MHB Simple Issue with Submodules - Northcott, Proposition 2, pages 7-8

[Math Amateur]

I am reading D. G. Northcott's book, Lessons on Rings, Modules and Multiplicities.

On pages 7 and 8, Northcott defines submodules in terms of the inclusion mapping, and then presents Proposition 2 and its proof as follows:
https://www.physicsforums.com/attachments/3500
View attachment 3501Under the definition of a submodule and before the statement and proof of Proposition 2, Northcott makes the following argument: (Note I have changed the notation to fit with Proposition 2)" ... ... Let $$a_1, a_2$$ belong to $$A$$ and let $$r$$ be an element of $$R$$. Since the image of the sum of $$a_1 \text{ and } a_2$$ is the sum of their separate images, we see that $$a_1 + a_2$$ is the same whether we regard $$a_1, a_2$$ as elements of $$A$$ or as elements of $$N$$. ... ... etc. etc. "Essentially, Northcott seems to be saying that

$$a_1, a_2 \in A $$

and

$$j(a_1 + a_2) = j(a_1) + j(a_2) = a_1 + a_2$$

means that (but how exactly does it follow?)

if $$a_1, a_2 \in A$$ then $$a_1 + a_2 \in A$$

Can someone demonstrate rigorously that this is in fact true - I do not completely follow Northcott's argument ...

Help will be appreciated ... ...

Peter

 

[Fallen Angel]

Hi Peter,

Essentially, Northcott seems to be saying that

$$a_1, a_2 \in A $$

and

$$j(a_1 + a_2) = j(a_1) + j(a_2) = a_1 + a_2$$

means that (but how exactly does it follow?)

if $$a_1, a_2 \in A$$ then $$a_1 + a_2 \in A$$

$$A$$ is a submodule, so it must be closed with respect to the sum, without needing any more conditions.
This is not the point of what the book is saying.

In the book you have three modules, $$A\subseteq M \subseteq N$$ (I'm not following the notation) and he is proving that, if the inclusion is a module homomorphism, then the sum in A and the sum in M are the same map over the elements in A.

I mean, as long as you have two modules you have defined
$$\begin{array}{cccc}+_{A}:& A \times A & \longrightarrow A \\ & (x,y)& \mapsto & x+_{A}y\end{array}$$

and

$$\begin{array}{cccc}+_{M}:& M \times M & \longrightarrow M \\ & (x,y)& \mapsto & x+_{M}y\end{array}$$

He proves that $$+_{M}|_{A}=+_{A}$$

 

[Math Amateur]

Hi Peter,

$$A$$ is a submodule, so it must be closed with respect to the sum, without needing any more conditions.
This is not the point of what the book is saying.

In the book you have three modules, $$A\subseteq M \subseteq N$$ (I'm not following the notation) and he is proving that, if the inclusion is a module homomorphism, then the sum in A and the sum in M are the same map over the elements in A.

I mean, as long as you have two modules you have defined
$$\begin{array}{cccc}+_{A}:& A \times A & \longrightarrow A \\ & (x,y)& \mapsto & x+_{A}y\end{array}$$

and

$$\begin{array}{cccc}+_{M}:& M \times M & \longrightarrow M \\ & (x,y)& \mapsto & x+_{M}y\end{array}$$

He proves that $$+_{M}|_{A}=+_{A}$$

Hi Fallen Angel,

Thanks for the help ... appreciate it ...

Still reflecting on what you have written ...

Thanks again,

Peter

***EDIT*** Just re-read your post ... ... yes, very clear ... most helpful ... thanks!