MHB Short Exact Sequences and Direct Sums .... Bland, Proposition 3.2.7 .... ....

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

Currently I am focused on Section 3.2 Exact Sequences in [FONT=MathJax_Main]Mod[FONT=MathJax_Math]R ... ...

I need some help in order to fully understand the proof of Proposition 3.2.7 ...

Proposition 3.2.7 and its proof read as follows:

https://www.physicsforums.com/attachments/8082
In the above proof we read the following:

"... ... Then $$M_2 \cong M/ \text{ Ker } g \cong N$$ and $$ \text{ Ker } g = \text{ I am } f \cong M_1$$ ... ... My questions regarding the above are as follows:Question 1I understand that $$M_2 \cong M/ \text{ Ker } g$$ by the First Isomorphism Theorem for Modules ... ... but why is $$M_2 \cong M/ \text{ Ker } g \cong N$$ ... ... ?Question 2


Why, exactly, is $$\text{ Ker } g = \text{ I am } f \cong M_1$$ ... ... ?

Help will be much appreciated ...

Peter
 
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Hi Peter,

For question 1, this is a particular case of $(A\oplus B)/A\simeq B$, with $A = \ker g$ and $B=N$. To prove that, you can use the second (or third, depending on the book) isomorphism theorem:
$$
\frac{A+B}{A}\simeq\frac{B}{A\cap B}
$$
taking into account the fact that $A\cap B=0$ if the sum is direct.

For question 2, $\mathrm{img}\: f = f(M_1) \simeq M_1$ because $f$ is an injective homomorphism. If you restrict the co-domain to the image, you get a bijection and therefore an isomorphism. In some cases, $M_1$ will actually be a submodule of $M$.
 
 

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