MHB 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 $$M \bigoplus N$$ 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

 

[steenis]

We claim that $$M \bigoplus N$$ is finitely generated ...

$$M \bigoplus N$$ = 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)

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 $y = \sum_Y y_i s_i$ ($s_i \in R$).

$$M \bigoplus N$$ finitely generated $$\Longrightarrow \exists$$ a finite subset $$S \subseteq M \bigoplus N$$ 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.