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Algebraic numbers which prescribed degree which does not belong to some fields

Proposition 2 is false, although perhaps only for $n = 4$ (and $t = 2$ or $t = 3$). If $t = 2$, every algebraic number $\gamma$ of degree $4$ is contained in $K_{4}$. Let $K$ be the Galois closure of $...
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5 votes

Algebraic numbers which prescribed degree which does not belong to some fields

Proposition 1 is true. Let $x$ be an element of degree $n$ whose Galois group is isomorphic to $S_n$. Let $F$ be a finite subset of algebraic elements of degree $<n$. I claim that $x$ is not in the ...
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1 vote
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Can a general quintic be solved using inverse beta regularized function?

Yes, it seems, it can. Any general quintic can be reduced to the Bring-Jerrard form $x^5+ax+b=0$. Then Tyma Gaidash found a solution for the Bring-Jerrard form via Inverse Beta Regularized function: $...
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6 votes
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Completion of infinite degree extension of perfectoid fields is perfectoid?

I'm assuming you mean infinite algebraic extensions, as otherwise there is no standard way of completing them. Let $K$ be a perfectoid field, let $L$ be an infinite algebraic extension. Then $L$ ...
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0 votes

Galois groups vs. fundamental groups

I found this introduction by David Corwin very accessible and well explained: https://math.berkeley.edu/~dcorwin/files/etale.pdf.
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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 ...
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24 votes

Absolute Galois group, number theory and the Axiom of Choice

In the absence of the axiom of choice, it is still possible to define the "usual" algebraic closure of $\mathbb{Q}$ because you can just explicitly enumerate all polynomials with integer ...
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