Intereting Posts

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the Nordhaus-Gaddum problems for chromatic number of graph and its complement
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How to select the right modulus to prove that there do not exist integers $a$ and $b$ such that $a^2+b^2=1234567$?
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The series $\sum\limits_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}$ is convergent

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.

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So what are my other options in general?

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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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