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Let $p$ be an odd prime. Compute the Galois group of $x^4 - t$ in $\mathbb{F}_p(t)$, distinguishing the cases $p \equiv 1 \pmod 4$ and $p \equiv 3 \pmod 4$.

For $p \equiv 1$ we have $s \in \mathbb{F}_p$ such that $s^2 = - 1$ and we get $(x-\alpha)(x+\alpha)(x-s\alpha)(x+s\alpha)$ where $\alpha^4 - t$, so the splitting field is $F_p(t)(\alpha)$ and the group should be $C_4$.

For $p\equiv 3$ however, we do not have $-1$ as a quadratic residue and I guess we get a factorization as above, but with $s \not\in \mathbb{F}_p(t)$. I suspect that the group should be $D_8$ but how to show it rigorously, i.e.: Why there is not a smaller splitting field than $\mathbb{F}_p(t)(\alpha, s)$ and that this degree is 8? (Or perhaps there is a better way?)

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Since $f(x) = x^4 - t$ is irreducible over $\mathbb{F}_p(t)$ and has degree $4$, if its splitting field has degree less than $8$ over $\mathbb{F}_p(t)$, it must have degree $4$.

Let $K = \mathbb{F}_p(t)(s)$, where $s^2 = -1$. A splitting field of $f$ over $\mathbb{F}_p(t)$ must contain $K$, so if the splitting field is of degree $4$ over $\mathbb{F}_p(t)$, then it must be of degree $2$ over $K$. In other words, $f$ must have non-trivial factors in $K[x]$. However, $f$ is irreducible over $K$, so this is not possible. Hence, the splitting field is of degree $8$ and is as you have computed.

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  • $\begingroup$ But how do we know that $f$ is irreducible over $K$? That's where we need $p\equiv 3 \pmod 4$, I guess :( $\endgroup$ – DesmondMiles Oct 11 '18 at 19:58
  • $\begingroup$ The factorization of $f$ into linear factors remains the same regardless of whether $s$ lies in the base field or not. You could check that there is no nontrivial factor of $f$ over $K$ by multiplying together the different bunches of factors, for instance. $\endgroup$ – Brahadeesh Oct 11 '18 at 20:01
  • $\begingroup$ I did not get that, where do we exactly use $p\equiv 3 \pmod 4$? $\endgroup$ – DesmondMiles Oct 11 '18 at 20:06
  • $\begingroup$ OK I did it. It's easier to define s via s^2 + 1 = 0 and show that x^2 + 1 is irreducible over $\mathbb{F}_p(t)(\alpha$. Indeed $s$ can only be a constant (otherwise is a poly of $t, \alpha$ of non-zero degree) and now this is impossible when $p \equiv 3 \pmod 4$. Thank you for the help! $\endgroup$ – DesmondMiles Oct 11 '18 at 20:37
  • $\begingroup$ @DesmondMiles Some parts of your previous comment are a little unclear to me. When you say that it is easier to define $s$ via $s^2 + 1 = 0$, does it mean you were using a different definition of $s$ before? In my understanding, this is the only possible way to define $s$, as a root of the polynomial $x^2 + 1 \in \mathbb{F}_p(t)[x]$. $\endgroup$ – Brahadeesh Oct 12 '18 at 11:11

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