Start on Bland Problem 1, Problem Set 4.1: Generating & Cogenerating Modules

[Math Amateur]

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

Currently I am focused on Section 4.2 Noetherian and Artinian Modules ... ...

I need some help in order to make a meaningful start on Problem 1, Problem Set 4.1 ...

Problem 1, Problem Set 4.1 reads as follows:
View attachment 8097
Can someone please help me to make a meaningful start on Problem 1 above ... as well as clarifying the definition of one module generating another ... see my notes following ... ...

I am having a bit of trouble pinning down a definition from Bland that gives me the meaning of a module \(\displaystyle M\) generating a module \(\displaystyle N\) ...

Now ... on page 104 of Bland (see first page of Bland Section 4.1 below) we read the following:

" ... ... If \(\displaystyle \mathscr{S}\) is a set of submodules of \(\displaystyle M\) such that \(\displaystyle M = \sum_{ \mathscr{S} } N\), then \(\displaystyle \mathscr{S}\) is said to span \(\displaystyle M\) ... ... "

Now if spanning \(\displaystyle M\) is the same as generating \(\displaystyle M\) ... so in Problem 1 we could take \(\displaystyle \mathscr{S} = \{ M \} \) as generating \(\displaystyle N\) ... but is that the correct interpretation of "span" and would the definition be useful in Problem 1 ... ?

Alternatively ... on page 105 we read in Definition 4.1.2 (see Bland's text displayed below ... )

" ... ... An R-module \(\displaystyle M\) is said to be generated by a set \(\displaystyle \{ M_\alpha \}_\Delta \) of R-modules if there is an epimorphism \(\displaystyle \bigoplus_\Delta M_\alpha \rightarrow M\). ... ... "

Maybe this is the definition to se in Problem 1 above ... with \(\displaystyle \{ M_\alpha \}_\Delta = \{ N \}\) ... ... is that the correct start on the problem ...?
Can someone please clarify the above ... and then, further, help me to make a meaningful start on the problem ...Help will be appreciated,

Peter==========================================================================================In order to give readers the definitions, notation and context of Bland's treatment of generating and cogenerating classes, I am providing access to Bland's Section 4.1 ... as follows ...
https://www.physicsforums.com/attachments/8098
https://www.physicsforums.com/attachments/8099
View attachment 8100
Hope that helps ...

Peter

 

[steenis]

Before you read section 4.1, you should read def.1.4.3. on p.28, prop.1.4.4. on p.29, and section 2.2 on free modules with def.2.2.1 on p.51, prop.2.2.2. on p.52 and prop.2.2.3 on p.53.

Let $M$ be a left R-module and $S \subset M$. If every element $x \in M$ can be written as a finite sum $x = \Sigma_{i=1,n} \text{ } x_ir_i$ with $x_i \in S$ and $r_i \in R$, then we say that $S$ is a set of generators of $M$, and $M$ is generated by $S$, and we write $M = \Sigma_S \text{ } xR$. ($S$ maybe finite, in that case we say that $M$ is finitely generated.)

Here we say that $M$ is generated by $S$, but we could also say: $M$ is spanned by all the $xR$, for $x \in S$. Just like $\mathbb{R}^3$ is spanned by the x, y, and z-axis. Here the x-axis = $xR$, the y-axis = $yR$, and the z-axis = $zR$, and $\mathbb{R}^3 =xR + yR + zR$. Of course $R = \mathbb{R}$ here.

Now the $xR$ are very special submodules of $M$ that “span” $M$. You could ask yourself: can $M$ be spanned by more general submodules of $M$ ? For instance, $M = A + B$, with $A, B \leq M$, when for each $m \in M$ there is a $a \in A$ and a $b \in B$ such that $m = a + b$.

In this light we can say that if $M = \Sigma N_\alpha$ and $N_\alpha \leq M$ then $M$ is spanned by the submodules $N_\alpha \leq M$. Furthermore for each $m \in M$ there are $n_\alpha \in N_\alpha$ (not necessarily unique) such that $m = \Sigma n_\alpha$.

I did not answer all your questions so maybe later more.

 

[Math Amateur]

Before you read section 4.1, you should read def.1.4.3. on p.28, prop.1.4.4. on p.29, and section 2.2 on free modules with def.2.2.1 on p.51, prop.2.2.2. on p.52 and prop.2.2.3 on p.53.

Let $M$ be a left R-module and $S \subset M$. If every element $x \in M$ can be written as a finite sum $x = \Sigma_{i=1,n} \text{ } x_ir_i$ with $x_i \in S$ and $r_i \in R$, then we say that $S$ is a set of generators of $M$, and $M$ is generated by $S$, and we write $M = \Sigma_S \text{ } xR$. ($S$ maybe finite, in that case we say that $M$ is finitely generated.)

Here we say that $M$ is generated by $S$, but we could also say: $M$ is spanned by all the $xR$, for $x \in S$. Just like $\mathbb{R}^3$ is spanned by the x, y, and z-axis. Here the x-axis = $xR$, the y-axis = $yR$, and the z-axis = $zR$, and $\mathbb{R}^3 =xR + yR + zR$. Of course $R = \mathbb{R}$ here.

Now the $xR$ are very special submodules of $M$ that “span” $M$. You could ask yourself: can $M$ be spanned by more general submodules of $M$ ? For instance, $M = A + B$, with $A, B \leq M$, when for each $m \in M$ there is a $a \in A$ and a $b \in B$ such that $m = a + b$.

In this light we can say that if $M = \Sigma N_\alpha$ and $N_\alpha \leq M$ then $M$ is spanned by the submodules $N_\alpha \leq M$. Furthermore for each $m \in M$ there are $n_\alpha \in N_\alpha$ (not necessarily unique) such that $m = \Sigma n_\alpha$.

I did not answer all your questions so maybe later more.

Thanks Steenis ...

... just studying your post now ...

I very much appreciate your help ...

Peter

 

[Math Amateur]

Before you read section 4.1, you should read def.1.4.3. on p.28, prop.1.4.4. on p.29, and section 2.2 on free modules with def.2.2.1 on p.51, prop.2.2.2. on p.52 and prop.2.2.3 on p.53.

Let $M$ be a left R-module and $S \subset M$. If every element $x \in M$ can be written as a finite sum $x = \Sigma_{i=1,n} \text{ } x_ir_i$ with $x_i \in S$ and $r_i \in R$, then we say that $S$ is a set of generators of $M$, and $M$ is generated by $S$, and we write $M = \Sigma_S \text{ } xR$. ($S$ maybe finite, in that case we say that $M$ is finitely generated.)

Here we say that $M$ is generated by $S$, but we could also say: $M$ is spanned by all the $xR$, for $x \in S$. Just like $\mathbb{R}^3$ is spanned by the x, y, and z-axis. Here the x-axis = $xR$, the y-axis = $yR$, and the z-axis = $zR$, and $\mathbb{R}^3 =xR + yR + zR$. Of course $R = \mathbb{R}$ here.

