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1d
awarded  Constituent
Dec
8
awarded  Caucus
Nov
29
comment How is half-contour integration possible?
@AlexR. I am. Apologies. I meant a contour that does not stop at any finite point (that is, goes off to infinity in both directions).
Nov
29
asked How is half-contour integration possible?
Nov
19
comment Is there a general way to prove series and products are modular?
@user45195 The mild moral qualm about reposting this question is stronger than my attachment to reputation.
Nov
16
awarded  Popular Question
Nov
15
revised Is there a general way to prove series and products are modular?
edited title
Nov
15
asked Is there a general way to prove series and products are modular?
Nov
9
answered How many subgroups or order 8 an abelian Group of order 72 can have
Oct
26
comment Extension field of finite degree
Do you mean finite?
Oct
25
comment condition for homeomorphism
Are $A,B$ open sets?
Oct
24
comment Path connected iff the action of $\pi_1(Y,y)$ on $p^{-1}(y)$ is transitive.
If I have been unclear, please ask for clarification or rephrasing of my answer.
Oct
24
answered Path connected iff the action of $\pi_1(Y,y)$ on $p^{-1}(y)$ is transitive.
Oct
24
comment Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$
@Walterr if $\alpha, \beta, \gamma$ are the roots, then $f(a_1 \alpha+a_2 \alpha^2 +b_1 \beta +b_2 \beta^2 +c_1 \gamma +c_2 \gamma^2+d)=a_1 f(\alpha)+a_2 f(\alpha)^2+b_1 f(\beta)+b_2 f(\beta)^2 +c_1 f(\gamma)+c_2 f(\gamma)^2+d.$ Thus $f(l)$ for any $l \in L$ is determined by the action of $f$ on the three roots.
Oct
24
comment Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$
@Walterr I mean that one of $f(\alpha)=\alpha, f(\alpha)=\omega \alpha, f(\alpha)=\omega^2 \alpha$ will occur, and for each of these options you have $2$ choices as to what $f(\alpha \omega)$ is (which, along with $f(\alpha)$, determines $f(\omega^2 \alpha)$).
Oct
24
answered Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$
Oct
23
comment How to visualise Bollobas' 1965 theorem?
I have crossposted to MO.
Oct
23
answered Why in this sense this homomorphism is injective?
Oct
21
comment So why isn't $\Bbb R^n = \oplus _{n = 1}^{m}\Bbb R^n$
Decomposed uniquely.
Oct
20
revised How to visualise Bollobas' 1965 theorem?
added 444 characters in body