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visits member for 2 years, 6 months
seen Feb 7 '13 at 18:18

Mar
21
accepted Can an algebraic group only have trivial elements over $k$
Mar
21
asked Can an algebraic group only have trivial elements over $k$
Mar
19
accepted What's so special about unipotent groups
Mar
19
comment What's so special about unipotent groups
So the algebraic group $\mathrm{GL}_n$ can be written as the semi-product of which reductive and which unipotent one? (This must have something to do with some well-known matrix decomposition.)
Mar
19
asked What's so special about unipotent groups
Mar
19
awarded  Supporter
Mar
18
comment discriminant of an étale cover of an elliptic curve
Just to be precise, the action of $(\mathbf{Z}/n\mathbf{Z})^\ast$ on $\mathbf{Q}(\zeta_n)$ is given by $a \cdot \zeta_n = \zeta_n^a$. So the field of invariants of $H$ is $\mathbf{Q}(\zeta_n)$ when $\Gamma = \Gamma_1(n)$ and $H$ is the group of matrices with first column $(1 \ 0)^t$ in $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$.
Mar
18
comment discriminant of an étale cover of an elliptic curve
I thought "the" field of definition of a modular curve given by a congruence subgroup $\Gamma$ can be "computed" as follows. Let $H$ be a subgroup of $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$ such that $\Gamma$ is the inverse image of $H$ under the natural map $\mathrm{SL}_2(\mathbf{Z})\to \mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$. Then $H$ acts on $\mathbf{Q}(\zeta_n)$ via the determinant map $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})\to (\mathbf{Z}/n\mathbf{Z})^\ast$. Then the modular curve $X_\Gamma$ can be defined over the field of invariants $\mathbf{Q}(\zeta_n)^H$. But maybe this isn't minimal?
Mar
18
comment discriminant of an étale cover of an elliptic curve
Is the modular curve $X_1(n)$ defined over $\mathbf{Q}$? I thought any field of definition should contain $\mathbf{Q}(\zeta_n)$.
Mar
18
comment Minimal polynomial of $\sqrt[3]{7-\sqrt{2}}$
this doesn't make sense to me. The constant term becomes $47^6 - 14\cdot 47^3+47$. How does one apply Eisenstein?
Mar
18
awarded  Teacher
Mar
18
answered What's wrong with this proof that $e^{i\theta} = e^{-i\theta}$?
Mar
17
awarded  Scholar
Mar
17
comment constructing finite extensions with prescribed ramification
This answers my questions. Thanks.
Mar
17
accepted constructing finite extensions with prescribed ramification
Mar
16
awarded  Student
Mar
16
asked constructing finite extensions with prescribed ramification