# ring endomorphisms vs module endomorphism of a ring R

this is quite strange to me. Let R be a unital (commutative for easier notions) ring. Considered as an R module, the module endomorphisms ring of R (as a module over itself) is isomorphic to R, and this is because a module endomorphism f is determined by f(1): f(r)=r*f(1). But when R is considered as a ring, this is not the case. I can't prove that a ring endomorphism is determined by f(1).

Well this is strage to me, because I can't see any structure difference between R considered as a ring, and R considered as an R-module, and yet the two endomorphism rings are different. Any ideas what's going on?

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If you cannot prove that ring endomorphisms are defined by $f(1)$, it is because $f(1)$ must always be $1$ but there exist nontrival ring endomorphisms (at least for some rings). Consider complex conjugation, or the substitution of any polynomial for $X$ in $R[X]$. – Marc van Leeuwen Jan 18 '13 at 3:58
If you look carefully to the definition of the ring/module homomorphisms will find the answer by yourself. I don't think this is particularly a deep question, but certainly the difference is deep enough. – user26857 Jan 18 '13 at 9:02

They are different because the homomorphism conditions are completely different:

For a module homomorphism of $R$ into $R$, $f(r)=f(1)r$

but for a ring homomorphism, $f(r)=f(r)f(1)$.

This is especially uninformative when (as usual) $f(1)$ is the identity of $R$: it's just a tautology.

In the first case, knowing $f(1)$ tells you every $f(r)$, but in the second case, you are still stuck wondering what $f(r)$ is. Being a multiplicative map and being and being $R$ linear are very different.

Perhaps a more enlightening way to think of ring and module homomorphisms is to think about their kernels. The kernels of ring homomorphisms have to be two-sided ideals, but the kernels of module homomorphisms only need to be one-sided.

The complicating factor here for you, it seems, is that the module multiplication and the ring multiplication match for the module $R_R$. Since they have the same multiplication, you might be tempted to think they have "the same structure".

However, this is simply not the case, because morphisms are intimately tied with the structure. If you've never heard of a category before, here is a rough summary. A category consists of a bunch of objects that you are talking about, along with morphisms between them.

So there is a category of groups (whose objects are groups and whose morphisms are group homomrphisms) and there is a category of rings (whose objects are rings and whose morphisms are ring homomorphisms.) There is also a category of $R$ modules (whose objects are right $R$ modules and morphisms are $R$ module homomorphisms.) There is probably no doubt in your mind that the category of rings and $R$ modules are different, but here we are trying to figure out how $R$ can be an object in the category of rings and the category of $R$ modules without "having the same structure".

Since morphisms are important, their kernels become important substructres of each object. In rings, these kernels are ideals, in modules, these kernels are submodules. Now in the case of $R_R$, a right submodule happens to be called a right ideal, because of the resemblance with being "half" the ideal axioms.

It's easy to think of a ring $R$ which has very different ideal and right submodule structures. If you take the matrix ring $M_n(F)$ over a field $F$, then we know that there are only two ideals, and so the only ring homomorphisms starting in $R$ are either the zero homorphism, or they are injective.

But the same can't be said about the module $R_R$, since it has lots of right $R$ submodules. The easiest ones to describe are the matrices which are zero outside of a fixed row. In fact if $F$ is infinite, there are infinitely many distinct right ideals.

So I hope by showing how different substructures can be will help you see why they do not really have "the same structure". Because the morphisms are different, the structures are different.

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Thanks! it is weird because if a homomorphism is function preserving structure, and the structure of R as a ring and as an R-module is exactly the same, then the homomorphisms being different is not something you would expact. – cruvadom Jan 18 '13 at 1:42
so still I'm looking for some deeper insight about the issue. – cruvadom Jan 18 '13 at 2:45
@user57076 I know what you mean, but I think you're hastily equating "having the same structure" to "having matching multiplication". I'm going to add another line to my answer above to try to help you see why they don't have the same structure. – rschwieb Jan 18 '13 at 14:16
@user57076, there is nothing deep to be found. – Mariano Suárez-Alvarez Jan 18 '13 at 18:09
@MarianoSuárez-Alvarez There is nothing deep for us, but clearly there is an explanable difference, and as I mentioned in my last comment, a plausible misconception. Wouldn't it be preferable to avoid language which 'writes off' the OP? – rschwieb Jan 18 '13 at 18:27