Can two monic irreducible polynomials over $\mathbb{Z}$, of coprime degrees, have the same splitting field? Let $f,g \in \mathbb{Z}[X]$ be monic polynomials. It is possible for distinct monic polynomials over $\mathbb{Z}$ to have the same splitting field. For example $f = x^4 - 2$ and $g= x^4+2$ both have splitting field
 $\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}4]{2}, i)$. Here $f,g$ are both irreducible with equal degrees.
Another example is $f=x^n - 1$ and $g=\Phi_n(x)$, the $n$'th cyclotomic polynomial. The splitting field of both is the $n$'th cyclotomic field. In this case, $g$ is irreducible and divides $f$. The degree of $g$ is of course $\varphi(n)$.
Question: Suppose $f,g$ are monic and irreducible, with coprime degrees. Say $\deg f = n$ and $\deg g = k$. Let us write $K_f$ and $K_g$ for their splitting fields.
Is it possible for these splitting fields to coincide, $K_f = K_g$? I am particularly interested in the case $k=n-1$.
I have attempted to approach the question using group theory. If we write $G$ for their common Galois group, then via the action of $G$ on the roots of $f$ and $g$ respectively, we would realize $G$ as a transitive subgroup of both $S_n$ and $S_k$. Therefore, as $k$ and $n$ are coprime


*

*$nk \mid |G|$.

*$|G| \leq k!$

*$[S_n : G] \geq n(n-1)\cdots(k+1)$.


Furthermore, if $\alpha$ is a root of $f$ and $\beta$ is a root of $g$, then their stabilizers $Stab(\alpha)$ and $Stab(\beta)$ are subgroups of index $n$ and $k$ respectively. As these are coprime by assumption we would have $G = Stab(\alpha)Stab(\beta)$. 
My conjecture is that this is not possible, but I am not able to show it. I seem to remember that, if $l < q$ then $S_l$ is never a transitive subgroup of $S_q$, except the case $S_5 \leq S_6$. Here $S_5$ acts transitively on it's $5$-Sylow subgroups by conjugation, and there are $6$ of them. However, I might be wrong as I am unable to recover a reference for this. Of course, if this was true, we would simply strengthen 2. and 3. above to strict inequalities, unless $n=6$ and $k=5$.
I am also unaware whether there exists such a $f$ with degree $6$ and Galois group $S_5$ acting on the roots in this exotic manner.
 A: Yes, this can happen. A number of the hypotheses in the problem are distractions.
(1) If $L/K$ is a finite Galois extension with Galois group $G$, and $X$ is a set with a transitive $G$-action, then there is always an element $\theta$ of $L$ such that $G$ acts on the Galois conjugates of $\theta$ as $G$ acts on $X$. Namely, let $H$ be the stabilizer of a point of $X$ and choose $\theta$ to lie in $L^H$ and not in any smaller subfield.
(2) The OP asks that the monic minimial polynomial of $\theta$ have integer coefficients, but we can always replace $\theta$ by $N \theta$ for some highly divisible integer $N$ to make this true.
Thus, the question comes down to the following:

Can we find a Galois group $G$ over $\mathbb{Q}$ which acts transitively on two sets of relatively prime sizes?