Now the $xR$ are very special submodules of $M$ that “span” $M$. You could ask yourself: can $M$ be spanned by more general submodules of $M$ ? For instance, $M = A + B$, with $A, B \leq M$, when for each $m \in M$ there is a $a \in A$ and a $b \in B$ such that $m = a + b$.

In this light we can say that if $M = \Sigma N_\alpha$ and $N_\alpha \leq M$ then $M$ is spanned by the submodules $N_\alpha \leq M$. Furthermore for each $m \in M$ there are $n_\alpha \in N_\alpha$ (not necessarily unique) such that $m = \Sigma n_\alpha$.

I did not answer all your questions so maybe later more.

Hi Steenis,

I think I have now understood what you have said ... but still feeling unsure of Problem 1, Problem Set 4.1 ...We seem to be dealing with a case of an R-module \(\displaystyle N\) generated by an R-Module \(\displaystyle M\) ...

So ... we seem to have a case where \(\displaystyle N = \sum_\Delta M_\alpha = M\) ... is that right?

How exactly do we deal with this ... ?Is it best to think of this as a case there exists an epimorphism \(\displaystyle \bigoplus_\Delta M_\alpha = M \rightarrow N\) ... ?Can you help ... ?

Peter

 

[steenis]

In my post #2, I tried to make clear the difference between “generating a module” and “spanning a module”.

A module $M$ is generated by a subset $S \subset M$ if every element $x \in M$ can be written as a finite sum $x = \Sigma_{i=1,n} \text{ } x_ir_i$ with $x_i \in S$ and $r_i \in R$, in that case $M = \Sigma_S \text{ } xR$.

A module $M$ is spanned by a family $(N_\alpha)_{\alpha \in \Delta}$ with $N_\alpha \leq N$ if $N = \Sigma_{\alpha \in \Delta} \text{ } N_\alpha $, that is, for each $m\in M$ there are $n_\alpha \in N_\alpha$ such that $m = \Sigma n_\alpha$.

[[[ I must say that the notation $\Sigma$ (defined in prop.1.4.4. p.29) is confusing here . I think Bland meant $\Sigma M_\alpha = \{ \text{finite sums } \Sigma m_\alpha | m_\alpha \in M_\alpha \}$, i.e., the direct sum $\bigoplus M_\alpha$, defined later in the book. I will read $\bigoplus$ for $\Sigma$ and $m = \Sigma n_\alpha$ is then a finite sum. ]]]

We see that if $M$ is generated by a set $S \subset M$ then $M$ is spanned by the family $\{ xR | x \in S \}$ because $M = \Sigma_S \text{ } xR$. Proposition 4.1.1 shows the converse way: when is a module that is spanned by submodules, also generated by a set.

Proposition. 4.1.1. learns in what way “spanning” and “generating” coincides. If $M$ is finitely generated then every spanning set containes a finite spanning set. And conversely if every spanning set containes a finite spanning set, then $M$ is finitely generated. It also learns what to do if the spanning sets $M_\alpha$ are not contained in $M$. In that case we have to construct epimorphisms $\{ f_\alpha:M_\alpha \to M \}$ and use $\text{im } f_\alpha$, in stead of $M_\alpha$.

Now Bland defines what it means for a module to be generated by a set of R-modules. It is a general definition. The definition 4.1.2. is this:
(1) An $R$-module $N$ is said to be generated by a set $(M_\alpha)_\Delta$ of R-modules [[[ $(M_\alpha)_\Delta$ generates $N$ ]]] if there is an epimorphism $\bigoplus_\Delta M_\alpha \to N$.

(2) An $R$-module $M$ is said to generate an $R$-module $N$ if there is an epimorphism $M^{(\Delta)} \to N$ for some set $\Delta$.

These are not two definitions, but it is one definition: one for the case the generating set is a family of modules $(M_\alpha)_\Delta$ and one for the case that the generating set is a singleton $\{ M \}$.

So, (2) reads that if the module $M$ generates the module $N$, then there exists a set $\Delta$ (finite or infinite) and an epimorphism $M^{(\Delta)} \to N$.

At this moment it is not clear to me how (1) implies (2), I take it for granted.

Peter, I hope you understand all this, including how to interprete definition 4.1.2. If not, we will try again.
I think your confusion lies in definition 4.1.2.

A meaningful start of solving problem 1 is to decide which part of definition 4.1.2. you are going to use.

Almost Craftsman.

 

[Math Amateur]

In my post #2, I tried to make clear the difference between “generating a module” and “spanning a module”.

A module $M$ is generated by a subset $S \subset M$ if every element $x \in M$ can be written as a finite sum $x = \Sigma_{i=1,n} \text{ } x_ir_i$ with $x_i \in S$ and $r_i \in R$, in that case $M = \Sigma_S \text{ } xR$.

A module $M$ is spanned by a family $(N_\alpha)_{\alpha \in \Delta}$ with $N_\alpha \leq N$ if $N = \Sigma_{\alpha \in \Delta} \text{ } N_\alpha $, that is, for each $m\in M$ there are $n_\alpha \in N_\alpha$ such that $m = \Sigma n_\alpha$.

[[[ I must say that the notation $\Sigma$ (defined in prop.1.4.4. p.29) is confusing here . I think Bland meant $\Sigma M_\alpha = \{ \text{finite sums } \Sigma m_\alpha | m_\alpha \in M_\alpha \}$, i.e., the direct sum $\bigoplus M_\alpha$, defined later in the book. I will read $\bigoplus$ for $\Sigma$ and $m = \Sigma n_\alpha$ is then a finite sum. ]]]

We see that if $M$ is generated by a set $S \subset M$ then $M$ is spanned by the family $\{ xR | x \in S \}$ because $M = \Sigma_S \text{ } xR$. Proposition 4.1.1 shows the converse way: when is a module that is spanned by submodules, also generated by a set.

Proposition. 4.1.1. learns in what way “spanning” and “generating” coincides. If $M$ is finitely generated then every spanning set containes a finite spanning set. And conversely if every spanning set containes a finite spanning set, then $M$ is finitely generated. It also learns what to do if the spanning sets $M_\alpha$ are not contained in $M$. In that case we have to construct epimorphisms $\{ f_\alpha:M_\alpha \to M \}$ and use $\text{im } f_\alpha$, in stead of $M_\alpha$.

Now Bland defines what it means for a module to be generated by a set of R-modules. It is a general definition. The definition 4.1.2. is this:
(1) An $R$-module $N$ is said to be generated by a set $(M_\alpha)_\Delta$ of R-modules [[[ $(M_\alpha)_\Delta$ generates $N$ ]]] if there is an epimorphism $\bigoplus_\Delta M_\alpha \to N$.

(2) An $R$-module $M$ is said to generate an $R$-module $N$ if there is an epimorphism $M^{(\Delta)} \to N$ for some set $\Delta$.

These are not two definitions, but it is one definition: one for the case the generating set is a family of modules $(M_\alpha)_\Delta$ and one for the case that the generating set is a singleton $\{ M \}$.

So, (2) reads that if the module $M$ generates the module $N$, then there exists a set $\Delta$ (finite or infinite) and an epimorphism $M^{(\Delta)} \to N$.

