Finitely Generated Modules - Bland Problem 1(a), Set 2.2

[Math Amateur]

Homework Statement

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.2 Free Modules ... ...

I need someone to check my solution to the first part of Problem 1(a) of Problem Set 2.2 ...

Problem 1(a) of Problem Set 2.2 reads as follows:

Can someone please critique my proof and either confirm it to be correct and/or point out the errors and shortcomings ...

Homework Equations

The Attempt at a Solution

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My solution/proof of the first part of Problem 1(a) is as follows:
We claim that $M \bigoplus N$ is finitely generated ...Now ...

$M \bigoplus N =$ the direct product $M \times N$ since we are dealing with the external direct sum of a finite number of modules ...

$M$ finitely generated $\Longrightarrow \exists$ a finite subset $X \subseteq M$ such that

$M = \sum_X x_i R = \{x_1 r_1 + \ ... \ ... \ x_m r_m \mid x_i \in X, r_i \in R \}$ ... ... ... ... (1)
$N$ finitely generated $\Longrightarrow \exists$ a finite subset $Y \subseteq N$ such that

$N = \sum_Y y_i R = \{y_1 r_1 + \ ... \ ... \ y_n r_n \mid y_i \in Y, r_i \in R \}$ ... ... ... ... (2)
$M \bigoplus N$ finitely generated $\Longrightarrow \exists$ a finite subset $S \subseteq M \bigoplus N$ such that

$M \bigoplus N = \sum_S ( x_i, y_i ) R = \{ (x_1, y_1) r_1 + \ ... \ ... \ ( x_s, y_s) r_s \mid (x_i, y_i) \in S , r_i \in R \}$$= \{ (x_1 r_1, y_1 r_1) + \ ... \ ... \ + ( x_s r_s, y_s r_s) \}$

$= \{ (x_1 r_1 + \ ... \ ... \ + x_s r_s , y_1 r_1 + \ ... \ ... \ + y_s r_s \}$ ... ... ... ... ... (3)
Now if we take $s \ge m, n$ in (3) ... ...

Then the sum $x_1 r_1 + \ ... \ ... \ + x_s r_s$ ranging over all $x_i$ and $r_i$ will generate all the elements in $M$ as the first variable in $M \bigoplus N$

... and the sum $y_1 r_1 + \ ... \ ... \ + y_s r_s$ ranging over all $y_i$ and $r_i$ will generate all the elements in $N$ as the second variable in $M \bigoplus N$

Since s is finite ... $M \bigoplus N$ is finitely generated ...
Can someone please critique my proof and either confirm it to be correct and/or point out the errors and shortcomings ...

Peter

 

[andrewkirk]

The easiest way to prove this is to construct a finite generating set. A natural candidate is
$$S\triangleq X^*\cup Y^* $$
where $X^*\triangleq X\times\{0_N\}$ and $Y^*\triangleq \{0_M\}\times Y$
Choose an arbitrary element $(a,b)\in M\oplus N$ and use the fact that there exist $r_1,...,r_m\in R$ and $s_1,...,s_n\in R$ such that
$a=\sum_{j=1}^m r_jx_j$ and $b=\sum_{j=1}^n s_jy_j$ to express $(a,b)$ as a finite sum of elements of $S$.

 

[Math Amateur]

The easiest way to prove this is to construct a finite generating set. A natural candidate is
$$S\triangleq X^*\cup Y^* $$
where $X^*\triangleq X\times\{0_N\}$ and $Y^*\triangleq \{0_M\}\times Y$
Choose an arbitrary element $(a,b)\in M\oplus N$ and use the fact that there exist $r_1,...,r_m\in R$ and $s_1,...,s_n\in R$ such that
$a=\sum_{j=1}^m r_jx_j$ and $b=\sum_{j=1}^n s_jy_j$ to express $(a,b)$ as a finite sum of elements of $S$.

Thanks Andrew ... indeed a much nicer way to do the proof ...

Peter