The theorem you are about to see, and its proof, is entirely constructed by me as part of my senior project in mathematics. Node your homework.

Let R be an integral domain whose characteristic is not 2. Let R0 be the set R without its additive identity (i.e., the zero of the ring). Let K be an ideal in R such that K={2*r|r ∈ R} (K is an ideal by the Characterization of Integral Domains, see integral domain). Let n and rR0, such that if K != R, nK.

Definition: r is divisible under n if there exist s, tR0 and units u and vR such that

r = n*s*t+u*s+v*t.

The theorem

"r is divisible under 2*n iff 2*r is divisible under n."

The proof

Now for the main theorem (note that '3*n' would mean 'n+n+n' under the ring's additive operation). Suppose r is divisible under 2*n (given the sets defined above). So there exist s, tR0 and units u and v such that

r = 2*n*s*t+u*s+v*t.

R has a characteristic other than 2, so it's not a meaningless operation to multiply both sides of this equation by 2:

2*r = 4*n*s*t+2*u*s+2*v*t.

After doing a little commutative and distributive work (allowable for integer operations on elements of an integral domain),

2*r = n*(2*s)*(2*t)+u*(2*s)+v*(2*t).

Again, the characteristic of R is not 2, so 2*s and 2*t are both in R0. Thus we have shown that r divisible under 2*n implies 2*r divisible under n.

Now suppose that 2*r is divisible under n. If K=R, then we note that there exist x and y such that s=2*x and t=2*y and we use our equation again:

2*r = n*s*t+u*s+v*t, or

2*r = n*(2*x)*(2*y)+u*(2*x)+v*(2*y).

With a little commutativity and cancellation, we get

r = 2*n*x*y+u*x+v*y,

which is exactly what we wanted, i.e., r is divisible under 2*n (note that the cancellation of the integer 2 is allowed by having the characteristic not be 2 and by Moore's theorem on integral domains).

Now suppose that K != R. Looking at the cosets of K under R, we see that 0R+K and 1R+K are the only possible cosets (by our definition of K). So each of s and t must fall into one of these two cosets. By the Characterization of Integral Domains(2), no unit of R can be in K in this case. Note also that we defined n to not be in K if possible. So in this case, n, u and v are in 1R+K. So we now look at our equation from a modular standpoint:

2*r+K = 0R+K = n*s*t+u*s+v*t+K = (1R+K)*(s+K)*(t+K)+(1R+K)*(s+K)+(1R+K)*(t+K) = s*t+K+s+K+t+K = s*t+s+t+K.

Now we have some more cases:

1) s and t are in the same coset but not in K. Then the end congruence winds up being 1R*1R+1R+1R+K = 1R+K. But this is not congruent to 0R+K (which is the coset that 2*r is in), so this cannot be the case.
2) s and t are not in the same coset. Since there are only two (possible) cosets, we will say (without loss of generality) that s is in K and t is not. Now our congruence looks like 0R*1R+0R+1R+K = 1R+K. Again, this is not congruent to 0R+K, so we must assume that
3) s and t are both in K. At last, our congruence looks like 0R*0R+0R+0R+K = 0R+K. This is exactly what we want.

Now we know that both s and t must be in K (in any case), so there must exist x and y in R0 (like before) such that s=2*x and t=2*y. And again we reach the conclusion in our equation that

2*r = n*s*t+u*s+v*t, or

2*r = n*(2*x)*(2*y)+u*(2*x)+v*(2*y).

Once again, we can pare this down to

r = 2*n*x*y+u*x+v*y,

so we are now certain that 2*r divisible under n implies r divisible under 2*n. QED


Yes, it does seem a bit silly. :-P However, this theorem is only true for 2. Characteristic, ideal, equations... they fall apart if any number other than 2 is used.

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