# “Standard” ways of telling if an irreducible quartic polynomial has Galois group C_4?

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.

So what are my other options in general?

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@quanta: $D_4$ is the dihedral group of order 8. en.wikipedia.org/wiki/Dihedral_group – Brandon Carter May 11 '11 at 17:17
If you're only going to be able to cite results from the course, then why do you even think you'd be asked for a C_4 example? After all, if the course has not developed methods to deal with the case (other than, say, the 5th cyclotomic field), then you wouldn't be asked about it. – KCd May 14 '11 at 6:59
@KCd: yes, I guess this is true. I asked in case there was something more standard I could use. – Qiaochu Yuan May 14 '11 at 7:09

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 $F(\sqrt{\Delta})$.

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Cool. I don't know how easy it is to test irreducibility over $\mathbb{Q}(\sqrt{\Delta})$, though, if the ring of integers of the latter doesn't have unique factorization... – Qiaochu Yuan May 11 '11 at 22:37
Qiaochu, didn't you read the appendix to the handout of mine which you cited? What Brandon writes above is Theorem A.1 there and this is the standard method of distinguishing D_4 and Z/4Z which you'll find (or at least I found) in books written before Kappe and Warren's paper. This method is hardly a practical option (esp. in an exam) because determining irred. of a quartic over a quadratic field is a pain even in simple cases. See the worked examples I wrote up in that file you cite. – KCd May 14 '11 at 7:10
@KCd: unfortunately I didn't; thanks for the pointer! I agree this is impractical. Generally the quartics we are asked about have only even terms in them and this is doable with standard tools. – Qiaochu Yuan May 14 '11 at 8:28
How did you know that $F(\sqrt{\Delta}, \alpha)$ has degree $8$ over F ? – Jack Sep 21 '15 at 17:27

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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Although my notes have a proof of this special case using the Kappe--Warren result, I do point out after the proof that this special case was known long before then (e.g., in a book of Kaplansky). I included that corollary more as an illustration of the Kappe--Warren result than for its own sake, as it can be handled by more basic methods. – KCd May 15 '11 at 9:09
@KCd: yes, I learned it from Artin's Algebra myself. I just wanted to know if I could reconstruct it from memory, and after having done so figured I might as well put it here for the sake of completeness. – Qiaochu Yuan May 15 '11 at 9:32