chracateristic of an integral domain.

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# chracateristic of an integral domain.

Submitted by shahab on Fri, 09/03/2004 - 11:07

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The characteristic of an integral domain D is the least positve integer such that n.1=0. Prove that if f: Z -------> D is defined so that f(n)=1+1+1....n times = n.1 ;f(0)=0 and f(-n) =-(n.1)then the characteristic of D is just the non negative generator of kernel f.

Any help will be appreciated.

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## Re: chracateristic of an integral domain.

Thank you both.

## Re: chracateristic of an integral domain.

First of all, if your map is an omomorphism, then the map you defined is the only possible (this is immediate, from the very definition of omomorphism).

Second, the definition of characteristic of a ring, is exactly the aim of your exercise, and from this point of view, if ker f = (n) \ne (0), then it is obvious that n is the smallest positive integer such that "n times 1" is 0.

Anyway if you want to take as the definition of characteristic the one you gave, do as follow:

you have THE map f: Z \to R; then if ker f is a nonzero subring of Z, ker f = nZ for some n (which of course can be chosen positive). Now, of course, 1_R+1_R+...+1_R n times = f(1)+f(1)+...+f(1) n times = f(1+1+...+1) = f(n) = 0 (1_R means the neutral element with respect the product of R).

If n wasn't the smallest such positive integer, then for the same reason, ker f would be bigger, contraddiction, and you are done.

## Re: chracateristic of an integral domain.

The kernel of f is exactly the set of all integers divisible by n.

It is very easy to see, because f(1),...,f(n-1) are not zero, but f(n) is.

This is true when n > 0, that is, if there is actually an integer for which n*1 = 0. If there is no positive integer such that n*1 = 0, then the characteristic must be (and is usually) defined as 0, this has been omitted in your question. The kernel consists then of 0 alone, and 0 is then the generator of this kernel.

Best regards

Wolfgang

## Re: chracateristic of an integral domain.

The kernel of f is exactly the set of all integers divisible by n.

It is very easy to see, because f(1),...,f(n-1) are not zero, but f(n) is.

This is true when n > 0, that is, if there is actually an integer for which n*1 = 0. If there is no positive integer such that n*1 = 0, then the characteristic must be (and is usually) defined as 0, this has been omitted in your question. The kernel consists then of 0 alone, and 0 is then the generator of this kernel.

Best regards

Wolfgang