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 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 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 rI where r Z and rI is an endomorphism that acts like
rI(m) = r(m).I(m) = rm for all m 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 Q and look at r rI we get
3/2.I(m) = 3/2(m).I(m) = 3/2.m for all m
Then looking at the endomorphism m 3/2m for all m M we get:
0 3/2.0 = 0
and 1 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 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?
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?