Intereting Posts

Why does polynomial factorization generalize to matrices
if $\int_1^\infty f(x)dx$ exist, then $\int_1^\infty f^2(x)dx$ exist?
Irreducible cyclotomic polynomial
How to deduce the area of sphere in polar coordinates?
Division polynomials of elliptic curves
Graph Theory: How quickly will triadic closure create a complete graph?
The center of a non-Abelian group of order 8
Linear Algebra, Vector Space: how to find intersection of two subspaces ?
Intuition of Addition Formula for Sine and Cosine
Is this proof for Theorem 16.4 Munkers Topology correct?
Principal value of the singular integral $\int_0^\pi \frac{\cos nt}{\cos t – \cos A} dt$
What is wrong in my proof that 90 = 95? Or is it correct?
An arbitrary product of connected spaces is connected
Can a matrix in $\mathbb{R}$ have a minimal polynomial that has coefficients in $\mathbb{C}$?
A question about sampling distribution

The following facts are standard: an irreducible quartic polynomial $p(x)$ can only have Galois groups $S_4, A_4, D_4, V_4, C_4$. Over a field of characteristic not equal to $2$, depending on whether or not the discriminant $\Delta$ is a square and whether or not the resolvent cubic $q(x)$ is irreducible, we can distinguish four cases:

- If $\Delta$ is not a square and the resolvent cubic is irreducible, then the Galois group is $S_4$.
- If $\Delta$ is a square and the resolvent cubic is irreducible, then the Galois group is $A_4$.
- If $\Delta$ is a square and the resolvent cubic is reducible, then the Galois group is $V_4$.
- If $\Delta$ is not a square and the resolvent cubic is reducible, then the Galois group is either $D_4$ or $C_4$.

How can we resolve the ambiguity in the last case? For $p$ a monic polynomial in $\mathbb{Z}[x]$, I know the following approaches:

- In the simplest cases one can work directly with the splitting field. But this is rare, although it can work if $p = x^4 + ax^2 + b$ for some $a, b$.
- If $p$ has two complex roots (equivalently, if the discriminant is negative), there is a transposition in the Galois group, so the Galois group is $D_4$.
- If $p$ factors as the product of two linear factors and an irreducible quadratic factor modulo a prime, there is a transposition in the Galois group, so the Galois group is $D_4$.

In practice, the last two often work to identify a Galois group of $D_4$ (and in principle it must eventually work by the Frobenius density theorem). But I don’t know a corresponding practical way to identify a Galois group of $C_4$. There is a criterion due to Kappe and Warren which I learned about from one of Keith Conrad’s expository notes here. However, in an upcoming exam I’m taking, I’ll only be able to cite results proven in the course, and this criterion isn’t one of them.

- Minimal Polynomial of $\zeta+\zeta^{-1}$
- Ramanujan-type trigonometric identities with cube roots, generalizing $\sqrt{\cos(2\pi/7)}+\sqrt{\cos(4\pi/7)}+\sqrt{\cos(8\pi/7)}$
- Embedding of a field extension to another
- What is the prerequisite knowledge for learning Galois theory?
- Construct generator matrix given generator polynomial?
- Show that $K = \mathbb{Q}(\sqrt{p} \ | \ \text{p is prime} \}$ is an algebraic and infinite extension of Q

So what are my other options in general?

- A certain inverse limit
- Galois group of a quartic
- What is $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q(\sin(2\pi k/n))$?
- Embedding fields into the complex numbers $\mathbb{C}$.
- Monic irreducible polynomials of degree 6 in $F_{5}$
- Cardinality of algebraic extensions of an infinite field.
- Structure of $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$?
- Field extensions, inverse limits, notation and roots of unity
- When is $X^n-a$ is irreducible over F?
- Show that $\mathbb{Q}(\sqrt{2 +\sqrt{2}})$ is a cyclic quartic field i.e. is a galois extension of degree 4 with cyclic galois group

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 $C_4$.

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