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Let $p \in \mathbb{P}$ and $n \in \mathbb{N}: p \nmid n$; the set of $n$th roots of unity be $W_n$; $\mathbb{F}$ be a field$: char(\mathbb{F}) \in \left\{ {0,p}\right\}, \mathbb{F} \supseteq W_n$ (set inclusion); $\alpha^n = a \in \mathbb{F}$ and $n$ is the minimal positive natural number such that $\alpha^n \in \mathbb{F}$; $\mathbb{E} = \mathbb{F}(\alpha)$ is a field; $\forall \sigma \in G = Gal(\mathbb{E} / \mathbb{F}), f: G \rightarrow W_n, \sigma \mapsto \sigma(\alpha)/ \alpha$. I have proven that $\mathbb{E}$ is a Galois extension over $\mathbb{F}$ with cyclic Galois group $G$.

Since it is apparently harder than I originally thought, how does one prove that $f$ is a (group?) isomorphism? Specifically, the surjectivity is causing problems. But proof of the homomorphicity and injectivity would not go amiss, in case I am misunderstanding. (I relied on the cyclicity of $G$ and the fact that the automorphism $\sigma$ would be uniquely determined by its action on $\alpha$ in order to work with $f$ in this situation.)

Why would $x^n - a$ be irreducible in $\mathbb{F}[x]$? I am probably just not thinking but all of the above information is kind of just confusing me.

Also, I must prove: $\beta \in \mathbb{E}: \beta^n = b \in \mathbb{F}, \mathbb{E} = \mathbb{F}(\beta) \implies \exists k \in \mathbb{N} \cup \left\{ {0}\right\}, \exists c \in \mathbb{F}: gcd(k,n)=1, b = a^k c^n$. I was thinking that for $G = \langle \sigma \rangle$, I could prove that $f(\sigma)$ generates $W_n$ (thus $W_n = \langle \sigma(\alpha)/ \alpha \rangle$); but I am not sure how to do so. From there, I think that I would consider $\sigma(\beta)/ \beta$ and possibly use the previously mentioned isomorphicity of $f$ or the minimality of $n$, but I am not sure how that would go. Any suggestions?

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  • $\begingroup$ How did you prove $f$ is surjective? You must have extra information about the field $\mathbb{F}$. Did you do the case of the field of rational numbers? There you can take $a$ to be an integer and apply Eisenstein criterion and conclude $x^n-a$ is irreducible. $\endgroup$ – P Vanchinathan Apr 8 '14 at 6:10
  • $\begingroup$ I do not think that we are supposed to restrict to a special case. $\endgroup$ – kevin Apr 8 '14 at 6:19
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    $\begingroup$ You may want to use standard English in questions in the future. Many users here are not native English speakers. $\endgroup$ – Potato Apr 8 '14 at 14:56

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