At this moment it is not clear to me how (1) implies (2), I take it for granted.

Peter, I hope you understand all this, including how to interprete definition 4.1.2. If not, we will try again.
I think your confusion lies in definition 4.1.2.

A meaningful start of solving problem 1 is to decide which part of definition 4.1.2. you are going to use.

Almost Craftsman.

Thanks so much for your help, steenis

I am still rather vague on how to go about the problem ... but will make a start ...

Peter

 

[Math Amateur]

In my post #2, I tried to make clear the difference between “generating a module” and “spanning a module”.

A module $M$ is generated by a subset $S \subset M$ if every element $x \in M$ can be written as a finite sum $x = \Sigma_{i=1,n} \text{ } x_ir_i$ with $x_i \in S$ and $r_i \in R$, in that case $M = \Sigma_S \text{ } xR$.

A module $M$ is spanned by a family $(N_\alpha)_{\alpha \in \Delta}$ with $N_\alpha \leq N$ if $N = \Sigma_{\alpha \in \Delta} \text{ } N_\alpha $, that is, for each $m\in M$ there are $n_\alpha \in N_\alpha$ such that $m = \Sigma n_\alpha$.

[[[ I must say that the notation $\Sigma$ (defined in prop.1.4.4. p.29) is confusing here . I think Bland meant $\Sigma M_\alpha = \{ \text{finite sums } \Sigma m_\alpha | m_\alpha \in M_\alpha \}$, i.e., the direct sum $\bigoplus M_\alpha$, defined later in the book. I will read $\bigoplus$ for $\Sigma$ and $m = \Sigma n_\alpha$ is then a finite sum. ]]]

We see that if $M$ is generated by a set $S \subset M$ then $M$ is spanned by the family $\{ xR | x \in S \}$ because $M = \Sigma_S \text{ } xR$. Proposition 4.1.1 shows the converse way: when is a module that is spanned by submodules, also generated by a set.

Proposition. 4.1.1. learns in what way “spanning” and “generating” coincides. If $M$ is finitely generated then every spanning set containes a finite spanning set. And conversely if every spanning set containes a finite spanning set, then $M$ is finitely generated. It also learns what to do if the spanning sets $M_\alpha$ are not contained in $M$. In that case we have to construct epimorphisms $\{ f_\alpha:M_\alpha \to M \}$ and use $\text{im } f_\alpha$, in stead of $M_\alpha$.

Now Bland defines what it means for a module to be generated by a set of R-modules. It is a general definition. The definition 4.1.2. is this:
(1) An $R$-module $N$ is said to be generated by a set $(M_\alpha)_\Delta$ of R-modules [[[ $(M_\alpha)_\Delta$ generates $N$ ]]] if there is an epimorphism $\bigoplus_\Delta M_\alpha \to N$.

(2) An $R$-module $M$ is said to generate an $R$-module $N$ if there is an epimorphism $M^{(\Delta)} \to N$ for some set $\Delta$.

These are not two definitions, but it is one definition: one for the case the generating set is a family of modules $(M_\alpha)_\Delta$ and one for the case that the generating set is a singleton $\{ M \}$.

So, (2) reads that if the module $M$ generates the module $N$, then there exists a set $\Delta$ (finite or infinite) and an epimorphism $M^{(\Delta)} \to N$.

At this moment it is not clear to me how (1) implies (2), I take it for granted.

Peter, I hope you understand all this, including how to interprete definition 4.1.2. If not, we will try again.
I think your confusion lies in definition 4.1.2.

A meaningful start of solving problem 1 is to decide which part of definition 4.1.2. you are going to use.

Almost Craftsman.

Hi steenis,

Thanks for the help so far ...

But ... not getting very far with the problem ...... we have ...

\(\displaystyle M\) generates \(\displaystyle N \Longrightarrow \exists\) a set \(\displaystyle H \subseteq \text{Hom}_R (M,N)\) such that \(\displaystyle N = \sum_H \text{Im } f\) Now ... assume \(\displaystyle M\) generates \(\displaystyle N\)

We have ...

\(\displaystyle M\) generates \(\displaystyle N\)

\(\displaystyle \Longrightarrow \exists\) an epimorphism \(\displaystyle \psi \: \ M^{ ( \Delta ) } = M \bigoplus M \bigoplus\) ... ... \(\displaystyle \rightarrow\) \(\displaystyle N\) for some set \(\displaystyle H\) ... ...

Define homomorphisms \(\displaystyle h_i\) as follows:

\(\displaystyle h_1 \ : \ ( x_1, 0, 0, \ ... \ ... ) \rightarrow n \ ... \ ... ? \ \ \ \ \text{ where } h_1 \in H\) ... ...

\(\displaystyle h_2 \ : \ ( 0, x_2, 0, \ ... \ ... ) \rightarrow n\) ... ... ? where \(\displaystyle \ \ \ h_2 \in H\) ... ...

etc ?? how do we progress ... ??
Then we assume ..

\(\displaystyle \exists set H \subseteq \text{Hom}_R (M,N)\) such that \(\displaystyle N = \sum_H \text{Im } f\)

Now \(\displaystyle N = \sum_H \text{Im } f \Longrightarrow \exists\) an epimorphism \(\displaystyle \sum f \rightarrow N\) ... ... ?
Can you help?

Peter

 

[steenis]

Peter, I sure hope you understand the theory of spanning and generating sets and how to use definition 4.1.2. To be honest, I did not understand your remarks in post #1, so I tried to explain the theory from the beginning. If there is something you do not understand, try to describe it bit by bit and we go through it.
PF (including you ?) ignores my remarks and continues on the wrong track. So you can post the solution of problem 1 on PF and amaze them.

Let’s prove problem 1
$M$ and $N$ are R-modules.
$M$ (finitely) generates $N$ $\Leftrightarrow$ there is a (finite) subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{im } f$.

$\Rightarrow$) $M$ generates $N$ so there is an (index-) set $\Delta$ and an epimorphism $f:M^{(\Delta)} = \bigoplus_\Delta M \to N$ by definition 4.1.2.(2), see post #5.
For each $\alpha \in \Delta$ define $f_\alpha = f \circ i_\alpha$, where $i_\alpha: M \to \bigoplus_\Delta M $ is the canonical inclusion.
Now $f_\alpha \in \text{Hom }(M, N)$.
Define $H = \{f_\alpha | \alpha \in \Delta \}$
All you have to do now is to prove that $N = \Sigma_H \text{ I am } f$, you can do that.$\Leftarrow$) There is a subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{ I am } f$
Define $h = |H|$ and $ f:M^{(h)} = \bigoplus_h M \to N$ by $f = \Sigma_H f_\alpha$, that is, $f((x_\alpha)_H) = \Sigma_H f(x_\alpha)$.
Prove that $f$ is an epimorpism, you can do that. And then you are ready.
Notice: I improved the notation in post #11, below.

