Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.