Direct Sums and Factor Modules .... Bland Problem 14, Problem Set 2.1

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help to make a meaningful start on Problem 14 of Problem Set 2.1 ...

Problem 14 of Problem Set 2.1 reads as follows:View attachment 8115I am somewhat overwhelmed by this problem ...

Can someone please hep me to make a meaningful start on the problem ...Hope someone can help ...

Peter

 

[steenis]

Define an R-homomorphism $$f:\bigoplus_\Delta M_\alpha \to \bigoplus_\Delta \frac{M_\alpha}{N_\alpha}$$

 

[Math Amateur]

Define an R-homomorphism $$f:\bigoplus_\Delta M_\alpha \to \bigoplus_\Delta \frac{M_\alpha}{N_\alpha}$$

Thanks for the help, Steenis ...

Hmm ... but still not sure how to proceed ...

Can you help a bit further ...

Peter

 

[steenis]

It is an obvious R-map.

Another hint: what is the obvious R-map between $M$ and $M/N$ ?

 

[Math Amateur]

It is an obvious R-map.

Another hint: what is the obvious R-map between $M$ and $M/N$ ?

Hi Steenis ...

The obvious R-map between $M$ and $M/N$ is the canonical surjection or natural map \(\displaystyle \eta \ : \ M \rightarrow M/N\) ...

... defined by \(\displaystyle \eta (x) = x + N = \overline{ x }\) ...
Now you suggest we consider ...

\(\displaystyle f:\bigoplus_\Delta M_\alpha \to \bigoplus_\Delta \frac{M_\alpha}{N_\alpha}\)... but how do we define this R-homomorphism ...?Well ... we can write \(\displaystyle x \in \bigoplus_\Delta M_\alpha\) as \(\displaystyle x = \sum_\Delta x_\alpha\) ...... but how do we define \(\displaystyle f\) ... is it something like ...\(\displaystyle f(x) = f ( \sum_\Delta x_\alpha ) = \sum_\Delta ( x_\alpha + N_\alpha )\) ...Unsure of this ... can you help further ...

... and anyway ... how do we proceed ...I am having trouble understanding the exact and explicit form of elements in \(\displaystyle ( \bigoplus_\Delta M_\alpha ) / ( \bigoplus_\Delta N_\alpha )\) and \(\displaystyle \bigoplus_\Delta M_\alpha / N_\alpha\) ...Hope you can help further ...

Peter

 

[steenis]

1) from the context it does not show that the direct sums are internal, but that they are external (compare the definitions), so the notations should be ...

2) if you have the notations right, your R-map is correct.

3) now determine the kernel of the R-map.

 

[Math Amateur]

1) from the context it does not show that the direct sums are internal, but that they are external (compare the definitions), so the notations should be ...

2) if you have the notations right, your R-map is correct.

3) now determine the kernel of the R-map.

Hi Steenis ... thanks ...

Have taken note of the points you have made ...We have to show that \(\displaystyle ( \bigoplus_\Delta M_\alpha ) / ( \bigoplus_\Delta N_\alpha )\) \(\displaystyle \cong\) \(\displaystyle \bigoplus_\Delta M_\alpha / N_\alpha\) ...So we define ...

\(\displaystyle f:\bigoplus_\Delta M_\alpha \to \bigoplus_\Delta M_\alpha / N_\alpha \)

where \(\displaystyle f( (x_\alpha) ) = \bigoplus_\Delta ( x_\alpha + N_\alpha ) \)Now f is a surjective homomorphism with kernel ...

\(\displaystyle \text{Ker } f = \bigoplus_\Delta ( 0 + N_\alpha ) = \bigoplus_\Delta N_\alpha\) ...Therefore ... by the First Isomorphism Theorem we have ...\(\displaystyle ( \bigoplus_\Delta M_\alpha ) / \text{Ker } f \) \(\displaystyle \cong\) \(\displaystyle \bigoplus_\Delta M_\alpha / N_\alpha\) ...That is ...\(\displaystyle ( \bigoplus_\Delta M_\alpha ) / ( \bigoplus_\Delta N_\alpha )\) \(\displaystyle \cong\) \(\displaystyle \bigoplus_\Delta M_\alpha / N_\alpha\) ...
Is that correct... ?

Peter

 

[steenis]

Yes all correct, one remark:

$\text{Ker } f = \bigoplus_\Delta ( n_\alpha + N_\alpha ) = \bigoplus_\Delta N_\alpha$, where $n_\alpha \in N_\alpha$

 

[Math Amateur]

Yes all correct, one remark:

$\text{Ker } f = \bigoplus_\Delta ( n_\alpha + N_\alpha ) = \bigoplus_\Delta N_\alpha$, where $n_\alpha \in N_\alpha$

Thanks for the correction, Steenis ...

... ... and thanks for all your help with this problem ...

Peter