# Thread: Natural Map from R into End (M)

1. ## 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

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

Originally Posted by Bernhard
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 $f_r\in \textrm{End}(M)$ .

3. ## 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

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

Originally Posted by Bernhard
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.

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

Originally Posted by NonCommAlg
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

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

Originally Posted by Bernhard
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.

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

Thanks for the help

Peter