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Established Domain

Thursday, September 1st, 2011

Established Domain
Isomorphisms, integral domain and characteristic zero help?

Part 1: Prove that every integral domain with characteristic zero contains a subring isomorphic to Z(the integers). (Hint: Use multiple of the identity to define the desired subring.)

Part 2: One of the hypotheses in the statement from Part 1 can be weakened, and the statement will still be true. State this improved result, and explain why your proof from part 1 is sufficient to establish it.

Any help appreciated,

Thanks!

An integral domain R must contain a multiplicative identity 1. For n > 0, define the ring element a_n to be 1 added to itself n times. Define a_0 = 0 and a_(-n) = -a_n. Let S = {a_k : k is in Z}. Then S is a subset of R. Clearly, S contains 0, -a_n = a_(-n) is in S, and a_n + a_m = a_(m+n) is in S, so S is a subgroup of R. To show S is a subring we only need to show a_m*a_n is in S for any m,n. We do this by proving the following claim.

Claim: a_m*a_n = a_(mn).
Proof:
First suppose m,n > 0. We prove the claim by induction on m. If m = 1 then the claim is 1*a_n = a_n, which is true. So assume the result for m. Then a_(m+1)*a_n = (a_m + 1)a_n = a_(mn) + a_n = a_(mn+n) = a_((m+1)n). So the result is true for m+1.
Now if one of m or n is 0 then a_m*a_n = 0 = a_(mn).
If m > 0 and n < 0 then a_m*a_n = a_m(-a_(-n)) = -a_m*a_(-n) = -a_(-mn) = a_mn
Similarly if m < 0 and n > 0.
If both m,n < 0 then a_m*a_n = (-a_(-m))(-a_(-n)) = (-1)^2(a_(mn)) = a_mn.

Alright so S is a subring of R. Define f: Z --> S such that f(n) = a_n.
Then f(m+n) = a_(m+n) = a_m + a_n = f(m) + f(n);
and f(mn) = a_(mn) = a_m*a_n = f(m)f(n).
So f is a ring homomorphism.
Clearly f is onto. So we only need to show that f is one to one, and this is the only part that uses the fact that R has characteristic zero. Suppose f(m) = f(n), without loss of generality say m-n >=0. Then a_m = a_n
i.e.
a_m – a_n = 0
i.e.
a_m – (-a_(-n)) = 0
i.e.
a_m + a_(-n) = 0
i.e.
a_(m-n) = 0
If m-n > 0 this says 1 added to itself a positive number of times is 0, which contradicts characteristic 0. Therefore m-n = 0 so m=n, so f is one to one.

Therefore S is a subring of R that is isomorphic to Z.

Notice that we never used that R was an integral domain. Characteristic 0 and nontrivial unity imply that there are no nonzero zero divisors. And the construction above does not require R to be commutative. (You can prove that a_m*a_n = a_n*a_m even if R is not commutative.) So your statement can be weakened to
“Every ring with nontrivial unity and characteristic zero contains a subring isomorphic to Z.”

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