How can I find the Galois group of $x^5-x-3$ over $\mathbb{F}_5$?
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1It would be easier to help if you briefly mentioned your attempts so far. To start how does the polynomial factor over $\mathbb{F}_5$? – Μάρκος Καραμέρης Sep 01 '20 at 15:23
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2It's an Artin-Schreier polynomial, so clearly irreducible. – Angina Seng Sep 01 '20 at 15:27
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Furthermore, the Wiki page says that if $\beta$ is a root,so are $\beta+i$, $1 \leq i \leq p$, so the splitting field is $\mathbb F_5(\beta)$. It doesn't give the details, but you can try using Fermat's little theorem to show this. – Sarvesh Ravichandran Iyer Sep 01 '20 at 15:33
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Several proofs of irreducibility here. – Jyrki Lahtonen Sep 02 '20 at 04:06
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Let $P=x^5-x-3$, the polynomial over $\mathbb{F}_5$.
For any $a\in \mathbb{F}_5$ we have $a^5=a\neq a+3$, so $a$ is not a root of $P$. If we attach a single root $\beta$ of $P$, we obtain a finite extension $\mathbb{F}_5[\beta]$, of degree $2,3$ or $5$, depending on whether or not the polynomial is irreducible. Let $F:\mathbb{F}_5[\beta]\to \mathbb{F}_5[\beta]$ denote the Frobenius automorphism, sending $\alpha\mapsto \alpha^5$.
Now for $a\in \mathbb{F}_5$ we have that $F(\beta+a)=F(\beta)+F(a)=F(\beta)+a$. Thus $$(\beta+a)^5-(\beta+a)-3=\beta^5-\beta-3+a-a=0.$$
Thus we know $P$ splits in $\mathbb{F}_5[\beta]$. Further, if $P$ were not irreducible it would have a quadratic factor with coefficients in $\mathbb{F}_5$. However:$$(x-\beta-a_1)(x-\beta-a_2)=x^2-(2\beta+a_1+a_2)x+(\beta-a_1)(\beta-a_2)$$ and $2\beta+a_1+a_2\notin \mathbb{F}_5$ for any $a_1,a_2\in \mathbb{F}_5$.
We conclude that $P$ is irreducible and has a splitting field $\mathbb{F}_5[\beta]$ of degree $5$.
Its Galois group is therefore generated by $F$ which has order $5$ (that is $\alpha^{5^5}=\alpha\alpha^{5^5-1}=\alpha\alpha^{|\mathbb{F}_5[\beta]^*|}=\alpha$, for every $\alpha\in \mathbb{F}_5[\beta]$). We conclude that the Galois group is $C_5$.
tkf
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A slicker argument would be to say that if $P$ factorised, then attaching a root of one irreducible factor would give you a splitting field for any other irreducible factor, so all irreducible factors have the same degree and the degrees add up to $5$. However $5$ is prime. – tkf Sep 01 '20 at 17:09