What is the difference between ring homomorphism and module homomorphism?

Question

I know that $\mathbb{Z}$ and $2\mathbb{Z}$ are isomorphic as Z-modules. But they are not isomorphic as ring. I used to think that I can look at isomorphism as being "equal". However, now it seems that it is not really expressing "equality" as there are different kinds of homomorphism. What then does isomorphism actually expresses. How can I think about it intuitively? I know by definition they are different, I just want to have an intuition of their difference.

Answer 1

The two types just preserve different structure. You may as well ask "what is the difference between an airplane pilot's exam and a driving exam?" Driving and flying have different requirements, and so naturally their examinations have different requirements. (The analogy is not very good, sorry.)

You could also ask "what is the difference between a function and a module homomorphism?" or "what is the difference between a continuous function and a homomorphism?" They are all just classes of functions with special requirements on their behavior.

As in your example, $f:\Bbb Z\to 2\Bbb Z$ given by $n\mapsto 2n$ is a $\Bbb Z$ module homomorphism because $f(mn)=2mn=m(2n)=mf(n)$, i.e. elements of $\Bbb Z$ can come outside of the parens for free.

On the other hand, this map is clearly not multiplicative, as a ring homomorphism would have to be. If it had that property, then $f(mn)=f(m)f(n)=4mn$, but actually $f(mn)=2mn$.

Here's another example to give you a feel for the difference. It's well-known that for a commutative ring $R$ and positive natural numbers $n,m$, you cannot have the module products $R^n\cong R^m$ isomorphic as $R$ modules if $m\neq n$. However, there are easy examples of commutative rings where the product rings $R^n\cong R^m$ for a particular pair $m,n$, $m\neq n$.

Answer 2

A module homomorphism between two rings ignores the multiplicative structure. There is a module homomorphism $\phi:\mathbb{Z}\to 2\mathbb{Z}$ given by $$\phi(n)=2n$$ You can verify that since it is a group homomorphism (additively) and it preserves scalar multiplication it is a homomorphism of modules. This is a bijective homomorphism and for modules this means it is an isomorphism. However, it is not a homomorphism of rings. If it were, then for all $a,b$ we'd have $\phi(ab)=\phi(a)\phi(b)$. However, $$\phi(1\cdot 1)=\phi(1)=2\neq 4=\phi(1)\phi(1)$$

Answer 3

To put it simply, isomorphism is always "isomorphism of structures", i.e. you need to specify the structure you are working with, not simply the "object".

For instance the bijection $$ f:\Bbb Z\longrightarrow\Bbb Z,\qquad f(n)=n+1 $$ is certainly an automorphism of the set $\Bbb Z$ (automorphism $=$ isomorphism with itself) as it has an inverse, but certainly not an automorphism of the additive group $\Bbb Z$ as it even fails to respect the operation in the sense that $$ f(m+n)\neq f(m)+f(n). $$ So the example you give is certainly an isomorphism of $\Bbb Z$-modules (aka abelian groups) but not of rings pretty much in the same way: the map is a bijection and respect the module structure, but does not respect the ring structure.

Answer 4

I hope this example will help you understand: $\mathbb Q[X]$ is a ring as well as a $\mathbb Q[X]$-module. Then a ring homomorphism over $\mathbb Q[X]$ is always of form $f(X)\mapsto f(g(X))$, $g$ an arbitrary element of $\mathbb Q[X]$; while a $\mathbb Q[X]$-module homomorphism is always of form $f(X)\mapsto f(X)g(X)$, $g$ an arbitrary element of $\mathbb Q[X]$.