Characterizing the A-module M/S I've been working through this for a little while, and I'm not 100% sure I understand what I'm supposed to be doing here, or maybe I'm not grasping correctly what they mean by "Characterize".
Characterize each of the following quotient $A$-modules $M/S$ (sorry if translation is a little bit off).


*

*$M = A^n$ and $S=\{(a_1,...,a_n) : \sum_{i=1} ^n a_i=0\}$.


I would imagine this one to be the easiest one, but I'm not sure how to approach it. I can imagine a few obvious elements of this module, but I can't really see it in a general sense.


*$M = A[X]$ and $S=\{f \in M : f(1) = 0 \}$.


Unless I'm missing something, $f$ is in the same class as 0 if $ x - 1 | f$ and the class of $f$ is made up by $f + p(x)(x-1)$. Is this "characterization" enough? 
I also understand that this very similar to the quotient in the first item, which I imagine is the point. 
Edit: Ok, so this is isomorphic to $A$ with $\phi : A[X] \rightarrow A$ so that $\phi(f) = f(1) $ since it's obvious that $Ker\phi = S$ and $Im \phi = A$.


*$M = M_n (A)$ and $S= \{ (a_{ij}) \in M : a_{i1} = 0 \forall 1 \leq i \leq n \}$


Here since $ \begin{bmatrix}
    0 & x_{12} & x_{13} & \dots  & x_{1n} \\
    0 & x_{22} & x_{23} & \dots  & x_{2n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & x_{n2} & x_{n3} & \dots  & x_{nn}
\end{bmatrix} $ is in the same class as $0$, $f$ and $g$ are in the same class if their first columns coincide. As I understand it, this is isomorphic to $A^n$ since it's just taking the projection of the first column. 
Any help would be greatly appreciated.
 A: The problem asks to find an $A$-module $N$ (eventually you already know from class) such that $M/S\simeq N$ (eventually by using a well known isomorphism theorem).
Hints.


*

*Define $f:A^n\to A$ by $f(a_1,\dots,a_n)=\sum_{i=1}^na_i$.

*Define $f:A[X]\to A$ by $f(p(X))=p(1)$.

*Define $f:M_n(A)\to A^n$ by $f((a_{ij})_{i,j})=(a_{11},\dots,a_{n1})$.
A: To augment user26857's answer, note that:


*

*$(a_1,a_2,\dots,a_n) + S$ 


$= (a_1+a_2+\cdots+a_n,0,\dots,0) + (a_1-a_2-\cdots-a_n,a_2,\dots,a_n) + S$
$= (a_1+a_2+\cdots+a_n,0,\dots,0) + S$.



*If $p(X) = q(X)(X - 1) + r$ (for $r \in A$), then $p(X) + S = r + S$.





*If $B = \begin{bmatrix}
    x_{11} & x_{12} & x_{13} & \dots  & x_{1n} \\
    x_{12} & x_{22} & x_{23} & \dots  & x_{2n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    x_{n1} & x_{n2} & x_{n3} & \dots  & x_{nn}
\end{bmatrix}$,


then $B + S = \begin{bmatrix}x_{11}&0&0&\dots&0\\x_{12}&0&0&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\x_{n1}&0&0&\dots&0\end{bmatrix} + \begin{bmatrix}
    0 & x_{12} & x_{13} & \dots  & x_{1n} \\
    0 & x_{22} & x_{23} & \dots  & x_{2n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & x_{n2} & x_{n3} & \dots  & x_{nn}
\end{bmatrix} + S$
$= \begin{bmatrix}x_{11}&0&0&\dots&0\\x_{12}&0&0&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\x_{n1}&0&0&\dots&0\end{bmatrix} + S$.
