"Standard" ways of telling if an irreducible quartic polynomial has Galois group C_4?

Let $F$ be a field of characteristic not equal to 2. One option is to can look at the reducibility of $p(x)$ over $F(\sqrt{\Delta})$. If $\Delta$ is not a square and the resolvent cubic is reducible, then we have the following classification: $$\text{Gal}(p/F) = D_4 \iff p(x)\text{ is irreducible in } F(\sqrt{\Delta}).$$

To see this, note that if $\alpha$ is a root of $p$ and $\text{Gal}(p/F) = D_4$, then $p$ splits over $F(\sqrt{\Delta},\alpha)$ as $F(\sqrt{\Delta},\alpha)$ sits inside the splitting field for $p$ and has degree 8 over $F$. But this implies that $[F(\sqrt{\Delta},\alpha):F(\sqrt{\Delta})] = 4$, from which we can derive that $p$ must be irreducible over $F(\sqrt{\Delta})$.

Alternatively, if $\text{Gal}(p/F) = C_4$, then the splitting field for $p$ over $F(\sqrt{\Delta})$ has degree $\#\text{Gal}(p/F)/[F(\sqrt{\Delta}):F]=2$, and thus $p$ must be reducible over $F(\sqrt{\Delta})$.

For what it's worth, here's a special case that can be done with standard tools. Again we restrict to a base field $k$ of characteristic not $2$. Suppose that $f(x) = (x^2 - a)^2 - b$ is irreducible. (This is Corollary 4.5 in the linked notes by Keith Conrad, but there it is proven as a corollary of Kappe-Warren.) We compute that $f'(x) = 4x(x^2 - a)$; letting the roots of $f$ be $\alpha_1, ... \alpha_4$, the discriminant is then

$$(-1)^{ {4 \choose 2} } \prod f'(\alpha_i) = 64 (a^2 - b) b^2$$

which is square if and only if $a^2 - b$ is. Since the splitting field of $f$ is given by a tower of quadratic extensions, the Galois group is one $V_4, D_4, C_4$. Casework:

• If $a^2 - b$ is a square, then the Galois group is $V_4$.
• If $a^2 - b$ is not a square and neither is $b(a^2 - b)$, then the splitting field contains distinct quadratic subfields $k(\sqrt{b})$ and $k(\sqrt{a^2 - b})$, hence the Galois group is $D_4$.
• If $a^2 - b$ is not a square but $b(a^2 - b)$ is, write $u = \sqrt{a + \sqrt{b}}$ in a splitting field for $f$. Then $u \sqrt{a - \sqrt{b}} = \sqrt{a^2 - b} = c \sqrt{b} \in k(u)$ for some $c \in k$, hence $u$ generates the splitting field, which must have degree $4$. Hence the Galois group is $C_4$.