 

[Math Amateur]

Peter, I sure hope you understand the theory of spanning and generating sets and how to use definition 4.1.2. To be honest, I did not understand your remarks in post #1, so I tried to explain the theory from the beginning. If there is something you do not understand, try to describe it bit by bit and we go through it.
PF (including you ?) ignores my remarks and continues on the wrong track. So you can post the solution of problem 1 on PF and amaze them.

Let’s prove problem 1
$M$ and $N$ are R-modules.
$M$ (finitely) generates $N$ $\Leftrightarrow$ there is a (finite) subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{im } f$.

$\Rightarrow$) $M$ generates $N$ so there is an (index-) set $\Delta$ and an epimorphism $f:M^{(\Delta)} = \bigoplus_\Delta M \to N$ by definition 4.1.2.(2), see post #5.
For each $\alpha \in \Delta$ define $f_\alpha = f \circ i_\alpha$, where $i_\alpha: M \to \bigoplus_\Delta M $ is the canonical inclusion.
Now $f_\alpha \in \text{Hom }(M, N)$.
Define $H = \{f_\alpha | \alpha \in \Delta \}$
All you have to do now is to prove that $N = \Sigma_H \text{ I am } f$, you can do that.

$\Leftarrow$) There is a subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{ I am } f$
Define $h = |H|$ and $ f:M^{(h)} = \bigoplus_h M \to N$ by $f = \Sigma_H f_\alpha$, that is, $f((x_\alpha)_H) = \Sigma_H f(x_\alpha)$.
Prove that $f$ is an epimorpism, you can do that. And then you are ready.

Thanks steenis ...

I am now studying your post carefully...

Thanks again ...

Peter

 

[Math Amateur]

Peter, I sure hope you understand the theory of spanning and generating sets and how to use definition 4.1.2. To be honest, I did not understand your remarks in post #1, so I tried to explain the theory from the beginning. If there is something you do not understand, try to describe it bit by bit and we go through it.
PF (including you ?) ignores my remarks and continues on the wrong track. So you can post the solution of problem 1 on PF and amaze them.

Let’s prove problem 1
$M$ and $N$ are R-modules.
$M$ (finitely) generates $N$ $\Leftrightarrow$ there is a (finite) subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{im } f$.

$\Rightarrow$) $M$ generates $N$ so there is an (index-) set $\Delta$ and an epimorphism $f:M^{(\Delta)} = \bigoplus_\Delta M \to N$ by definition 4.1.2.(2), see post #5.
For each $\alpha \in \Delta$ define $f_\alpha = f \circ i_\alpha$, where $i_\alpha: M \to \bigoplus_\Delta M $ is the canonical inclusion.
Now $f_\alpha \in \text{Hom }(M, N)$.
Define $H = \{f_\alpha | \alpha \in \Delta \}$
All you have to do now is to prove that $N = \Sigma_H \text{ I am } f$, you can do that.

$\Leftarrow$) There is a subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{ I am } f$
Define $h = |H|$ and $ f:M^{(h)} = \bigoplus_h M \to N$ by $f = \Sigma_H f_\alpha$, that is, $f((x_\alpha)_H) = \Sigma_H f(x_\alpha)$.
Prove that $f$ is an epimorpism, you can do that. And then you are ready.

Hi Steenis ...

Thanks again for all your help ...I think I have read and understood all of your posts on this problem (including the posts on PF ... ) ... and the posts are very helpful to me ...

Just now trying to fully understand your proof of Problem 1 ... First trying to get an idea of the nature and explicit form of the mappings involved ...

For \(\displaystyle i_\alpha\) ... to try to make things easier assume for a moment that \(\displaystyle \Delta = \mathbb{N}\) ...

Now \(\displaystyle i_\alpha \ : \ M \rightarrow \bigoplus_\Delta M = \bigoplus_\mathbb{N} M\) ... ...

Then for \(\displaystyle m \in M\), the explicit form of \(\displaystyle i_\alpha\) is as follows:

\(\displaystyle i_1 (m) = (m, 0, 0, 0, \ ... \ ... \ ... )\)

\(\displaystyle i_2 (m) = (0, m, 0, 0, \ ... \ ... \ ... )\)

\(\displaystyle i_3 (m) = (0, 0, m, 0, \ ... \ ... \ ... )\)

... ... etc ...Is that correct?
Now consider \(\displaystyle f \ : \ M^{ ( \Delta ) } = M^{ ( \mathbb{N} ) } \rightarrow N \text{ and } f_\alpha\) where

\(\displaystyle f_\alpha = f \circ i_\alpha\) ... so ... where \(\displaystyle m \in M\) ...

\(\displaystyle f_1(m) = f \circ i_1 (m)= f( (m, 0, 0, 0, \ ... \ ... \ ... ) )\) where \(\displaystyle f\) is an epimorphism onto \(\displaystyle N\) ... ...

\(\displaystyle f_2(m) = f \circ i_2 (m)= f( (0, m, 0, 0, \ ... \ ... \ ... ) )\) where \(\displaystyle f\) is an epimorphism onto \(\displaystyle N\) ... ...

... ... etc ... ...Is that correct?Note that \(\displaystyle f_\alpha\) is a homomorphism since it is the composition of two homomorphisms, namely \(\displaystyle f\) and \(\displaystyle i_\alpha\) ... ... But ... how do I show that \(\displaystyle N = \sum_H \text{Im } f\) ... ...

... or do you mean \(\displaystyle N = \sum_H \text{Im } f_\alpha\) ... ...
Hope that you can give some feedback on the above and also give some further help with the proof ...

Peter

 

[steenis]

Yes, correct.
Yes, correct.
But remember that $\Delta$ is not equal to $\mathbb{N}$

The notation is somewhat sloppy, I agree, but I used Bland’s notation:
$N = \Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \} = \Sigma_\Delta \text{ I am } f_\alpha$

You also have to figure out what is $f((x_\alpha)_\Delta)$ for $(x_\alpha)_\Delta \in M^{(\Delta)} = \bigoplus_\Delta M$.
You need prop.2.1.5 for this.

I also improved the notation of the $\Leftarrow )$ direction:

$\Leftarrow$) There is a subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{ I am } g$. Let $H = \{h_\alpha | \alpha \in \Delta \}$.
Define $ f:M^{(\Delta)} = \bigoplus_\Delta M \to N$ by $f = \Sigma_H \text{ } g = \Sigma_\Delta \text{ } h_\alpha$, that is, $f((x_\alpha)_\Delta) = \Sigma_\Delta h_\alpha(x_\alpha)$.
Prove that $f$ is an epimorpism, you can do that. And then you are ready.

 

[Math Amateur]

Yes, correct.
Yes, correct.
But remember that $\Delta$ is not equal to $\mathbb{N}$

The notation is somewhat sloppy, I agree, but I used Bland’s notation:
$N = \Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \} = \Sigma_\Delta \text{ I am } f_\alpha$

You also have to figure out what is $f((x_\alpha)_\Delta)$ for $(x_\alpha)_\Delta \in M^{(\Delta)} = \bigoplus_\Delta M$.
You need prop.2.1.5 for this.

