Results 1 to 4 of 4

Thread: Unique R-Module Homomorphism

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    594
    Thanks
    2

    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 $\displaystyle \mapsto $$\displaystyle f_a $ where $\displaystyle f_a $ is the function which is 1 at a and zero elsehwhere"

    Thus $\displaystyle f_a $ $\displaystyle \in $ F(A) - but how can A be a subset of F(A)? The members of F(A) are functions $\displaystyle f_a $ , $\displaystyle f_b $ , $\displaystyle 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 $\displaystyle \mapsto $$\displaystyle f_a $ used as the proof progresses from this point?

    2. Following the quote above regarding $\displaystyle 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 $\displaystyle r_1 a_1 $ + $\displaystyle r_2 a_2 $ + ..... + $\displaystyle r_n a_n $ where f takes the value $\displaystyle r_i $ at $\displaystyle 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 $\displaystyle \in $ F(A) and also my understanding of $\displaystyle f_a $ - see pdf attached called sketches of functions

    Given that D&F describe f as taking on the value $\displaystyle r_i $ at $\displaystyle 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 $\displaystyle r_1 a_1 $ + $\displaystyle r_2 a_2 $ + ..... + $\displaystyle r_n a_n $.

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

    Can anyone help clarify this theorem for me?

    Peter
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: Unique R-Module Homomorphism

    Quote Originally Posted by Bernhard View Post
    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 $\displaystyle \mapsto $$\displaystyle f_a $ where $\displaystyle f_a $ is the function which is 1 at a and zero elsehwhere"

    Thus $\displaystyle f_a $ $\displaystyle \in $ F(A) - but how can A be a subset of F(A)? The members of F(A) are functions $\displaystyle f_a $ , $\displaystyle f_b $ , $\displaystyle 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 $\displaystyle \mapsto $$\displaystyle f_a $ used as the proof progresses from this point?

    2. Following the quote above regarding $\displaystyle 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 $\displaystyle r_1 a_1 $ + $\displaystyle r_2 a_2 $ + ..... + $\displaystyle r_n a_n $ where f takes the value $\displaystyle r_i $ at $\displaystyle 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 $\displaystyle \in $ F(A) and also my understanding of $\displaystyle f_a $ - see pdf attached called sketches of functions

    Given that D&F describe f as taking on the value $\displaystyle r_i $ at $\displaystyle 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 $\displaystyle r_1 a_1 $ + $\displaystyle r_2 a_2 $ + ..... + $\displaystyle r_n a_n $.

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

    Can anyone help clarify this theorem for me?

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

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

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

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

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

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

    $\displaystyle r_j=f(a_j)=\sum_{i=1}^nr_if_{a_i}(a_j)=\sum_{i=1}^ n s_if_{a_i}(a_j)=s_j.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    594
    Thanks
    2

    Re: Unique R-Module Homomorphism

    Thanks NonCommAlg

    I would have never got that working by myself!

    Peter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    594
    Thanks
    2

    Re: Unique R-Module Homomorphism

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

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

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

    and so

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

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

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

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

    Peter
    Last edited by Bernhard; Sep 2nd 2011 at 09:30 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Z-module homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Nov 28th 2009, 03:55 AM
  2. simple question on R module homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Nov 23rd 2009, 07:12 AM
  3. projective, module homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 13th 2009, 05:57 PM
  4. module homomorphism, ring, subset
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 18th 2008, 10:00 PM
  5. R-module homomorphism surjectivity
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 9th 2008, 12:32 AM

Search Tags


/mathhelpforum @mathhelpforum