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A field extension $E\supseteq F$ is said to be normal if for any irreducible polynomial $f\in F[X]$, with a root in $E$, splits over $E$.

Using this, I was wondering, how does one show that $\mathbb{C}(X)$ is a normal extension of $\mathbb{C}(X^5)$?

The only idea I got is to probably use the fact that $\mathbb{C}$ is algebraically closed, but other than that, I have no idea.

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    A splitting field is always a normal extension. The proof is a bit delicate, but is covered in all the courses/texts on field extensions. So it remains to show that $E=\Bbb{C}(X)$ is the splitting field of $p(T)=T^5-X^5\in F[T]$ which has coefficients in the smaller field $F=\Bbb{C}(X^5)$. – Jyrki Lahtonen Nov 26 '21 at 14:31
  • @JyrkiLahtonen a splitting field is a normal extension provided the degree is finite, right? –  Nov 26 '21 at 16:48
  • I think that the extra condition (finite degree) would not be necessary. Not really relevant here, though. After all, $[E:F]=5$. – Jyrki Lahtonen Nov 26 '21 at 16:51
  • @JyrkiLahtonen How do you see that $[E:F]=5$? –  Nov 26 '21 at 16:57
  • The polynomial $p(T)$ in my first comment is the minimal polynomial of $X$ over $F$. Many ways to see that it is irreducible in $F[T]$. Eisenstein is one way. Another way is here. – Jyrki Lahtonen Nov 26 '21 at 17:55
  • Actually it is not really important, if you don't see that $p(T)$ is irreducible over $F$. What matters is that $E$ is the splitting field of $p(T)$. Do you see why? – Jyrki Lahtonen Nov 26 '21 at 18:35
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    I guess (from teaching this material for a number of years) that your difficulties come from thinking about those $X$ and $X^5$ as variables rather than constants. When studying function fields this takes a little while, but you'll get the hang of it soon enough. You may want to compare $E/F$ first with the extension $\Bbb{Q}(\pi)/\Bbb{Q}(\pi^5)$. However, that extension is NOT normal, but you hopefully see that it is algebraic of degree five. What we have here is very similar to $K(\pi)/K(\pi^5)$, where $K=\Bbb{Q}(e^{2\pi i/5})$. That extension IS normal. – Jyrki Lahtonen Nov 28 '21 at 05:42

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Any time you adjoin one $n$-th root of an element downstairs, your extension will be normal if the downstairs field contains all the $n$-th roots of unity. That’s the case here, with $n=5$.

Lubin
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  • downstairs? what does that mean? –  Nov 26 '21 at 16:33
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    @monoidaltransform In an extension $E/F$, the field $E$ is upstairs and $F$ is downstairs. When explaining field extensions on a blackboard we more or less always place the extension field higher in diagrams. Connected with a solid line to the subfield. – Jyrki Lahtonen Nov 26 '21 at 17:57
  • Thanks, @JyrkiLahtonen. I was in a car with an iPhone, and wanted to use a minimum of TeX. – Lubin Nov 26 '21 at 23:17