I also improved the notation of the $\Leftarrow )$ direction:

$\Leftarrow$) There is a subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{ I am } g$. Let $H = \{h_\alpha | \alpha \in \Delta \}$.
Define $ f:M^{(\Delta)} = \bigoplus_\Delta M \to N$ by $f = \Sigma_H \text{ } g = \Sigma_\Delta \text{ } h_\alpha$, that is, $f((x_\alpha)_\Delta) = \Sigma_\Delta h_\alpha(x_\alpha)$.
Prove that $f$ is an epimorpism, you can do that. And then you are ready.[/QUO
Hi Steenis ...

Studying your post now ...

You are asking me to consider \(\displaystyle f( ( x_\alpha ) )\) and refer me to Proposition 2.1.5 on page 43 ...

Now ... on page 43 in the paragraph just above Proposition 2.1.5 \(\displaystyle f\) is defined by \(\displaystyle f(( x_\alpha )) = \sum_\Delta f_\alpha (x)\) ... but why is it defined this way ... and is it the same \(\displaystyle f\) in Proposition 2.1.5 and also in the problem ... and what exactly is \(\displaystyle x\) ...

Is \(\displaystyle f(( x_\alpha )) = \sum_\Delta f_\alpha (x)\) the answer to the question you posed when you wrote: " ... ... You also have to figure out what is $f((x_\alpha)_\Delta)$ for $(x_\alpha)_\Delta \in M^{(\Delta)} = \bigoplus_\Delta M$.
You need prop.2.1.5 for this. ... ... " ?

Are you able to clarify things a bit for me ... sorry if I'm being a bit slow ...

Peter

 

[steenis]

$i_\alpha: M_\alpha \to \bigoplus_\Delta M_\alpha $ are the canonical inclusions for all $\alpha \in \Delta$.

If you have R-maps $h_\alpha:M_\alpha \to N$, then prop.2.1.5 says there is a unique $h: \bigoplus_\Delta M_\alpha \to N$ such that $h \circ i_\alpha = h_\alpha$ and this R-map $h$ is defined in the proof of prop.2.1.5. by $h((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } h_\alpha (x_\alpha)$

In your case you have given R-maps $f_\alpha:M_\alpha \to N$. So by prop.2.1.5. there is a unique $g: \bigoplus_\Delta M_\alpha \to N$ such that $g \circ i_\alpha = f_\alpha$ and this R-map is defined in the proof of prop.2.1.5. by $g((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$
But your map $f:\bigoplus_\Delta M_\alpha \to N$ also satisfies $f \circ i_\alpha = f_\alpha$. Therefore there is only one possibility: $f = g$, because $g$ is unique.

 

[Math Amateur]

$i_\alpha: M_\alpha \to \bigoplus_\Delta M_\alpha $ are the canonical inclusions for all $\alpha \in \Delta$.

If you have R-maps $h_\alpha:M_\alpha \to N$, then prop.2.1.5 says there is a unique $h: \bigoplus_\Delta M_\alpha \to N$ such that $h \circ i_\alpha = h_\alpha$ and this R-map $h$ is defined in the proof of prop.2.1.5. by $h((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } h_\alpha (x_\alpha)$

In your case you have given R-maps $f_\alpha:M_\alpha \to N$. So by prop.2.1.5. there is a unique $g: \bigoplus_\Delta M_\alpha \to N$ such that $g \circ i_\alpha = f_\alpha$ and this R-map is defined in the proof of prop.2.1.5. by $g((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$
But your map $f:\bigoplus_\Delta M_\alpha \to N$ also satisfies $f \circ i_\alpha = f_\alpha$. Therefore there is only one possibility: $f = g$, because $g$ is unique.

Thanks for the clarification Steenis ...

Appreciate your help ...

But ... getting late here in Tasmania ...

Will start work on this again tomorrow morning ...

Thanks again ...

Peter

 

[Math Amateur]

$i_\alpha: M_\alpha \to \bigoplus_\Delta M_\alpha $ are the canonical inclusions for all $\alpha \in \Delta$.

If you have R-maps $h_\alpha:M_\alpha \to N$, then prop.2.1.5 says there is a unique $h: \bigoplus_\Delta M_\alpha \to N$ such that $h \circ i_\alpha = h_\alpha$ and this R-map $h$ is defined in the proof of prop.2.1.5. by $h((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } h_\alpha (x_\alpha)$

In your case you have given R-maps $f_\alpha:M_\alpha \to N$. So by prop.2.1.5. there is a unique $g: \bigoplus_\Delta M_\alpha \to N$ such that $g \circ i_\alpha = f_\alpha$ and this R-map is defined in the proof of prop.2.1.5. by $g((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$
But your map $f:\bigoplus_\Delta M_\alpha \to N$ also satisfies $f \circ i_\alpha = f_\alpha$. Therefore there is only one possibility: $f = g$, because $g$ is unique.

Hi Steenis ...

Still working on Problem 1, Problem Set 4.1 ... BUT ... have an issue/problem that is stopping my progress ...

My issue/problem relates to Bland initial treatment of external direct sums including Proposition 2.1.5 ... especially Bland's definition of the sum of a family of mappings ...

Bland's text on this is as follows:View attachment 8111
In the above text by Bland we read the following:

" ... ... We now need the concept of a family of mappings. If \(\displaystyle f_\alpha \ : \ M_\alpha \rightarrow N\) is an R-linear mapping for each \(\displaystyle \alpha \in \Delta\), where \(\displaystyle N\) is a fixed R-module, then \(\displaystyle f \ : \ \bigoplus_\Delta M_\alpha \rightarrow N\) defined by \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x)\) ... ... "

But in the last sentence of the proof of Proposition 2.1.5 ( ... again, see above text by Bland ... ) we read:

" ... ... If \(\displaystyle ( x_\alpha ) \in \bigoplus_\Delta M_\alpha\), then \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x_\alpha)\) ... ... "So ... in the text above the Proposition we have ... \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x)\) ... ... and in the proof of the proposition we have \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x_\alpha)\) ...

... which of these is correct ... or in some strange way, are they both correct ...

... I note that x is mentioned in the definition of the canonical injections above ..I do suspect ... from your last post that \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x_\alpha)\) is correct because that is how you express \(\displaystyle f\) ...

Can you clarify ... ?
But ... if \(\displaystyle f\) is defined by \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x_\alpha)\) ... then I have a further problem ...

... we know that \(\displaystyle f_\alpha \ : \ M_\alpha \rightarrow N\) ... that is the domain of \(\displaystyle f_\alpha\) is \(\displaystyle M_\alpha\) ... BUT ...PROBLEM ... \(\displaystyle ( x_\alpha ) \in \bigoplus_\Delta M_\alpha\) and \(\displaystyle ( x_\alpha ) \notin M_\alpha\) ...

... can you please clarify ?
Hope you can help ...

Peter
***EDIT***Just a further couple of thoughts ...

1. Defining \(\displaystyle f\) by \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x_\alpha)\) seems a little circular since the\(\displaystyle f_\alpha\) are defined in terms of \(\displaystyle f\) ... so how do we get a clear and explicit idea of the form of \(\displaystyle \sum_\Delta f_\alpha (x_\alpha)\) ... when it is defined in terms of \(\displaystyle f\) ...

