K-algebras, endomorphisms and k-linear mappings

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings Exercise 2.1 (1) reads as follows:"Let A be a k-algebra, where k is a field.

Given an A-module M, verify that any endomorphism of M as an A-module is a k-linear mapping."

My thinking on this follows.I would very much appreciate it if someone critiques my argument … and hopefully confirms it is OK ...
Consider \(\displaystyle A\) to be a right \(\displaystyle A\)-module where the action of \(\displaystyle A\) on \(\displaystyle M\) is denoted as \(\displaystyle xa\) where \(\displaystyle a \in A\) and \(\displaystyle x \in M\)

Now an endomorphism of \(\displaystyle M\) as an \(\displaystyle A\)-module is defined as a mapping \(\displaystyle f \ : \ M \to M\) such that \(\displaystyle f\) is a homomorphism of the additive groups and further, that:

\(\displaystyle (xf)a =(xa)f\) for all \(\displaystyle x \in M, a \in A\)Now, … … if we identify any element \(\displaystyle a\) in the
field \(\displaystyle k\) with the element \(\displaystyle a.1\) in the algebra A then we can write:

\(\displaystyle (xf)a =(xa)f\) for all \(\displaystyle x \in M, a \in k\)

and so \(\displaystyle f\) is then a \(\displaystyle k\)-linear mapping.

Can someone please confirm that the argument above is valid?

Peter

 

[Euge]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings Exercise 2.1 (1) reads as follows:"Let A be a k-algebra, where k is a field.

Given an A-module M, verify that any endomorphism of M as an A-module is a k-linear mapping."

My thinking on this follows.I would very much appreciate it if someone critiques my argument … and hopefully confirms it is OK ...
Consider \(\displaystyle A\) to be a right \(\displaystyle A\)-module where the action of \(\displaystyle A\) on \(\displaystyle M\) is denoted as \(\displaystyle xa\) where \(\displaystyle a \in A\) and \(\displaystyle x \in M\)

Now an endomorphism of \(\displaystyle M\) as an \(\displaystyle A\)-module is defined as a mapping \(\displaystyle f \ : \ M \to M\) such that \(\displaystyle f\) is a homomorphism of the additive groups and further, that:

\(\displaystyle (xf)a =(xa)f\) for all \(\displaystyle x \in M, a \in A\)Now, … … if we identify any element \(\displaystyle a\) in the
field \(\displaystyle k\) with the element \(\displaystyle a.1\) in the algebra A then we can write:

\(\displaystyle (xf)a =(xa)f\) for all \(\displaystyle x \in M, a \in k\)

and so \(\displaystyle f\) is then a \(\displaystyle k\)-linear mapping.

Can someone please confirm that the argument above is valid?

Peter

I think you should put in more detail in your argument that $(xf)a = (xa)f$ for all $x\in M$ and $a\in k$. This is how I would argue the last part:

Take $x\in M$ and $a\in k$. Then

$\displaystyle (xf)a = (xf)(a\cdot 1) = (x(a\cdot 1))f = (xa)f$.

The first and third equalities follows from the distributive property and the middle equality follows from the $A$-homomorphism property of $f$.

 

[Math Amateur]

I think you should put in more detail in your argument that $(xf)a = (xa)f$ for all $x\in M$ and $a\in k$. This is how I would argue the last part:

Take $x\in M$ and $a\in k$. Then

$\displaystyle (xf)a = (xf)(a\cdot 1) = (x(a\cdot 1))f = (xa)f$.

The first and third equalities follows from the distributive property and the middle equality follows from the $A$-homomorphism property of $f$.

Thanks for the critique, Euge ... Most helpful to me ...

Peter