MHB Split Short Exact Sequences .... Bland, Proposition 3.2.6 .... ....

[Math Amateur]

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

Currently I am focused on Section 3.2 Exact Sequences in $$\text{Mod}_R$$ ... ...

I need some help in order to fully understand the proof of Proposition 3.2.6 ...

Proposition 3.2.6 and its proof read as follows:
View attachment 8078
In the above proof of Proposition 3.2.6 we read the following:"... ... Then $$x - x'\in \text{Ker } g = \text{Im } f$$, so $$(x - f( f' (x))) - (x' - f( f' (x'))) = ( x - x') - ( f ( f'(x) ) - f ( f'(x') ) )$$

$$= ( x - x') - f ( f' ( x - x') )$$

$$\in \text{Ker } f' \cap \text{Im } f = 0$$ ... ...

Thus it follows that $$g'$$ is well-defined ... ... "Can someone please explain exactly why/how $$( x - x') - f ( f' ( x - x') ) \in \text{Ker } f' \cap \text{Im } f = 0$$ ... ... Further, can someone please explain in some detail how the above working shows that $$g'$$ is well-defined ...
Help will be much appreciated ...

Peter

 

[castor28]

Hi Peter,

By a previous remark, each of $x-f(f'(x))$ and $x'-f(f'(x'))$ belongs to $\ker f'$. As $\ker f'$ is a sub-module, this shows that
$$u = (x-f(f'(x))) - (x'-f(f'(x'))) \in\ker f'$$

on the other hand, we have $x-x'\in\ker g=\mathrm{img}\:f$, and, obviously, $f(\ldots)\in\mathrm{img}\:f$. This shows that we also have $u\in\mathrm{img}\:f$, and, as $\ker f'\cap\mathrm{img}\:f=0$, $u=0$.

This shows that, if you use $x'$ instead of $x$ (subject to $g(x')=g(x)$) in the definition of $g'(y)$, the difference will be $u=0$, which means that you will get the same value for $g'(y)$; this is what "g' is well-defined" means.

This proof looks a little like black magic, but there is a trick that allows you to see what happens. You will end up proving that $M$ is isomorphic to $M_1\times M_2$. Of course, you cannot use that in the proof, but you can use it to understand what happens in the proof.

Knowing that, we can write any element of $M$ as $(a,b)$, with $a\in M_1$ and $b\in M_2$. You can define:
$$\begin{align*}
f(a) &= (a,0)\\
f'(a,b) &= a\\
g(a,b) &= b\\
\end{align*}$$
and you are trying to define $g'(b)$ as $(0,b)$. If $x$ is any element that maps to $b$, like $(c,b)$, you cannot simply define $g'(b)=(c,b)$, because there can be many possible elements $(c,b)$ in the pre-image of $b$. The trick is to use $f$ and $f'$ to get rid of $c$. Specifically, you have, using the formula in the text : $(c,b) - f(f'(c,b))= (c,b) - f(c) = (c,b) - (c,0) = (0,b)$, and this is what you want.

 

[Math Amateur]

Hi Peter,

By a previous remark, each of $x-f(f'(x))$ and $x'-f(f'(x'))$ belongs to $\ker f'$. As $\ker f'$ is a sub-module, this shows that
$$u = (x-f(f'(x))) - (x'-f(f'(x'))) \in\ker f'$$

on the other hand, we have $x-x'\in\ker g=\mathrm{img}\:f$, and, obviously, $f(\ldots)\in\mathrm{img}\:f$. This shows that we also have $u\in\mathrm{img}\:f$, and, as $\ker f'\cap\mathrm{img}\:f=0$, $u=0$.

This shows that, if you use $x'$ instead of $x$ (subject to $g(x')=g(x)$) in the definition of $g'(y)$, the difference will be $u=0$, which means that you will get the same value for $g'(y)$; this is what "g' is well-defined" means.

".

Thanks castor28 ... most helpful ...

Most interesting and enlightening is when you write: " ... ... This proof looks a little like black magic, but there is a trick that allows you to see what happens. You will end up proving that $M$ is isomorphic to $M_1\times M_2$. Of course, you cannot use that in the proof, but you can use it to understand what happens in the proof.

Knowing that, we can write any element of $M$ as $(a,b)$, with $a\in M_1$ and $b\in M_2$. You can define:
$$\begin{align*}
f(a) &= (a,0)\\
f'(a,b) &= a\\
g(a,b) &= b\\
\end{align*}$$
and you are trying to define $g'(b)$ as $(0,b)$. If $x$ is any element that maps to $b$, like $(c,b)$, you cannot simply define $g'(b)=(c,b)$, because there can be many possible elements $(c,b)$ in the pre-image of $b$. The trick is to use $f$ and $f'$ to get rid of $c$. Specifically, you have, using the formula in the text : $(c,b) - f(f'(c,b))= (c,b) - f(c) = (c,b) - (c,0) = (0,b)$, and this is what you want. ... ... Still reflecting on these ideas ...

Thanks again ...

Peter