Galois cohomology and cyclic and abelian extensions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 01 February 2012

Galois cohomology and cyclic and abelian extensions

Let $𝔽$ be a field.

• A cyclic extension of $𝔽$ is a Galois extension $𝔼$ of $𝔽$ such that $\mathrm{Gal}(𝔼/𝔽)$ is cyclic.
• An abelian extension of $𝔽$ is a Galois extension $𝔼$ of $𝔽$ such that $\mathrm{Gal}(𝔼/𝔽)$ is abelian.
• An abelian extension of exponent dividing $n$ is an abelian extension $𝔼$ of $𝔽$ such that $g^n=1$ for all $g\in \mathrm{Gal}(𝔼/𝔽)$.

(Hilbert's Theorem 90) Let $𝔼$ be a cyclic extension of $𝔽$ and let $\sigma$ be a generator of $\mathrm{Gal}(𝔼/𝔽)$.

1. (a) Let $x\in {𝔼}^{*}$. $$N_{𝔼/𝔽}\left(x\right)=1if\; and\; only\; ifthere\; exists\; an\; element\; y\in {𝔼}^{*} \; such\; that\; x=\frac{y}{\sigma \left(y\right)}.$$
2. (a') Let $x\in {𝔼}^{*}$. If there exists an element $y\in {𝔼}^{*}$ such that $x=\frac{y}{\sigma \left(y\right)}$ then it is unique up to multiplication by elements of ${𝔽}^{*}$.
3. (b) Let $x\in 𝔼$. $${\mathrm{Tr}}_{𝔼/𝔽}\left(x\right)=0if\; and\; only\; ifthere\; exists\; an\; element\; z\in 𝔼 \; such\; that\; x=z-\sigma \left(z\right).$$
4. (b') Let $x\in 𝔼$. If there exists an element $z\in 𝔼$ such that $x=z-\sigma \left(z\right)$ then it is unique up to adding elements of $𝔽$.

(Kummer theory) Assume $𝔽$ contains a primitive $n^{th}$ root of unity.

1. There is a bijection $$\begin{array}{ccc}\left\{abelian\; extensions\; of\; exponent\; dividing\; n\right\}& \leftrightarrow & \{subgroups\; H\subseteq {𝔽}^{*} \; such\; that\; H\supseteq {\left({𝔽}^{*}\right)}^n\}\\ 𝔼& \mapsto & {𝔼}^n\cap {𝔽}^{*}\\ 𝔽\left(H^{\frac{1}{n}}\right)= {\alpha \in \stackrel{\_}{𝔽}|{\alpha }^n\in H} & \leftarrow & H\end{array}$$
2. If $H\supseteq {𝔽}^{*}$ is a subgroup such that $H\supseteq {\left({𝔽}^{*}\right)}^n$ then $$\mathrm{Gal}\left(𝔽\right(H^{\frac{1}{n}})/𝔽)\to \mathrm{Hom}\left(H/{\left({𝔽}^{*}\right)}^n,\left\{primitive\; n^{th} \; roots\; of\; unity\right\}\right)$$ and $$[𝔽\left(H^{\frac{1}{n}}\right):𝔽] = [H:{\left({𝔽}^{*}\right)}^n] and {{\theta }_{\stackrel{\_}{\alpha }}\in \stackrel{\_}{𝔽}|\stackrel{\_}{\alpha }\in H/{\left({𝔽}^{*}\right)}^n, {\theta }_{\stackrel{\_}{\alpha }}^n=\alpha \in H}$$ is a basis of $𝔽\left(H^{\frac{1}{n}}\right)$ over $𝔽$.

Let $𝔽$ be a field with $\mathrm{char}\left(𝔽\right)=0$ and let incomplete sentence

Notes and References

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