \alpha_j,$$
however, judging from a quick glance, they do not appear to make the claim that $[a_{ij}]$ is the matrix of multiplication by $\alpha$.
A: isn't this just the usual situation where the matrix that transforms the basis elements looks transposed from the point of view of the coordinates? (something that must be badly taught since it causes endless confusion). using the summation convention, with $a$ as coordinates, $e$ as basis elements we have:
$$
Ma = Ma_ie_i = a_i M e_i = a_i(m_{ij}e_j) = (a_im_{ij})e_j
$$
A: It may be a question of convention; some authors define matrices so that operation of the matrix must be obtained by right-multiplication of a row vector by the matrix rather than left-multiplication of a column vector by the matrix. Such authors typically define companion matrices transposed, with the entries $1$ just above the diagonal and minus the polynomial coefficients in the final row.
With this convention, the image by a linear map $g$ of source basis vector $e_i$ is expressed in the destination basis $[f_1,\ldots,f_n]$ by row $i$ of the matrix $A$ representing $g$, which gives $g(e_i)=\sum_j A_{i,j}f_j$, and this would correspond to what you wrote, with $g$ multiplication by $x$ and both bases equal to $[y_1,\ldots,y_n]$. If however Samuel should not adhere to this right-multiplying convention then you are right and his formula is wrong, as it should involve a column of $A$ instead. 
