Direct Sums of Noetherian Modules .... Bland Proposition 4.2.7 .... ....

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.7 ... ...

Proposition 4.2.7 reads as follows:https://www.physicsforums.com/attachments/8208In the above proof by Paul Bland we read the following:

" ... ... and since each $$M_i$$ is isomorphic to a submodule of $$\bigoplus_{ i = 1 }^n M_i$$ ... ... "Can someone please explain to me how/why each $$M_i$$ is isomorphic to a submodule of $$\bigoplus_{ i = 1 }^n M_i$$ ... ... ?

Do we know what the submodule is ... ?
Help will be appreciated ...

Peter

 

[caffeinemachine]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.7 ... ...

Proposition 4.2.7 reads as follows:In the above proof by Paul Bland we read the following:

" ... ... and since each $$M_i$$ is isomorphic to a submodule of $$\bigoplus_{ i = 1 }^n M_i$$ ... ... "Can someone please explain to me how/why each $$M_i$$ is isomorphic to a submodule of $$\bigoplus_{ i = 1 }^n M_i$$ ... ... ?

Do we know what the submodule is ... ?
Help will be appreciated ...

Peter

What you are asking is the following: If $M$ and $N$ are $R$-modules, then how is $M$ as submodule of $M\oplus N$? Just unravel the definition of $M\oplus N$. $M\oplus N$ as a set is just $M\times N$, where you add and scale coordinate-wise. So a natural embedding of $M$ into $M\oplus N$ is $i:M\to M\oplus N$ given by $i(x)=(x, 0)$. Do you see why $i$ is an injective $R$-module homomorphism?

 

[Math Amateur]

What you are asking is the following: If $M$ and $N$ are $R$-modules, then how is $M$ as submodule of $M\oplus N$? Just unravel the definition of $M\oplus N$. $M\oplus N$ as a set is just $M\times N$, where you add and scale coordinate-wise. So a natural embedding of $M$ into $M\oplus N$ is $i:M\to M\oplus N$ given by $i(x)=(x, 0)$. Do you see why $i$ is an injective $R$-module homomorphism?

Thanks for the help, caffeinemachine ... ...

You write:

" ... ... Do you see why $i$ is an injective $R$-module homomorphism? ... ... "

Well ...

$$i( x + y ) = i(x) + i(y)$$ holds because ...

$$i(x+y) = (x+y, 0 ) = (x,0) + (y,0) = i(x) + i(y)$$ ... ...

... and ...

$$i(xa) = i(x) a$$ holds where $$a \in$$ ring $$R$$ holds since ...

$$i(xa) = (xa,0) = (x,0) a = i(x) a$$ ...

So $$i$$ is a homomorphism ...

... and ...

if $$i(x) = i(y)$$ ... then ...

$$(x,0) = (y,0) $$

$$\Longrightarrow x = y$$

... so $$i$$ is injective ...Peter