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Ok, ok, I'm back. Keep your yappers shut, I know you all missed me.
So I'm taking Ring and Field Theory this term, which is a branch of group theory.
Here is a proof that has been stumping me for a bit now. I think it's clever more than anything...

Suppose that R is a commutative ring and |R|=30, with an ideal I, where |I|=10.
Claim: I is a maximal ideal of R.
Sidebar: The order of R is 30, and, by Lagrangre's Theorem, the order of any subset divides the order of the group. Then, there are supgroups of order 10 and order 15 (call them I and J). But we aren't talking about groups, we're talking about rings. We know that there is a subring of order 10 (I). So, I need to show that I is not a subset of J or that J is not a subring (although it is a subgroup). A third possibility is to show that all groups of order 30 are isomorphic to Z(30), at which point I can fairly easily prove that I is maximal. No, I just Googled it: not every group of order 30 is isomorphic to Z(30). Wait. Wait. Wait. In I, there are elements of order 2 (because |I|=10=5*2), whereas in J, there are no elements of order 2 (because |J|=15=5*3). So, I is not a subset of J. And, by Lagrange's Theorem, there can be no other subgroups. So, I is maximal. Incidentally, J is maximal, too. Now, the formal proof.
Pf: Let R be a commutative ring, with |R|=30, and two ideas, I and J, where |I|=10 and |J|=15. By Lagrange's Theorem, there can be subgroups (and, so, subrings) of orders that divide the order of the group (or, rather, ring). Then, there can be no proper subring of order greater than 15. So, either I is maximal or J is maximal (or both). If I is not maximal, then I is a subset of J. However, I contains elements of order 2 (by Lagrange's Theorem and because 10=2*5), whereas J does not (by Lagrange's Theorem again, and because 15=3*5). So, I cannot be a subset of J. So, I is maximal. Incidentally, if there even exists a subring of order 15, then J is also maximal. QED.

Here's another that's been stumping me:
Question: Is the ring 2Z isomorphic to the ring 3Z?
Answer: Assume that 2Z is isomporphic to 3Z, where P is such an isomorphism. Then, by principal, 3Z is isomorphic to 2Z, where P(inv) is such an isomorphism. The ring 3Z has unity 1, so P(inv)(1)=1*, where 1* represents the unity of 2Z. However, no such unity exists. Then, 1 cannot map to 1*, so there is no such isomorphism P(inv). So there is no such isomorphism P, which contradicts our initial assumption.
Question: Is the ring 2Z isomorphic to the ring 4Z?
Answer: This one is a little more tricky. It feels obvious to me, frankly, that they are. They are both infinitely countable, so any mapping can be one-to-one and onto. Hmmm, let's see. Let P be such an isomorphism. Then, P(2)=4n, n in Z. Then, P(2*2)=P(2+2), implies that P(2)*P(2)=P(2)+P(2), which implies that 4n*4n=4n+4n, which implies that 16n^2=8n, which implies that 2n=1, which is impossible. So, we have a contradiction, so P cannot exist. Well, well, well, my intuition was wrong. The two aren't isomorphic. Incidentally, I can use this same method for the previous problem.

That all took me way too long. Time to get it formally written up, then off to read more ethics.