u/WMe6

I created this article recently, taking the liberty to name this reaction after its discoverer.

I'll just leave it to y'all to appreciate.

Am I the only one who finds the standard notation for polynomials annoying? Like, you have to have a dummy variable, and different people use different ones, like k[x], k[X], k[T], etc.

It's annoying that we still treat polynomials notationally like functions that you sub into to get a number and you have to specify the variable. I guess for individual polynomials, you can treat it as a sequence of ring elements with all but finitely many elements zero, following certain rules for how they add and multiply, but that still doesn't solve the problem if you want to talk about a polynomial ring. I guess you could write k[] or k[·] or k[-] for k[x]?

But then what do you do for the ring in two indeterminates?

>In many cases, it will be easier for AI to convince humans it has a proof than to come up with a correct mathematical argument, and I believe that we as mathematicians are not sufficiently prepared for this.

Given how persuasive LLM's can be, maybe they become better at exploiting certain subtle weaknesses in the abilities of humans to spot flaws in an argument faster than they become better at math. That is very worrying.

Must everything by AI be put into Lean then? Mecha-Mochizuki when???

Neukirch uses a smaller or maybe lower-case(?) version of calligraphic O as a general notation for Dedekind domains while using the upper-case version for the integral closure of the former.

Is that just his notational idiosyncrasy, or is this a convention that others also follow? I was only aware that \mathcal{O}_{K} is usually used to denote the ring of integers of a number field K.

It seems hard to show the difference between curly O and curly o on the board, and I don't know how you would even produce the symbol on TeX, since \mathcal is upper case only.

Kind of an idle question, but I figure Neukirch's algebraic number theory book is influential enough that maybe others also use this weird letter?

I've read that A/(I+J) \cong (A/I)/((I+J)/I). The way I understand that is that the projection map from A to A/I is supposed to map J to (I+J)/I. That way, it's like you're saying modding out by I+J is the same as modding out by I then by J.

I'm thinking about the example \mathbb{F}_9 \cong \mathbb{Z}[x]/(3,x^2+1), and it obviously works, but I'm still having trouble understanding why J \mapsto (I+J)/I under the projection map at an intuitive level. What's a quick way of seeing this?

I'm beginning to learn about number fields and the concept of ramification. I read something like this (from K. Conrad's notes):

In Z[i], the only prime that ramifies is 2: (2) = (1 + i)^(2).

Does he mean that 2 is considered prime in Z[i], or does this just refer to the fact that 2 is prime in the rational integers Z? On the face of it, being ramified means it can't be prime in Z[i], right, or is the word "prime" used differently because it's a kind of special case?

My understanding is that is p is prime means that p is non-zero and non-unit, and p|ab implies p|a or p|b. In this case, 2 factors into (1+i)*(1-i)=(-i)*(1+i)^(2), and 2 is nonzero and not a unit in Z[i], but 2 does not divide 1+i or 1-i even though 2 divides their product 2.

For those unfamiliar, this is an infamous problem: if a, b are integers and (a^2+b^2)/(1+ab) is also an integer, then it is in fact a perfect square.

Among those who solved it correctly (only 11 students) are Nicușor Dan (current president of Romania, scoring a 42/42 that year), Ravi Vakil, and Ngô Bảo Châu (also a perfect score, later Fields medalist for work in the Langlands program), while Terence Tao (only 13 at the time) received a 1/7 on this problem, but aced the rest and still ended up with a gold in 1988.

It must be so weird having an extremely smart person as a head of state.

This may be a really dumb question! Is there a simple description of the integers with only multiplication defined? So basically, take the ring (\mathbb{Z},+,\cdot) and ignore addition +. What you're left with should be a commutative monoid. Is that structure isomorphic to anything easy to describe?

I guess I was thinking along the lines of the positive rationals, whose multiplicative structure makes them isomorphic to the free abelian group on a countably infinite number of generators, essentially using the prime numbers as generators via unique factorization.

For the integers, you would not have anything raised to negative powers, so you obviously don't have a group. In addition, you have units, +1 and -1, as well as 0. But otherwise, the structure should also be described by the unique factorization of the integers.