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Math Help - Natural Map from R into End (M)

  1. #1
    Super Member Bernhard's Avatar
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    Natural Map from R into End (M)

    I am a math hobbyist working by myself and reading Dummit and Foote "Abstract Algebra CHapter 10 on Module Theory. I am currently studying Section 10.2 on Module Homomorphisms (see attached pdf for section 10.2 pages 345 to 348)

    In Section 10.2 D&F define an R-Module Homomorphism and go on to define Hom (M,M), the endomorphism ring of M denoted End(M).

    On the top of page 345 D&F write:

    When R is commutative, there is a natural map from R into End(M) given by r  \rightarrow rI, where the latter endomorphism of M is just multiplication by r on M"

    and then they write:

    "The ring homomorphism from R to End(M) may not be injective since for some r we may have rm = 0 for all m  \in M (e.g. R = Z, M = Z/2Z and r = 2"

    OK then - so looking at the situation for R = Z, M = Z/2Z we are looking at a natural map r  \rightarrow rI where r  \in Z and rI is an endomorphism that acts like

    rI(m) = r(m).I(m) = rm for all m  \in M

    so for example for r = 0

    0I(m) = 0(m).I(m) = 0.m = 0 for all m (that is for m = 0 and 1

    and for r = 1

    1I(m) = 1(m).I(m) = 1.m for all m
    i.e. for m=0, 1I(0) = 1.0 = 0 and for m=1, 1I(1) = 1.1 = 1

    and for r =2

    2I(m) = 2(m)I(m) = 2.m = 0.m = 0 for all m - this is the same endomorphism of M as for r = o and so clearly the ring hoimomorphism in this case is not injective.

    That is OK. The D&F claim

    "When R is a field, however, this map is injective ..."

    So for concreteness, I tried to come up with an example and thus changed the example above to R=Q, M = Z/2Z

    If we then take for example 3/2  \in Q and look at r  \rightarrow rI we get

    3/2.I(m) = 3/2(m).I(m) = 3/2.m for all m

    Then looking at the endomorphism m  \rightarrow 3/2m for all m  \in M we get:

    0  \rightarrow 3/2.0 = 0

    and 1  \rightarrow 3/2.1 = 3/2 which does not belong to M ???

    Obviously something is wrong! But what?

    I do not seem to have done anything illegal

    But the statement: " When R is commutative, there is a natural map from R into End(M) given by r  \rightarrow rI, where the latter endomorphism of M is just multiplication by r on M" - does not seem to be working out

    Can anyone help me in this mattter?

    Bernhard
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Natural Map from R into End (M)

    Quote Originally Posted by Bernhard View Post
    But the statement: " When R is commutative, there is a natural map from R into End(M) given by r  \rightarrow rI, where the latter endomorphism of M is just multiplication by r on M" - does not seem to be working out
    For an R-module M the "most" natural map \phi :R\to \textrm{End}(M) is \phi(r)=f_r where f_r(m)=rm for all m\in M. You can easily prove that f_r\in \textrm{End}(M) .
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Natural Map from R into End (M)

    Fernando,

    Is my formal working above sensible? [I hope it is not nonsense :-) ]

    Can you see what is wrong in my example of R=Q, M = Z/2Z

    Peter
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  4. #4
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    Re: Natural Map from R into End (M)

    Quote Originally Posted by Bernhard View Post
    I am a math hobbyist working by myself and reading Dummit and Foote "Abstract Algebra CHapter 10 on Module Theory. I am currently studying Section 10.2 on Module Homomorphisms (see attached pdf for section 10.2 pages 345 to 348)

    In Section 10.2 D&F define an R-Module Homomorphism and go on to define Hom (M,M), the endomorphism ring of M denoted End(M).

    On the top of page 345 D&F write:

    When R is commutative, there is a natural map from R into End(M) given by r  \rightarrow rI, where the latter endomorphism of M is just multiplication by r on M"

    and then they write:

    "The ring homomorphism from R to End(M) may not be injective since for some r we may have rm = 0 for all m  \in M (e.g. R = Z, M = Z/2Z and r = 2"

    OK then - so looking at the situation for R = Z, M = Z/2Z we are looking at a natural map r  \rightarrow rI where r  \in Z and rI is an endomorphism that acts like

    rI(m) = r(m).I(m) = rm for all m  \in M

    so for example for r = 0

    0I(m) = 0(m).I(m) = 0.m = 0 for all m (that is for m = 0 and 1

    and for r = 1

    1I(m) = 1(m).I(m) = 1.m for all m
    i.e. for m=0, 1I(0) = 1.0 = 0 and for m=1, 1I(1) = 1.1 = 1

    and for r =2

    2I(m) = 2(m)I(m) = 2.m = 0.m = 0 for all m - this is the same endomorphism of M as for r = o and so clearly the ring hoimomorphism in this case is not injective.

    That is OK. The D&F claim

    "When R is a field, however, this map is injective ..."

    So for concreteness, I tried to come up with an example and thus changed the example above to R=Q, M = Z/2Z

    If we then take for example 3/2  \in Q and look at r  \rightarrow rI we get

    3/2.I(m) = 3/2(m).I(m) = 3/2.m for all m

    Then looking at the endomorphism m  \rightarrow 3/2m for all m  \in M we get:

    0  \rightarrow 3/2.0 = 0

    and 1  \rightarrow 3/2.1 = 3/2 which does not belong to M ???

    Obviously something is wrong! But what?

    I do not seem to have done anything illegal

    But the statement: " When R is commutative, there is a natural map from R into End(M) given by r  \rightarrow rI, where the latter endomorphism of M is just multiplication by r on M" - does not seem to be working out

    Can anyone help me in this mattter?

    Bernhard
    well, your example is wrong because \mathbb{Z}/2\mathbb{Z}, with usual addition and scalar multiplication, is obviously not a \mathbb{Q}-module.
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  5. #5
    Super Member Bernhard's Avatar
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    Re: Natural Map from R into End (M)

    Quote Originally Posted by NonCommAlg View Post
    well, your example is wrong because \mathbb{Z}/2\mathbb{Z}, with usual addition and scalar multiplication, is obviously not a \mathbb{Q}-module.

    OK let me be clear on this - part of D&F's definition of a module on Page 337 states that a left R-module is a set M together with
    (1) a binary operation + on M under which M is an abelian group and
    (2) An action of R on M (that is a map R x M --> M) denoted by rm, for all r belonging to R and for al m belonging to M etc etc

    Are you saying that I have not followed (2) in constructing the example ie that R x M must map into M?

    Is that where my example went astray?

    Is that what you are saying?

    Peter
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  6. #6
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    Re: Natural Map from R into End (M)

    Quote Originally Posted by Bernhard View Post
    OK let me be clear on this - part of D&F's definition of a module on Page 337 states that a left R-module is a set M together with
    (1) a binary operation + on M under which M is an abelian group and
    (2) An action of R on M (that is a map R x M --> M) denoted by rm, for all r belonging to R and for al m belonging to M etc etc

    Are you saying that I have not followed (2) in constructing the example ie that R x M must map into M?

    Is that where my example went astray?

    Is that what you are saying?

    Peter
    yes.
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  7. #7
    Super Member Bernhard's Avatar
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    Re: Natural Map from R into End (M)

    Thanks for the help

    Peter
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