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

[Math Amateur]

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:
View attachment 8067
My solution/proof of the first part of Problem 1(a) is as follows:
We claim that \(\displaystyle M \bigoplus N\) is finitely generated ... Now ...

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

\(\displaystyle M\) finitely generated \(\displaystyle \Longrightarrow \exists\) a finite subset \(\displaystyle X \subseteq M\) such that

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

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

\(\displaystyle M \bigoplus N = \sum_S ( x_i, y_i ) R = \{ (x_1, y_1) r_1 + \ ... \ ... \ ( x_s, y_s) r_s \mid (x_i, y_i) \in S , r_i \in R \} \)\(\displaystyle = \{ (x_1 r_1, y_1 r_1) + \ ... \ ... \ + ( x_s r_s, y_s r_s) \} \)\(\displaystyle = \{ (x_1 r_1 + \ ... \ ... \ + x_s r_s , y_1 r_1 + \ ... \ ... \ + y_s r_s \}\) ... ... ... ... ... (3)
Now if we take \(\displaystyle s \ge m, n \) in (3) ... ...Then the sum \(\displaystyle x_1 r_1 + \ ... \ ... \ + x_s r_s\) ranging over all \(\displaystyle x_i\) and \(\displaystyle r_i\) will generate all the elements in \(\displaystyle M\) as the first variable in \(\displaystyle M \bigoplus N \)

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

Since \(\displaystyle s\) is finite ... \(\displaystyle M \bigoplus N \) 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

 

[steenis]

We claim that \(\displaystyle M \bigoplus N\) is finitely generated ...

\(\displaystyle M \bigoplus N\) = the direct product \(\displaystyle M \times N\) since we are dealing with the external direct sum of a finite number of modules ...
\(\displaystyle M\) finitely generated \(\displaystyle \Longrightarrow \exists\) a finite subset \(\displaystyle X \subseteq M\) such that
\(\displaystyle M = \sum_X x_i R = \{x_1 r_1 + \ ... \ ... \ x_m r_m \mid x_i \in X, r_i \in R \}\) ... ... ... ... (1)

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

So each $x \in M$ can be written as a finite sum $x = \sum_X x_i r_i$ ($r_i \in R$).
So each $y \in N$ can be written as a finite sum $y = \sum_Y y_i s_i$ ($s_i \in R$).

\(\displaystyle M \bigoplus N\) finitely generated \(\displaystyle \Longrightarrow \exists\) a finite subset \(\displaystyle S \subseteq M \bigoplus N\) such that

This must be: To prove that $M \bigoplus N$ is finitely generated, so we have to prove that there exists a finite subset $S \subseteq M \bigoplus N$ such that $M \bigoplus N = \sum_S z_i R$.

I think it has to be this way:
Define $X'=\{ (x_i, 0) | x_i \in X \}$ and $Y'=\{ (0, y_i) | y_i \in Y \}$ and $Z=X' \cup Y'$. Then $Z \subset M \bigoplus N$ and $Z$ is finite.

Take $z \in M \bigoplus N$, then $z=(m,n)$ with $m \in M$ and $n \in N$.
$m$ can be written as $m = \sum_X x_i r_i$ and $n$ can be written as $n = \sum_Y y_i s_i$.

Then $z = (m,n) = (\sum_X x_i r_i, \sum_Y y_i s_i) = (\sum_X x_i r_i, 0) + (0, \sum_Y y_i s_i) = \sum_X (x_i, 0)r_i + \sum_Y (0, y_i)s_i = \sum_{X'} x'_i r_i +\sum_{Y'} y'_i s_i = \sum_Z z_i t_i$

in which $x'_i=(x_i, 0) \in X'$ and $y'_i=(0, y_i) \in Y'$ and $t_i \in R$.

This proves that $M \bigoplus N$ is finitely generated.