# Galois group of $x^4 - t$ in $F_p(t)$

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?)

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.

• But how do we know that $f$ is irreducible over $K$? That's where we need $p\equiv 3 \pmod 4$, I guess :( – DesmondMiles Oct 11 '18 at 19:58
• 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. – Brahadeesh Oct 11 '18 at 20:01
• I did not get that, where do we exactly use $p\equiv 3 \pmod 4$? – DesmondMiles Oct 11 '18 at 20:06
• 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! – DesmondMiles Oct 11 '18 at 20:37
• @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 $x^2 + 1 \in \mathbb{F}_p(t)[x]$. – Brahadeesh Oct 12 '18 at 11:11