Two examples of modules: quotienting and direct sum, need some clarification.

127 Views Asked by At
  1. Any ring $A$ has the natural structure of a left, resp. right, module over itself. A submodule $J \subset A$ is just the same thing as a left, resp. right, ideal of $A$. Therefore, the set $A/J$ also has the natural structure of a left, resp. right, $A$-module.

  2. For any ring $A$ and an integer $n > 0$, the abelian group $A^n = A \oplus \ldots \oplus A$ of column, resp. row, vectors has the structure of a left, resp. right, $\text{M}_n(A)$-module.

I don't quite follow this, as it's quite terse. Can anybody help clarify what is being said here? Specifically:

  1. Can anybody explain to me what the multiplication $R×M \to M$ is here, in both these cases?
  2. How do we verify the module axioms here, in both cases?
  3. What is the intuition behind working with the $A$-module $A/J$, and what is the intuition behind working with the $\text{M}_n(A)$-module $A^n$?
1

There are 1 best solutions below

0
On BEST ANSWER
  1. In $A/J$ the elements are of the from $a+J$ for some $a\in A$. Now what is the natural choice to multiply this by $a'\in A$? Well the easiest thing to do is to take $a'a+J$ which does the job. For the direct sum remember that you can define them as functions from $\{1, \dots , n\}$ to $A$, here the natural choice for an action of $A$ on a function $f$ is given by $(a.f)(n)=a(f(n))$.

  2. I assume that you know that both constructions in 1. have a group structure. Then you just have to check that this scalar multiplication is well-defined and that the axioms for the scalar multiplication are satisfied.

  3. Considering $A^n$ as $\text{M}_n(A)$-module is just like in linear algebra that a $n\times n$ matrix with values in $A$ gives you a linear map on $A^n$ to $A^n$. For the other you may want some geometric interpretation like taking the ideal (continuous, smooth or something you want)functions which vanish at a point $x$, when you take the quotient you get functions which agree on a neighborhood of $x$. Then this tells you can multiply those local functions by global functions. An other nice thing is when $A$ is a PID then any finely generated module $M$ is of the form $A^n \oplus A/J_1 \oplus \dots \oplus A/J_m$. This is the elementary divisor theorem.