1. ## Unique R-Module Homomorphism

I am reading Dummit and Foote Ch 10: Modules and in particulat section 10.3 Generation of Modules, Direct Sums and Free Modules.

I am having problems understanding and interpreting Theorm 6 - see attached pdf for the two relevant pages of D&F.

Problems I have are as follows:

1. We are told to "identify A as a subset of F(A) by a $\mapsto$ $f_a$ where $f_a$ is the function which is 1 at a and zero elsehwhere"

Thus $f_a$ $\in$ F(A) - but how can A be a subset of F(A)? The members of F(A) are functions $f_a$ , $f_b$ , $f_c$, ... etc. The entries are functions not just simple points or elements of A. The nature of the elements of A and F(A) seem different, so can we "identify" the elements of A as a subset of F(A)?

Further to this point where is the function a $\mapsto$ $f_a$ used as the proof progresses from this point?

2. Following the quote above regarding $f_a$, we read:

"We can, in this way, think of F(A) as all finite R-linear combinations of elements of A by identifying each function f with the sum $r_1 a_1$ + $r_2 a_2$ + ..... + $r_n a_n$ where f takes the value $r_i$ at $a_i$ and is zero at all other elements of A. Moreover, each element of F(A) has a unique expression as such a formal sum."

So you can see what assumptions I am working on, I have sketched my understanding of functions f, g $\in$ F(A) and also my understanding of $f_a$ - see pdf attached called sketches of functions

Given that D&F describe f as taking on the value $r_i$ at $a_i$ and being zero at all other elements, how can we show that each element of F(A) has a unique expression as a formal sum of the form $r_1 a_1$ + $r_2 a_2$ + ..... + $r_n a_n$.

[I am also somewhat confused by the fact that the functional values of f are $r_1$, $r_2$, etc and not $r_1 a_1$ , $r_2 a_2$, etc ]

Can anyone help clarify this theorem for me?

Peter

2. ## Re: Unique R-Module Homomorphism

Originally Posted by Bernhard
I am reading Dummit and Foote Ch 10: Modules and in particulat section 10.3 Generation of Modules, Direct Sums and Free Modules.

I am having problems understanding and interpreting Theorm 6 - see attached pdf for the two relevant pages of D&F.

Problems I have are as follows:

1. We are told to "identify A as a subset of F(A) by a $\mapsto$ $f_a$ where $f_a$ is the function which is 1 at a and zero elsehwhere"

Thus $f_a$ $\in$ F(A) - but how can A be a subset of F(A)? The members of F(A) are functions $f_a$ , $f_b$ , $f_c$, ... etc. The entries are functions not just simple points or elements of A. The nature of the elements of A and F(A) seem different, so can we "identify" the elements of A as a subset of F(A)?

Further to this point where is the function a $\mapsto$ $f_a$ used as the proof progresses from this point?

2. Following the quote above regarding $f_a$, we read:

"We can, in this way, think of F(A) as all finite R-linear combinations of elements of A by identifying each function f with the sum $r_1 a_1$ + $r_2 a_2$ + ..... + $r_n a_n$ where f takes the value $r_i$ at $a_i$ and is zero at all other elements of A. Moreover, each element of F(A) has a unique expression as such a formal sum."

So you can see what assumptions I am working on, I have sketched my understanding of functions f, g $\in$ F(A) and also my understanding of $f_a$ - see pdf attached called sketches of functions

Given that D&F describe f as taking on the value $r_i$ at $a_i$ and being zero at all other elements, how can we show that each element of F(A) has a unique expression as a formal sum of the form $r_1 a_1$ + $r_2 a_2$ + ..... + $r_n a_n$.

[I am also somewhat confused by the fact that the functional values of f are $r_1$, $r_2$, etc and not $r_1 a_1$ , $r_2 a_2$, etc ]

Can anyone help clarify this theorem for me?

Peter
"identifying" here means that the map $\psi:A \longrightarrow F(A)$ defined by $\psi(a)=f_a$ is one-to-one. this is clear because if $f_a=f_b$ and $a \neq b,$ then $1=f_a(a)=f_b(a)=0,$ which is non-sense.

so, using the above identification, $r_1a_1 + \ldots + r_na_n$ will mean $r_1f_{a_1} + \ldots r_nf_{a_n}.$

now let $f \in F(A).$ then, by definition, $f(a)=0$ for all but finitely many $a \in A$. so suppose that $B=\{a_1, \ldots , a_n\} \subseteq A$ is the set of all elements at which $f$ is non-zero. let $f(a_i)=r_i, \ 1 \leq i \leq n.$ let $g = \sum_{i=1}^n r_i f_{a_i}.$ we claim that $f = g.$ to see this consider two cases:
case 1. $x \in B:$ in this case, $x = a_j,$ for some $1 \leq j \leq n,$ and thus

$f(x)=f(a_j)=r_j=r_jf_{a_j}(a_j)= \sum_{i=1}^n r_if_{a_i}(a_j)=g(a_j)=g(x).$

case 2. $x \notin B :$ in this case $f(x)=0=\sum_{i=1}^n r_if_{a_i}(x)=g(x).$

to see that why the representation is unique, suppose that $f = \sum_{i=1}^n r_if_{a_i}=\sum_{i=1}^n s_i f_{a_i}.$ then, for any $1 \leq j \leq n:$

$r_j=f(a_j)=\sum_{i=1}^nr_if_{a_i}(a_j)=\sum_{i=1}^ n s_if_{a_i}(a_j)=s_j.$

3. ## Re: Unique R-Module Homomorphism

Thanks NonCommAlg

I would have never got that working by myself!

Peter

4. ## Re: Unique R-Module Homomorphism

Thanks again ... but your clarification still left me wondering ...

Your clarification seems to indicate that for x $\in$ A that

f(x) = f( $a_j$ ) = $r_j$

and so

g(x) = $\Sigma$ $r_i f_a_i$ (x) = $r_j$

Does this mean all sums $r_1 a_1$ + $r_2 a_2$ + ..... + $r_n a_n$ collapse to $r_j$ for some index j

How do we get some 'genuine' sums $r_1 f_a_1$ + $r_2 f_a_2$ + ..... + $r_n f_a_n$ where $f_a_i$ is not equal to zero for many i or indeed for all i between 1 and n.

so that $r_1 f_a_1$ + $r_2 f_a_2$ + ..... + $r_n f_a_n$ can be identified with a sum $r_1 a_1$ + $r_2 a_2$ + ..... + $r_n a_n$ where $a_i$ is not equal to zero for many or all i

Peter