2. In terms of Problem 1, Problem Set 4.1 ... given \(\displaystyle f( ( x_\alpha ) ) = \sum_\Delta f_\alpha (x_\alpha)\) we could argue that given that \(\displaystyle f\) is an epimorphism that \(\displaystyle \sum_\Delta f_\alpha\) is also a surjection and so \(\displaystyle N = \sum_H \text{Im } f_\alpha\) where \(\displaystyle f_\alpha \in H\) ... is that valid ...?
Can you comment?

Peter

 

[steenis]

$f((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x)$ is a typo.
In cases like this I always compare it with another textbook that I have available, for instance, the book of Anderson, on p.83.
So it has to be $f((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$.

Notice the double parenthesis in $f((x_\alpha)_\Delta)$ and the single parenthesis in $f_\alpha (x_\alpha)$.

The R-map $f:\bigoplus_\Delta M_\alpha \to N$ works on the elements $(x_\alpha)_\Delta$ of $\bigoplus_\Delta M_\alpha$ giving $f((x_\alpha)_\Delta)$; the R-maps $f_\alpha: M_\alpha \to N$ for $\alpha \in \Delta$ work on the elements $x_\alpha$ of $M_\alpha$, giving $f_\alpha (x_\alpha)$.
$f_\beta$ automatically picks the $\beta$-th coordinate of $(x_\alpha)_\Delta$ to work on, i.e., the elements of $M_\beta$. It should be because $f_\beta$ is only defined on $M_\beta$.

Suppose you have $M = M_1 \oplus M_2$ with elements $(x_1, x_2)$ then by definition $x_1$ is an element of $M_1$ and $x_2$ is an element of $M_2$. Furthermore, $f_1$ automatically picks the $x_1$ coordinate to work on and $f_2$ automatically pick the $x_2$ coordinate. And $f$ works on $(x_1, x_2)$.Given the R-maps $(f_\alpha)_\Delta$, you can define $f:\bigoplus_\Delta M_\alpha \to N$ by $f((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$, see prop.2.1.5: $f$ and $(f_\alpha)_\Delta$ must satisfy $f_\alpha = f \circ i_\alpha$.

On the other hand, given $f:\bigoplus_\Delta M_\alpha \to N$, you can define $(f_\alpha)_\Delta$ using $f_\alpha = f \circ i_\alpha$.

Using these last defined $(f_\alpha)_\Delta$, you can define an R-map $g:\bigoplus_\Delta M_\alpha \to N$ by $g((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$, see prop.2.1.5: $g$ and $(f_\alpha)_\Delta$ must satisfy $f_\alpha = g \circ i_\alpha$.
But now, by prop.2.1.5. f and g must coincide, $f=g$, using the facts: $f_\alpha = f \circ i_\alpha$ and $f_\alpha = g \circ i_\alpha$.
You think you are working circular, but as long as $f_\alpha = f \circ i_\alpha$ nothing can go wrong by prop.2.1.5.

 

[Math Amateur]

$f((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x)$ is a typo.
In cases like this I always compare it with another textbook that I have available, for instance, the book of Anderson, on p.83.
So it has to be $f((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$.

Notice the double parenthesis in $f((x_\alpha)_\Delta)$ and the single parenthesis in $f_\alpha (x_\alpha)$.

The R-map $f:\bigoplus_\Delta M_\alpha \to N$ works on the elements $(x_\alpha)_\Delta$ of $\bigoplus_\Delta M_\alpha$ giving $f((x_\alpha)_\Delta)$; the R-maps $f_\alpha: M_\alpha \to N$ for $\alpha \in \Delta$ work on the elements $x_\alpha$ of $M_\alpha$, giving $f_\alpha (x_\alpha)$.
$f_\beta$ automatically picks the $\beta$-th coordinate of $(x_\alpha)_\Delta$ to work on, i.e., the elements of $M_\beta$. It should be because $f_\beta$ is only defined on $M_\beta$.

Suppose you have $M = M_1 \oplus M_2$ with elements $(x_1, x_2)$ then by definition $x_1$ is an element of $M_1$ and $x_2$ is an element of $M_2$. Furthermore, $f_1$ automatically picks the $x_1$ coordinate to work on and $f_2$ automatically pick the $x_2$ coordinate. And $f$ works on $(x_1, x_2)$.Given the R-maps $(f_\alpha)_\Delta$, you can define $f:\bigoplus_\Delta M_\alpha \to N$ by $f((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$, see prop.2.1.5: $f$ and $(f_\alpha)_\Delta$ must satisfy $f_\alpha = f \circ i_\alpha$.

On the other hand, given $f:\bigoplus_\Delta M_\alpha \to N$, you can define $(f_\alpha)_\Delta$ using $f_\alpha = f \circ i_\alpha$.

Using these last defined $(f_\alpha)_\Delta$, you can define an R-map $g:\bigoplus_\Delta M_\alpha \to N$ by $g((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } f_\alpha (x_\alpha)$, see prop.2.1.5: $g$ and $(f_\alpha)_\Delta$ must satisfy $f_\alpha = g \circ i_\alpha$.
But now, by prop.2.1.5. f and g must coincide, $f=g$, using the facts: $f_\alpha = f \circ i_\alpha$ and $f_\alpha = g \circ i_\alpha$.
You think you are working circular, but as long as $f_\alpha = f \circ i_\alpha$ nothing can go wrong by prop.2.1.5.

Thanks Steenis ... your post was most helpful ...

In particular your advice:

" ... Notice the double parenthesis in $f((x_\alpha)_\Delta)$ and the single parenthesis in $f_\alpha (x_\alpha)$. ... "

was very clarifying ...I think I understood most of your post ... BUT ... still cannot make further progress on Problem 1 ...

Can you help further ...

Peter

 

[Math Amateur]

Yes, correct.
Yes, correct.
But remember that $\Delta$ is not equal to $\mathbb{N}$

The notation is somewhat sloppy, I agree, but I used Bland’s notation:
$N = \Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \} = \Sigma_\Delta \text{ I am } f_\alpha$

You also have to figure out what is $f((x_\alpha)_\Delta)$ for $(x_\alpha)_\Delta \in M^{(\Delta)} = \bigoplus_\Delta M$.
You need prop.2.1.5 for this.

I also improved the notation of the $\Leftarrow )$ direction:

$\Leftarrow$) There is a subset $H \subset \text{Hom }(M, N)$ such that $N = \Sigma_H \text{ I am } g$. Let $H = \{h_\alpha | \alpha \in \Delta \}$.
Define $ f:M^{(\Delta)} = \bigoplus_\Delta M \to N$ by $f = \Sigma_H \text{ } g = \Sigma_\Delta \text{ } h_\alpha$, that is, $f((x_\alpha)_\Delta) = \Sigma_\Delta h_\alpha(x_\alpha)$.
Prove that $f$ is an epimorpism, you can do that. And then you are ready.

Hi Steenis ...

