Alyosha
Reputation
2,314
Top tag
Next privilege 2,500 Rep.
Create tag synonyms
 Oct 13 comment Ramification of prime ideal in Kummer extension @Lubin In case it makes a difference, this i.gyazo.com/86bd186242014b07ed7b4a28d82bba30.png halfway down is where the quesion arises from. Thank you for thinking about this. Oct 13 comment Power residue symbol @franzlemmermeyer Yes, you're right. I'm actually reading page 359 of your book on reciprocity laws now. Oct 13 comment Ramification of prime ideal in Kummer extension @Lubin I do mean that. Oct 13 comment Ramification of prime ideal in Kummer extension @Lubin Yes. I've edited to remove ambiguity. Oct 13 comment Ramification of prime ideal in Kummer extension MO asked me to reask this here. Sep 21 comment Why does totally ramified'' lift as a property? @Lubin If you're interested, my confusion basically arises from trying to understand the last sentence of the 2nd paragraph of math.lsa.umich.edu/~kartikp/teaching/678-Fall2009/L9.pdf. Sep 21 comment Why does totally ramified'' lift as a property? @Lubin No, you aren't. Sep 18 comment Inverse Elliptic function question @AlonsoDelfín Apologies for the confusion, I'll edit. Is it clear now? Sep 18 comment Inverse Elliptic function question @AlonsoDelfín Yes, though giving an explicit basis of the lattice is of course unnecessary. Sep 15 comment Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak… @draks... Have you resolved the confusion yet? Sep 15 comment Prime ideal factorisation degree @QiaochuYuan Okay, I'll edit to include only the more restrictive case. Thank you. Sep 15 comment Prime ideal factorisation degree This gyazo.com/f7f7710010512a2cdd35c550614d359a is what has caused my confusion. Sep 3 comment Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak… Are you sure that $\Omega$ is always a group? Any group where every element has order two has order $2^n$ for some $n$. Jul 29 comment Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak… I have constructed an injective homomorphism from a group with $H+1$ elements to one of $2^n$ elements. Thus $H+1$ must be a power of two. Jul 23 comment Proving that if $p^2 = a^2 + 2b^2$ then also $p$ can be written in form $a^2 +2b^2$ With hindsight splitting into the three cases was unnecessary, you only need that $(p)$ splits into one or two prime ideals, but maybe it's instructional to leave it in. Jul 11 comment For what functions is this theorem correct? @joriki The cardinality. One of the conditions for the theorem to be true is surely that preimages are finite. Jul 11 comment For what functions is this theorem correct? @joriki Apologies, I meant preimage of an element in the image. I thought you were being sarcastic. Jul 11 comment For what functions is this theorem correct? @joriki How is that relevant? Jul 9 comment For what functions is this theorem correct? That is correct. Jun 23 comment Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$. Aha, thanks. I see.