# proving - ring theory and homomorphisms

So I am having a lot of trouble proving this. It was for an assignment due tuesday, but the prof said I can have a bit of extra time with this question.

Please do not give me the answer because I am sure my prof is on this website :P

okay so the question is as follows:

Let $R$ be a ring with identity. Show that the map $f: \mathbb{Z} \rightarrow R$ given by $f(k) = k1_R$ is a homomorphism.

So obviously I went about proving the axioms thinking the $k$ and $1_R$ are integer products. BUT as my prof pointed out, there not. Here is the explanation on his course webpage.

A couple of remarks about Assignment 8 problems: In exercise 25, when H'ford writes something like $k1_R$ , where k is an integer, he of course is not speaking of a PRODUCT of elements of the ring R. Given an element a of R, the inductive definition of ka, k in ZZ, in connection with the definition of the characteristic of a ring, about which you folks had an assignment problem, was this:

0a := 0_R,

and, for all k in |N, (k+1)a := ka + a.

Also, if k is a negative integer, then ka := -((-k)a).

I believe we defined finite sums inductively a long time ago. One defines 0a as above, and we could then use our earlier inductive definition and just say that, for each k in ZZ^+, $$ka=\sum_{j=1}^k a$$

And then define ka for negative integers k as above.

so how do I use his definition of what $k$ is and how do I apply it to my proof?

-

You have to show that $(m\cdot n)1_R=(m1_R)\cdot(n1_R)$ for all integers $m$ and $n$.

If $m$ and $n$ are non-negative this follows from your definition $ka:=\sum_{j=1}^k a$.

The general case is then an easy consequence of the axioms. You should try to work this out.

-

So, I guess that for $m$ and $n$ positive you need to prove that $$\left(\sum_{i=1}^m\,1_R\right)\left(\sum_{i=1}^n\,1_R\right)= \left(\sum_{i=1}^{mn}\,1_R\right).$$ How many addends you have when you apply the distributive property on the LHS ?

Then you have to come up with something for when $m$ and $n$ are not both positive.

-