Regarding the above post ... you write:

" ... ... $N = \Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \} = \Sigma_\Delta \text{ I am } f_\alpha$ ... ... "Can you explain why/how \(\displaystyle \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}\) ... ?You also write:

" ... ... You also have to figure out what is $f((x_\alpha)_\Delta)$ for $(x_\alpha)_\Delta \in M^{(\Delta)} = \bigoplus_\Delta M$. ... ... "

Well, we know from page 43 that \(\displaystyle f(( x_\alpha )) = \sum_\Delta f_\alpha ( x_\alpha )\) ... but how do we use this to solve the problem ...

... do we in fact use the fact that \(\displaystyle f\) is an epimorphism onto \(\displaystyle N\) to assert that \(\displaystyle \sum_\Delta f_\alpha ( x_\alpha )\) is an epimorphism and hence \(\displaystyle N\) is the sum of the images of \(\displaystyle f_\alpha\) ... Can you comment ...

If that is not a valid approach can you give some help on solving the problem ...

Peter

 

[steenis]

$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$
because $H = \{f_\alpha | \alpha \in \Delta \}$

You have to prove that $N = \Sigma_H \text{ I am } f$,
you know that $\Sigma_H \text{ I am } f \subset N$,
so left to prove that $N \subset \Sigma_H \text{ I am } f$.
Take $x \in N$, now use the fact that $f$ is an epimorphism …

 

[Math Amateur]

$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$
because $H = \{f_\alpha | \alpha \in \Delta \}$

You have to prove that $N = \Sigma_H \text{ I am } f$,
you know that $\Sigma_H \text{ I am } f \subset N$,
so left to prove that $N \subset \Sigma_H \text{ I am } f$.
Take $x \in N$, now use the fact that $f$ is an epimorphism …

Thanks Steenis ... but just a clarification ...

You write:

" ... ... \(\displaystyle \Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \}\) ... ... "

The above makes it look as if there are many \(\displaystyle f \in H\) ... but isn't \(\displaystyle f\) the unique mapping \(\displaystyle f \ : \ \bigoplus_\Delta M_\alpha \rightarrow N \)... ... ?

Peter***EDIT***

Just a further clarification ... how do we know $\Sigma_H \text{ I am } f \subset N$ ... indeed why/how is this true ...

Hope you can help ...

Peter

 

[steenis]

In:

$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$,

$f$ is a dummy variable that takes all the values in $H$. It has nothing to do with “the unique $f$” because this $f$ is not an element of $H$. Confusing, yes, but I did it on purpose because you have to be able interpret these notations correctly. So if you want, use another dummy variable.

It does remind me of my old promoter who panicked because I used the letter $x$ as a variable, for instance in $f(x)$, for him $x$ was always the big unknown who has to be solved.

 

[Math Amateur]

In:

$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$,

$f$ is a dummy variable that takes all the values in $H$. It has nothing to do with “the unique $f$” because this $f$ is not an element of $H$. Confusing, yes, but I did it on purpose because you have to be able interpret these notations correctly. So if you want, use another dummy variable.

It does remind me of my old promoter who panicked because I used the letter $x$ as a variable, for instance in $f(x)$, for him $x$ was always the big unknown who has to be solved.

Hi Steenis ... thanks for explaining that ... so if I'm right we have an epimorphism f and a dummy variable f ...

Peter

 

[steenis]

NO, we do not have a dummy variable $f$.
$f$ is a dummy variable in:
$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$
the same way $\alpha$ is a dummy variable.
In stead of $f$, you can use $g$, $z$, $\kappa$, or $M_{oon}$.
After you used it in this equation, $f$ is gone.

You do have an epimorphism $f$, which you need to prove the theorem.

 

[Math Amateur]

NO, we do not have a dummy variable $f$.
$f$ is a dummy variable in:
$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$
the same way $\alpha$ is a dummy variable.
In stead of $f$, you can use $g$, $z$, $\kappa$, or $M_{oon}$.
After you used it in this equation, $f$ is gone.

You do have an epimorphism $f$, which you need to prove the theorem.

Hi Steenis,

Just a further clarification ... how do we know $\Sigma_H \text{ I am } f \subset N$ ... indeed why/how is this true ...

Hope you can help ...

Peter

 

[steenis]

$\Sigma_H \text{ I am } k = \Sigma \{ \text{ I am } k | k \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \} = \Sigma_\Delta \text{ I am } f_\alpha$
How is $f_\alpha$ defined ?

 

[Math Amateur]

$\Sigma_H \text{ I am } k = \Sigma \{ \text{ I am } k | k \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \} = \Sigma_\Delta \text{ I am } f_\alpha$
How is $f_\alpha$ defined ?

\(\displaystyle f_\alpha = f \circ i_\alpha\) so \(\displaystyle f_\alpha \ : \ M_\alpha \rightarrow N\) ... so \(\displaystyle f_\alpha\) maps \(\displaystyle M_\alpha = M\) into \(\displaystyle N\) ...

... hence I think we can then say ...

$\Sigma_H \text{ I am } f = \Sigma \{ \text{ I am } f | f \in H \} = \Sigma \{ \text{ I am } f_\alpha | \alpha \in \Delta \}= \Sigma_\Delta \text{ I am } f_\alpha$ \(\displaystyle \subseteq N\) ...

Is that correct?

Peter

 

[steenis]

Yes because $\text{ I am } f_\alpha \subset N$,
and now take an $x \in N$ and ...

 

[Math Amateur]

Yes because $\text{ I am } f_\alpha \subset N$,
and now take an $x \in N$ and ...

Getting late here in Tasmania ...

Will start again on this tomorrow morning ...

Thank you for all your explanations, guidance and help ...

I wouldn't have half the understanding of the concepts we have been discussing without your help ...

So ... thanks again ...

Peter

 

[steenis]

I am sorry that I did not make it easy for you. But you have to be able to automatically interpret the notations.

Sleep well.

O, tomorrrow I have less time ...

 

[Math Amateur]

I am sorry that I did not make it easy for you. But you have to be able to automatically interpret the notations.

Sleep well.

O, tomorrrow I have less time ...

Let \(\displaystyle M\) and \(\displaystyle N\) be R-modules ...

We want to show that:

\(\displaystyle M\) generates \(\displaystyle N \Longleftrightarrow \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } f\)In this post we show that

\(\displaystyle M\) generates \(\displaystyle N \Longrightarrow \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } f\)So we assume that \(\displaystyle M\) generates \(\displaystyle N\) ...

Now ... \(\displaystyle M\) generates \(\displaystyle N\)

\(\displaystyle \Longrightarrow \exists\) a set \(\displaystyle \Delta\) and an epimorphism \(\displaystyle g \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N\) ... ... by definition 4.1.2 ...

Now, for each \(\displaystyle \alpha \in \Delta\) define \(\displaystyle g_\alpha = g \circ i_\alpha\) where \(\displaystyle i_\alpha \ : \ M \rightarrow \bigoplus_\Delta M \) ... that is, \(\displaystyle i_\alpha\) is the canonical inclusion/injection ... ...

