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Let $f(x) \in F[x]$ be an irreducible separable polynomial of degree p, with distinct roots $\alpha_1, \ldots, \alpha_p$. I want to show that if $F(\alpha_1) = F(\alpha_1, \alpha_2)$, then $F(\alpha_1) = F(\alpha_1, \ldots, \alpha_p)$, ie. $F(\alpha_1)$ is the splitting field of $f(x)$.

So far, what I know is that $F(\alpha_1) \simeq F(\alpha_i)$ for all $1 \leq i \leq p$, can I somehow extend this to show that $(F(\alpha_1))(\alpha_1) \simeq (F(\alpha_1))(\alpha_i)$? Thanks

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    $\begingroup$ The title is not quite consistent with the question $\endgroup$
    – lhf
    Dec 1, 2017 at 1:38

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Let $G = \operatorname{Gal}(f) \subset S_p$ be the galois group of the splitting field. By a standard argument, $G$ contains a cycle of length $p$ (Since any element of order $p$ in $S_p$ is such a cycle).

Let $\sigma \in G$ be a cycle of length $p$. Replacing it by a suitable power, we can assume $\sigma(\alpha_1) = \alpha_2$.

Let $L = F(\alpha_1)$. In particular $\sigma(L) \subset L$ holds, by iterating this we get that $\sigma^n(L) \subset L$ holds for any $n \geq 1$. But since $\sigma$ is a cycle of length $p$, we get that all the $\alpha_i$ occur among $\sigma^n(\alpha_1), n \geq 1$. Hence $L$ contains all roots and is thus the splitting field.


Some more details on the "standard argument":

If $K$ is the splitting field, we have $F \subset F(\alpha_1) \subset K$, hence $$|G| = [K:F] = [K:F(\alpha_1)] \cdot [F(\alpha_1):F] = [K:F(\alpha_1)] \cdot p$$ is a multiple of $p$. By Cauchy's Theorem, $G$ has an element of order $p$.

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  • $\begingroup$ Thanks. Since posting the question, I've actually figured out all the above except for the "By a standard argument" part. How can we be sure that $G$, which (turns out to be) a strict subgroup of $S_p$, contains an element of order $p$? I've thought about this for a long time and it is not obvious to me. $\endgroup$
    – shmth
    Dec 3, 2017 at 12:37
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    $\begingroup$ This is Cauchy's Theorem. The order of the group is divisible by $p$, hence there is an element of order $p$. $\endgroup$
    – MooS
    Dec 3, 2017 at 12:49
  • $\begingroup$ @ihmth I have added some details in my answer. $\endgroup$
    – MooS
    Dec 3, 2017 at 13:58

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