MHB 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 $$A$$ to be a right $$A$$-module where the action of $$A$$ on $$M$$ is denoted as $$xa$$ where $$a \in A$$ and $$x \in M$$

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

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

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

and so $$f$$ is then a $$k$$-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 $$A$$ to be a right $$A$$-module where the action of $$A$$ on $$M$$ is denoted as $$xa$$ where $$a \in A$$ and $$x \in M$$

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

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

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

and so $$f$$ is then a $$k$$-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