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Let $a\in \mathbb{Q}$. Suppose $G_a$ is the Galois group of $f(x)=x^4+x^3+x^2+x+a$. Find integers $a_1, a_2, a_3$ so that if $i\ne j$, then $G_{a_i} \ne G_{a_j}$.

Here are some thoughts:

If $a=0$, then $f$ is reducible.

If $a=1$, then $f$ is irreducible

So when $a=0$ and $a=1$ we can get two different Galois groups.

But from here, I'm not sure what to do.

Thanks in advance!

Korn
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  • Using reduction modulo $p$ and Dedekind it is not difficult to show that with $a=2$ the degree of the splitting field is divisible by three, making the Galois group distinct from the other cases. Another choice is probably easier. Possibly the simplest way is to look at several cases where the polynomial is reducible. – Jyrki Lahtonen Apr 07 '23 at 17:47

1 Answers1

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It's true that $f$ is reducible when $a = 0$ but is irreducible when $a = 1$. However, we still need to justify why the Galois groups are different in these two cases.

  • In the $a = 0$ case, you can easily factorise the polynomial explicitly into irreducible factors. Once you've done that, it will be obvious what the splitting field and the Galois group are.
  • The $a = 1$ case is quite special because $x^4 + x^3 + x^2 + x + 1$ is the $5$th cyclotomic polynomial. For any $n \in \mathbb N$, it is well known that its splitting field for the $n$th cyclotomic polynomial is $\mathbb Q(\zeta_n)$ (where $\zeta_n = e^{2\pi i / n}$), and the Galois group consists of automorphisms of the form $\zeta_n \mapsto \zeta_n^{k}$ where $k$ is a unit in $\mathbb Z_n$. This information should give you a handle on the Galois group in the $a = 1$ case.

To come up with a third example, probably the easiest thing to do is to choose $a$ such that $f(x)$ factorises into an irreducible cubic factor times a linear factor.

It's very easy to arrange $f(x)$ so that it has a particular linear factor. For example, if you want $f(x)$ to have $(x + 1)$ as a linear factor, then just pick $a$ such that $f(-1) = 0$.

To demonstrate that the left-over cubic factor is irreducible (assuming that it is), it may be viable to use the Eisenstein criterion. Whether this is straightforward or not depends on which $a$ you picked. If Eisenstein doesn't work for you, another way to demonstrate irreducibility is simply to verify that the cubic factor has no integer roots, using some kind of calculus argument. By Gauss' lemma, once you know that the cubic factor is irreducible over $\mathbb Z$, you know that it's irreducible over $\mathbb Q$ too.

The Galois group of $f(x)$ is then the Galois group of this irreducible cubic. Since the Galois group of an irreducible polynomial of degree $n$ is a subgroup of $S_n$ that acts transitively on the polynomial's roots, this considerably narrows down the possibilities for what the Galois group can be. (I don't think you need to determine exactly what the Galois group is up to isomorphism for the purposes of this problem. But if you want to give this a go, then perhaps calculate its discriminant and check if the discriminant is a square - see here.)

I hope this is enough for you get going with.

Kenny Wong
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