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3 votes
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Galois action on ultrapowers

This is quite trivial but not enough for a comment: $F^G$ is not always $K^\mathbb N/\mathcal U$. Let $x_1,x_2,\dots$ be elements of $K$ such that no product of finitely many of them is a perfect squ …
Will Sawin's user avatar
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3 votes
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non-continuous inverse Galois problem

Q1: No. Suppose there were such a subgroup. Then there would certainly be a $K$ such that $G/K$ is $\mathbb Z$. $K$ would have to contain the commutator subgroup. The quotient of $G$ by the commutato …
Will Sawin's user avatar
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17 votes

Absolute Galois group, number theory and the Axiom of Choice

There would be no consequences, for two reasons: As Timothy Chow points out, if we define $\overline{\mathbb Q}$ as the set of complex numbers that are roots of a nonzero polynomial with rational coe …
Will Sawin's user avatar
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8 votes
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Galois action on automorphisms of a curve

$K$-defined would mean that the subgroup itself is defined over $K$, not the elements. Formally, this means that there are equations over $K$ which are satisfied by the coefficients of polynomials def …
Will Sawin's user avatar
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17 votes

Subgroups of GL(2,q)

We can generalize Ralph's answer to find the abelian groups that are $p$-groups for odd $p$ that cannot be subgroups of $GL_2$. if $p|(q^2-1)(q^2-q)$ then exactly one of $p|(q+1)$, $p|(q-1)$ and $p|q …
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