Now, since \(\displaystyle g\) and \(\displaystyle i_\alpha\) are homomorphisms, it follows that each \(\displaystyle g_\alpha \in \text{ Hom }_R (M, N)\) ... ...Define \(\displaystyle H = \{ g_alpha \mid \alpha \in \Delta \}\)

We now claim that \(\displaystyle N = \sum_H \text{Im } g_\alpha\) where \(\displaystyle g_\alpha \ : \ M \rightarrow N\) is as defined above ...

But, we have that \(\displaystyle \sum_H \text{Im } g_\alpha \subseteq N\) since each \(\displaystyle g_\alpha\) is a mapping from \(\displaystyle M\) to \(\displaystyle N\) ...

So we need to show \(\displaystyle N \subseteq \sum_H \text{Im } g_\alpha\) So ... let \(\displaystyle x \in N\) ...

Now, \(\displaystyle g\) is an epimorphism from \(\displaystyle M^{ ( \Delta ) } = \bigoplus_\Delta M to N\) ...

... so ... \(\displaystyle \exists\) some \(\displaystyle ( y_\alpha ) \in M^{ ( \Delta ) } = \bigoplus_\Delta M\) such that \(\displaystyle g( ( y_\alpha ) ) = x\) ... ... ... (1)

But \(\displaystyle g( ( y_\alpha ) ) = \sum_\Delta g_\alpha ( y_\alpha )\) ... ... ... (2) (1)(2) \(\displaystyle \Longrightarrow g_\alpha ( y_\alpha ) = x\)

Therefore \(\displaystyle x \in \sum_H \text{Im } g_\alpha \)Is that correct?

(Note: most unsure of last step ... )

Peter

 

[steenis]

Very good.

Mention that (2) follows from prop.2.1.5

Mention that $g_\alpha(y_\alpha) \in \text{ I am } g_\alpha$

therefore $x = g((y_\alpha)) = \Sigma_\Delta g_\alpha(y_\alpha) \in \Sigma_\Delta \text{ I am } g_\alpha$Very good that you changed the polluted notation to avoid confusion.

 

[Math Amateur]

Very good.

Mention that (2) follows from prop.2.1.5

Mention that $g_\alpha(y_\alpha) \in \text{ I am } g_\alpha$

therefore $x = g((y_\alpha)) = \Sigma_\Delta g_\alpha(y_\alpha) \in \Sigma_\Delta \text{ I am } g_\alpha$Very good that you changed the polluted notation to avoid confusion.

Thanks for those points of improvement ...

The previously polluted notation ... :) ... was a helpful lesson for me ...

Thanks for all your help ... I learned a lot ...

Peter

 

[steenis]

Ok, now prove the converse direction.

I have to go now, I have other things to do.
I will check your posts this evening or tomorrow, dutch time.

 

[Math Amateur]

Ok, now prove the converse direction.

I have to go now, I have other things to do.
I will check your posts this evening or tomorrow, dutch time.

Oh Lord ... forgot the converse ... !

Had gone onto something else ... see new problem ...

But will return to converse...

Peter

 

[Math Amateur]

Let \(\displaystyle M\) and \(\displaystyle N\) be R-modules ...

We want to show that:

\(\displaystyle M\) generates \(\displaystyle N \Longleftrightarrow \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } f\)In this post we show that

\(\displaystyle M\) generates \(\displaystyle N \Longrightarrow \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } f\)So we assume that \(\displaystyle M\) generates \(\displaystyle N\) ...

Now ... \(\displaystyle M\) generates \(\displaystyle N\)

\(\displaystyle \Longrightarrow \exists\) a set \(\displaystyle \Delta\) and an epimorphism \(\displaystyle g \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N\) ... ... by definition 4.1.2 ...

Now, for each \(\displaystyle \alpha \in \Delta\) define \(\displaystyle g_\alpha = g \circ i_\alpha\) where \(\displaystyle i_\alpha \ : \ M \rightarrow \bigoplus_\Delta M \) ... that is, \(\displaystyle i_\alpha\) is the canonical inclusion/injection ... ...

Now, since \(\displaystyle g\) and \(\displaystyle i_\alpha\) are homomorphisms, it follows that each \(\displaystyle g_\alpha \in \text{ Hom }_R (M, N)\) ... ...Define \(\displaystyle H = \{ g_alpha \mid \alpha \in \Delta \}\)

We now claim that \(\displaystyle N = \sum_H \text{Im } g_\alpha\) where \(\displaystyle g_\alpha \ : \ M \rightarrow N\) is as defined above ...

But, we have that \(\displaystyle \sum_H \text{Im } g_\alpha \subseteq N\) since each \(\displaystyle g_\alpha\) is a mapping from \(\displaystyle M\) to \(\displaystyle N\) ...

So we need to show \(\displaystyle N \subseteq \sum_H \text{Im } g_\alpha\) So ... let \(\displaystyle x \in N\) ...

Now, \(\displaystyle g\) is an epimorphism from \(\displaystyle M^{ ( \Delta ) } = \bigoplus_\Delta M to N\) ...

... so ... \(\displaystyle \exists\) some \(\displaystyle ( y_\alpha ) \in M^{ ( \Delta ) } = \bigoplus_\Delta M\) such that \(\displaystyle g( ( y_\alpha ) ) = x\) ... ... ... (1)

But \(\displaystyle g( ( y_\alpha ) ) = \sum_\Delta g_\alpha ( y_\alpha )\) ... ... ... (2) (1)(2) \(\displaystyle \Longrightarrow g_\alpha ( y_\alpha ) = x\)

Therefore \(\displaystyle x \in \sum_H \text{Im } g_\alpha \)Is that correct?

(Note: most unsure of last step ... )

Peter

Let \(\displaystyle M\) and \(\displaystyle N\) be R-modules ...

We want to show that:

\(\displaystyle M\) generates \(\displaystyle N \Longleftrightarrow \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } g\)In this post we show that

\(\displaystyle \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } g \Longrightarrow M\) generates \(\displaystyle N\) So ... assume that \(\displaystyle \exists\) a subset \(\displaystyle H \subseteq \text{ Hom }_R (M, N)\) such that \(\displaystyle N = \sum_H \text{Im } g \) ...

Let \(\displaystyle H = \{ h_\alpha \mid \alpha \in \Delta \}\)

Define \(\displaystyle f \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N\) where \(\displaystyle f = \sum_H g = \sum_\Delta h_\alpha\) ...

But ... given this we know that then \(\displaystyle f(( x_\alpha )) = \sum_\Delta h_\alpha ( x_\alpha )\) ...Now ,,, we claim that \(\displaystyle f\) is an epimorphism ...To demonstrate this let \(\displaystyle x \in N\) ...

Then \(\displaystyle x \in \sum_H \text{Im } g = \sum_\Delta \text{Im } h_\alpha\) ...

Therefore \(\displaystyle \exists \ y \in M^{ ( \Delta ) }\) such that \(\displaystyle f(y) = \sum_\Delta h_\alpha (y) = x \) ...

So \(\displaystyle f\) is an epimorphism ... and hence \(\displaystyle M\) generates \(\displaystyle N\) ...

Is that correct?

Peter