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 **R**_{0} 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 **r** ∈ **R**_{0}, such that if **K** != **R**, **n** ∉ **K**.

Definition: **r** is *divisible under ***n** if there exist **s, t** ∈ **R**_{0} and units **u** and **v** ∈ **R** such that

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

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

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, t** ∈ **R**_{0} 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 **R**_{0}. 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 0_{R}+**K** and 1_{R}+**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 1_{R}+**K**. So we now look at our equation from a modular standpoint:

**2*r+K = 0**_{R}+K = n*s*t+u*s+v*t+K = (1_{R}+K)*(s+K)*(t+K)+(1_{R}+K)*(s+K)+(1_{R}+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 1_{R}*1_{R}+1_{R}+1_{R}+**K** = 1_{R}+**K**. But this is not congruent to 0_{R}+**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 0_{R}*1_{R}+0_{R}+1_{R}+**K** = 1_{R}+**K**. Again, this is not congruent to 0_{R}+**K**, so we must assume that

3) **s** and **t** are both in **K**. At last, our congruence looks like 0_{R}*0_{R}+0_{R}+0_{R}+**K** = 0_{R}+**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 **R**_{0} (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

#### Notes

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.