# Tag Info

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### Why is the subfield of $\mathbb{Q}(\zeta_p)$ of index $2$ expressible in terms of the sum of $\zeta_p$ to the power of all quadratic residues mod $p$?

It is called a subfield of degree $2$ (over $\mathbf Q$), or more commonly a quadratic subfield. To see a number $\alpha$ in $\mathbf Q(\zeta_p)$ is quadratic over $\mathbf Q$, we can do two things: (...
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• 332k
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### When normal extensions are normal

Neither normal nor "finite and Galois" have property $\mathcal{P}$. (Note that there is such a thing as an infinite Galois extension...) Note: I misread the problem initially. What is below ...
• 402k
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### Degree of field extension $\Bbb Q(\sum\limits_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}):\Bbb Q$

The answer to your question is yes when $n$ is squarefree. Here is a general result. When $L/K$ is Galois with Galois group $G$, $L = K(\alpha)$, and $H$ is a subgroup of $G$, then we can ask ...
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### Inverse problem of the field extension $K/\Bbb{Q}$ of given number of prime above $p$

It is a theorem that for every number field, infinitely many primes split completely in it. So let $K$ be an arbitrary number field with degree $g$ over $\mathbf Q$, e.g., $\mathbf Q(\sqrt[g]{2})$ and ...
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### Do I need the assumption that $\operatorname{char}(k) = p$ here?

I'd like to add more details to your proof of $\{t^au^b\}_{0\le a,b<p}$ are linearly independent. Assume $\sum_{a,b}c_{a,b}t^au^b=0$, we want to show $c_{a,b}=0$. By multiplying the product of all ...
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### What is the fixed field of $\mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{5}\zeta_3)$ with 3-cycle group?

The fixed field is $\mathbb{Q}(\zeta_3)$. To verify that $\zeta_3$ lies in the fixed field, note that $$\zeta_3 = \frac{\sqrt[3]{5}\zeta_3}{\sqrt[3]{5}}.$$ So the image of $\zeta_3$ under the ...
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### Annihilator basis after extension of scalars

Note that $S_a$ is the kernel of the map $a : A \to A \oplus A$ given $$x \mapsto (ax, xa).$$ Thus, we have an exact sequence $$0 \to S_a \to A \xrightarrow{a} A \oplus A.$$ By exactness of the ...
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