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

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The discussion centers on Proposition 3.2.7 from Paul E. Bland's "Rings and Their Modules," specifically regarding the proof of exact sequences in module theory. Participants clarify the application of the First Isomorphism Theorem for Modules, demonstrating that \(M_2 \cong M/\text{Ker } g \cong N\) through the isomorphism theorem. Additionally, it is established that \(\text{Ker } g = \text{img } f \cong M_1\) due to the injectivity of the homomorphism \(f\). Errors in Bland's proposition are also noted, prompting further exploration of the topic.

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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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