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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.