The OP suggests trying $G=S_5$ acting on a $5$-element set and on a $6$-element set. Indeed, this works. Take a polynomial with Galois group $S_5$ and roots $\rho_i$. Let $L$ be the splitting field of this polynomial, and let $S_5$ act on the $\rho_i$ by acting on their subscripts.
Inside $S_5$, let $H$ be the $20$ element group of maps of the form $x \mapsto ax+b \bmod 5$, for $a \not\equiv 0 \bmod 5$. I'll construct an element of $L$ which is fixed by $H$ and no larger subgroup of $S_5$. Namely, let
$$\theta = \sum_{h \in H} \rho_{h(1)}^{\phantom{1}} \rho_{h(2)}^2 \rho_{h(3)}.$$
This is manifestly $H$-fixed; its Galois conjugates can be obtained by further acting on the subscripts by $S_5$. If the $6$ Galois conjugates we obtain are distinct, we win. (I tried some polynomials of lesser degree in the $\rho_i$ first, but they didn't give distinct conjugates in the examples I tried.)
As an example, I took the quintic $x^5-4x-1$ with roots roughly given by
$$-1.34325,\ -0.250245,\ 1.47082,\ 0.0613366 \pm 1.42087 i .$$
I numerically computed that the Galois conjugates of $\theta$ are the six distinct numbers
$$0.0134815 \pm 31.9642 i,\ 12.4594 \pm 4.4798 i,\ 19.5271 \pm 2.60181 i.$$
I computed the coefficients of a polynomial with these roots numerically. Since these numbers are Galois conjugate algebraic integers, the coefficients of the polynomial will be integers, so I just had to keep enough digits to round my final result to the nearest integer. I obtained:
$$x^6-64 x^5+2560 x^4-81920 x^3+ 1638400x^2- 16877216x+ 69508864.$$
In short, this sextic and $x^5-4x-1$ have the same splitting field.

It is interesting to think about the group theory problem: When can a group $G$ acts transitively and without kernel on two sets $A$ and $B$ of relatively prime sizes $a$ and $b$?
Lemma If $G$ is a group with such an action, then $G$ has no nontrivial normal $p$-subgroup
Proof Suppose that $Q$ is a $p$-subgroup of $G$. Then $Q$ is contained in some $p$-Sylow but, since all $p$-Sylows are conjugate and $Q$ is normal, we see that $Q$ is contained in every $p$-Sylow. Now, at least one of $a$ and $b$ is not divisible by $p$; say $a$ is. Then the stabilizer of every point of $A$ contains a $p$-Sylow of $G$, and hence contains $Q$, so $Q$ acts trivially on $A$. $\square$
Corollary  A nontrivial solvable group $G$ cannot have such actions.
Proof Suppose it did. Let $G=G^0 \supset G^1 \supset G^2 \supset \cdots$ be the derived series of $G$ and let $G^j$ be the last nontrivial term. Then $G^j$ is a canonical abelian subgroup of $G$. Let $p$ be a prime dividing $|G^j|$. Then the $p$-torsion of $G^j$ is a normal (and even canonical) $p$-subgroup of $G$. $\square$.
Thus, the first possibility is the alternating group $A_5$ and, indeed, this works:
$A_5$ acts on a $5$-element set, but also the usual action of the dihedral group of order $10$ on a $5$-element set lands in $A_5$, so $A_5$ acts on the $6$ element set $A_5/D_{10}$.
I was also able to construct such actions for $PGL_2(\mathbb{F}_p)$ and $PSL_2(\mathbb{F}_p)$ when $p \equiv 3 \bmod 4$: These act on the projective line, which has $p+1$ points. Let $\mathcal{J} = \{ J \in GL_2(\mathbb{F}_p) : J^2 = -\mathrm{Id} \}$, and let $\mathcal{K}$ be the quotient of $\mathcal{J}$ by $J \sim (-J)$. Then $PGL_2(\mathbb{F}_p)$ and $PSL_2(\mathbb{F}_p)$ act on $\mathcal{K}$ by conjugation, and I claim that $\mathcal{K}$ has $\tfrac{p(p-1)}{2}$ elements. Since $p \equiv 3 \bmod 4$, we have $GCD(p+1, \tfrac{p(p-1)}{2})=1$. To see that $\mathcal{J}$ has $p(p-1)$ elements, note that we can parametrize  it as $\left[ \begin{smallmatrix} x & \tfrac{-1-x^2}{y} \\ y & -x \end{smallmatrix} \right]$ for $y \neq 0$. (We are using that $p \equiv 3  \bmod 4$ to see that $x^2+1 \neq 0$, otherwise there would be additional solutions where $J$ is upper triangular with diagonal entries $\pm x$ for $x^2+1=0$.)
