Let $f(x)=x^3-5\in\mathbb{F}_7[x]$. What is the splitting field of $f$? I'm unsure how to approach splitting over a finite field.
I know $x^3-5$ is irreducible and that $\mathbb{F}_7[x]/(f(x))$ has at least one root.
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why are people downvoting my question – linus Apr 12 '20 at 03:49
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Can you prove that if $\alpha$ is one root, then $\alpha^7$ is another? – Jyrki Lahtonen Apr 12 '20 at 06:35
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Also, this follows from the general fact that every extension of finite fields is normal. Assuming that you know how normal extensions and splitting fields go together. – Jyrki Lahtonen Apr 12 '20 at 06:41
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Actually this may help you even more. – Jyrki Lahtonen Apr 12 '20 at 06:43
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1The downvotes may have come from the question being a bit lacking in context. You did explain how you can find an extension with one root. That surely counts. May be you could have also explained how you deduced that the cubic is irreducible over $\Bbb{F}_7$ :-). To avoid similar mishaps later, do take a look at our guide for new askers. – Jyrki Lahtonen Apr 12 '20 at 07:08
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The question is also close to being a FAQ in the context of finite fields. It may well be that it is so much of a FAQ that nobody who knows this well never thought there would not be a good answer already. I like to think we have better ones than those I linked to. May be they don't have [tag:finite-fields] tag? I recall having wasted a bit of time searching for those earlier :-) – Jyrki Lahtonen Apr 12 '20 at 07:12
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It is clear that, if $\alpha$ is a root of $f$ is an algebraic closure of $\mathbb{F}_7$ , then the roots of $f$ are $z\alpha,$ where $z\in\bar{\mathbb{F}}_7$ satisfies $z^3=1$ (easy exercice).
Now the polynomial $x^3-1$ splits in $\mathbb{F}_7$: $x^3-1=(x-1)(x-2)(x-4)$. Hence the splitting field $L$ of $f$ is the field extension generated by $\alpha, 2\alpha$ and $4\alpha$. Finally, we get $L=\mathbb{F}_7(\alpha)$, that is $L=\mathbb{F}_{7^3}.$
GreginGre
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