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?

Bernhard