| Version current |
Version 1 |
| \begin{proof} Let $\zeta\in K$ be a primitive $n^{\mathrm{th}}$ root of unity, and denote by $\boldsymbol{\mu}_n$ the subgroup of $K^{\star}$ generated by $\zeta$. |
\begin{proof} Let $\zeta\in K$ be a primitive $n^{\mathrm{th}}$ root of unity, and denote by $\boldsymbol{\mu}_n$ the subgroup of $K^{\star}$ generated by $\zeta$. |
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| (1) Let $L=K(\sqrt[n]{a})$; then $L/K$ is Galois since $K$ contains all $n^{\mathrm{th}}$ roots of unity and thus is a splitting field for $x^n-a$, which is separable since $n\neq 0$ in $K$. Thus the elements of $\Gal(L/K)$ permute the roots of $x^n-a$, which are |
(1) Let $L=K(\sqrt[n]{a})$; then $L/K$ is Galois since $K$ contains all $n^{\mathrm{th}}$ roots of unity and thus is a splitting field for $x^n-a$, which is separable since $n\neq 0$ in $K$. Thus the elements of $\Gal(L/K)$ permute the roots of $x^n-a$, which are |
| \[\sqrt[n]{a},\ \zeta \sqrt[n]{a},\ \zeta^2 \sqrt[n]{a},\ \dotsc,\ \zeta^{n-1}\sqrt[n]{a}\] |
\[\sqrt[n]{a},\ \zeta \sqrt[n]{a},\ \zeta^2 \sqrt[n]{a},\ \dotsc,\ \zeta^{n-1}\sqrt[n]{a}\] |
| and thus for $\sigma\in \Gal(L/K)$, we have $\sigma(\sqrt[n]{a}) = \zeta_{\sigma} \sqrt[n]{a}$ for some $\zeta_{\sigma}\in\boldsymbol{\mu}_n$. |
and thus for $\sigma\in \Gal(L/K)$, we have $\sigma(\sqrt[n]{a}) = \zeta_{\sigma} \sqrt[n]{a}$ for some $\zeta_{\sigma}\in\boldsymbol{\mu}_n$. |
| Define a map |
Define a map |
| \[p:\Gal(L/K)\to \boldsymbol{\mu}_n:\sigma\mapsto\zeta_{\sigma}\] |
\[p:\Gal(L/K)\to \boldsymbol{\mu}_n:\sigma\mapsto\zeta_{\sigma}\] |
| We will show that $p$ is an injective homomorphism, which proves the result. |
We will show that $p$ is an injective homomorphism, which proves the result. |
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| Since $\boldsymbol{\mu}_n\subset K$, each $n^{\mathrm{th}}$ root of unity is fixed by $\Gal(L/K)$. Then for $\sigma,\tau\in\Gal(L/K)$, |
Since $\boldsymbol{\mu}_n\subset K$, each $n^{\mathrm{th}}$ root of unity is fixed by $\Gal(L/K)$. Then for $\sigma,\tau\in\Gal(L/K)$, |
| \[ |
\[ |
| \zeta_{\sigma\tau}\sqrt[n]{a} =\sigma\tau(\sqrt[n]{a}) = \sigma(\zeta_{\tau}\sqrt[n]{a}) |
\zeta_{\sigma\tau}\sqrt[n]{a} =\sigma\tau(\sqrt[n]{a}) = \sigma(\zeta_{\tau}\sqrt[n]{a}) |
| = \zeta_{\tau}(\sigma(\sqrt[n]{a})) |
= \zeta_{\tau}(\sigma(\sqrt[n]{a})) |
| = \zeta_{\sigma}\zeta_{\tau}\sqrt[n]{a} |
= \zeta_{\sigma}\zeta_{\tau}\sqrt[n]{a} |
| \] |
\] |
| so that $\zeta_{\sigma\tau} = \zeta_{\sigma}\zeta_{\tau}$ and $p$ is a homomorphism. The kernel of the map consists of all elements of $\Gal(L/K)$ which fix $\sqrt[n]{a}$, so that $p$ is injective and we are done. |
so that $\zeta_{\sigma\tau} = \zeta_{\sigma}\zeta_{\tau}$ and $p$ is a homomorphism. The kernel of the map consists of all elements of $\Gal(L/K)$ which fix $\sqrt[n]{a}$, so that $p$ is injective and we are done. |
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| (2) Note that $\N_{L/K}(\zeta) = 1$ since $\zeta$ is a root of $x^n-1$, so that by Hilbert's Theorem 90, |
(2) Note that $\N_{L/K}(\zeta) = 1$ since $\zeta$ is a root of $x^n-1$, so that by Hilbert's Theorem 90, |
| \[\zeta = \sigma(u)/u,\quad\text{for some }u\in L\] |
\[\zeta = \sigma(u)/u,\quad\text{for some }u\in L\] |
| But then $\sigma(u) = \zeta u$ so that $\sigma(u^n) = \sigma(u)^n = \zeta^n u^n = u^n$ and $a=u^n\in K$ since it is fixed by a generator of $\Gal(L/K)$. Then clearly $K(u)$ is a splitting field of $x^n-a$, and the elements of $\Gal(L/K)$ send $u$ into distinct elements of $K(u)$. Thus $K(u)$ admits at least $n$ automorphisms over $K$, so that $[K(u):K]\geq n = [L:K]$. But $K(u)\subset L$, so $K(\sqrt[n]{a})=K(u)=L$. |
But then $\sigma(u) = \zeta u$ so that $\sigma(u^n) = \sigma(u)^n = \zeta^n u^n = u^n$ and $a=u^n\in K$ since it is fixed by a generator of $\Gal(L/K)$. Then clearly $K(u)$ is a splitting field of $x^n-a$, and the elements of $\Gal(L/K)$ send $u$ into distinct elements of $K(u)$. Thus $K(u)$ admits at least $n$ automorphisms over $K$, so that $[K(u):K]\geq n = [L:K]$. But $K(u)\subset L$, so $K(\sqrt[n]{a})=K(u)=L$. |
| \end{proof} |
\end{proof} |
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| \begin{thebibliography}{10} |
\begin{thebibliography}{10} |
| \bibitem{bib:df} |
\bibitem{bib:df} |
| Dummit,~D.,~Foote,~R.M., \emph{Abstract Algebra, Third Edition}, Wiley, 2004. |
Dummit,~D.,~Foote,~R.M., \emph{Abstract Algebra, Third Edition}, Wiley, 2004. |
| \bibitem{bib:kap} |
\bibitem{bib:kap} |
| Kaplansky,~I., \emph{Fields and Rings}, University of Chicago Press, 1969. |
Kaplansky,~I., \emph{Fields and Rings}, University of Chicago Press, 1969. |
| \end{thebibliography} |
\end{thebibliography} |
