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Alex B.'s user avatar
Alex B.
  • Member for 14 years, 3 months
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Irreducible and faithful $\operatorname{PSL}_2(q)$-module
@DerekHolt Ok, I have edited (I don't think I can ping Benjamin in the same comment, alas).
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Irreducible and faithful $\operatorname{PSL}_2(q)$-module
Clarified the actual answer to the question
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Irreducible and faithful $\operatorname{PSL}_2(q)$-module
@BenjaminSteinberg right, but only in the "obvious" way of just realising "the same" representations in different characteristics. The point is that you cannot have larger more complicated modules, e.g. ones of exponent $p^n$ for some $n\geq 2$. But yes, the answer to the literal question as asked is "no".
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Irreducible and faithful $\operatorname{PSL}_2(q)$-module
@LSpice: the group ${\rm PSL}(2,2^f)$ is simple, so if the action is not faithful, it is trivial, in which case either $N$ is simple or not a minimal normal subgroup. Also, in this case the question is answered by the theory of Schur multipliers.
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Homomorphisms of abelian varieties and Tate modules
For every possible kernel $K$ of a morphism $A\to B$, the map $\phi_K\colon {\rm Hom}(A/K,B)\otimes_{\mathbb{Z}}\mathbb{Z}_{\ell}\to {\rm Hom}(T_{\ell}(A/K),T_{\ell}B)$ is injective by the Tate conjecture; and the map ${\rm Hom}(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_{\ell}\to {\rm Hom}(T_{\ell}A,T_{\ell}B)$ when restricted to the image of ${\rm Hom}(A/K,B)$ in ${\rm Hom}(A,B)$, is just the composition of $\phi_K$ with the canonical map ${\rm Hom}(T_{\ell}(A/K),T_{\ell}B)\to {\rm Hom}(T_{\ell}A,T_{\ell}B)$, which is injective, because the image of $T_{\ell}A\to T_{\ell}(A/K)$ has finite index.
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answered
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Autobiographies and correspondences of mathematicians
Thanks, LSpice, corrected. I originally read it as a kid in Russian translation, where there are no articles, so I actually checked the Amazon page before posting. The title there has no article, but on the photo of the cover there is an "a", so this must be a typo on the Amazon page. I don't know whether this echo of Halmos is deliberate.
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What is the ring $A_{\Gamma}$ in the Cohen-Lenstra Heuristics?
@Melanka, that just shows that $\mathbb{Z}[G]$ projects surjectively onto the maximal order of $\mathbb{Q}[\Gamma]/\sum g$, not that it itself is maximal. You could inspect the case $p=2$ more closely, perhaps draw a picture to see what is going on. I am not sure what the hang-up is, but one possibility is that you are confusing subrings and quotient rings.
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What is the ring $A_{\Gamma}$ in the Cohen-Lenstra Heuristics?
...needs to be modified to take into account genus theory. This was done by Frank Gerth III some years after the original paper of C-L.
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What is the ring $A_{\Gamma}$ in the Cohen-Lenstra Heuristics?
@Melanka: if $\Gamma$ is cyclic of order $p$, then a maximal order in $\mathbb{Q}[\Gamma]$ will have denominators, when expressed in terms of the standard basis $g\in \Gamma$. When you pass to a quotient, there is no well-defined meaning of "having denominators", because it depends on the basis. Of course you can pick a $\mathbb{Z}$-basis for your maximal order, and then by definition the elements of that order will have no denominators w.r.t. that basis. Having said that, there is a well-defined action of $\mathbb{Z}[\zeta_p]$ on the $p$-class group in your example, but the C-L heuristic...
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Is there any conditions on a finite abelian group so that it cannot be class group of any number field?
Fixed two typos, updated a reference, plus a few other minor tweaks
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$x^2+7y^2=2^n$ and sums of four squares
@Zhi-WeiSun: this looks to me like a pretty exhaustive answer to your actual question "What nontrivial things can we say about the diophantine equation $x^2+7y^2=2^n$.
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