4 ms·

That sounds so baller. But one question. You say that he constructed a field that "behaves like C." Granted, you put that phrase in quotes for a reason, but it

That sounds so baller. But one question. You say that he constructed a field that "behaves like C." Granted, you put that phrase in quotes for a reason, but it seems to me as though if you pinned down the behavior of multiplication in this artificial field, then conjugation wouldn't be too far away. So what does it mean to "behave like C"? Or better yet, do you have a link to you friend's thesis or other papers?

ot 13y ago
Of course the details are very tricky :) "behaves like C" means that you construct a first-order theory that all the properties of C that you want, plus the conjecture. I don't know much about this, but I think that there is no conjugation in the language of exponential fields, and if you try to add it everything becomes intractable (for one, every proof would first require to prove Schanuel's conjecture). The Wikipedia has a few references: https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber.27s_pseudo-exponentiation https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber...