Simplest unsolvable quintic with one real root

Question

I am aware that $t^5-t-1$ is unsolvable, but the proof I have seen involves a theorem linking its Galois group with the Galois group of its reduction mod $p$. If I wish to have a simpler proof (that does not invoke that theorem), is there another quintic with only one real root that can be easily shown to be unsolvable?

For comparison, there is an elementary proof that $t^5-6t+3$ is unsolvable. In particular, it has $3$ real roots and so complex conjugation immediately furnishes a swap in the Galois group $G$. And $G$ acts transitively on the roots $R$, so letting $c{∈}R$ we have that $\{ \ (g,g(c)) : g{∈}G \ \}$ is a $k$-to-$1$ relation for some $k{∈}ℕ^+$, because for any $x,y{∈}R$ we have $h(x) = y$ for some $h{∈}G$ and so $\{ \ g : g{∈}G ∧ g(c) = x \ \}$ $= \{ \ g : g{∈}G ∧ (hg)(c) = y \ \}$ $= \{ \ h^{-1}g' : g'{∈}G ∧ g'(c) = y \ \}$, which is in obvious bijection with $= \{ \ g : g{∈}G ∧ g(c) = y \ \}$, and hence $\#(G) = k·\#(R)$ since for any $x{∈}R$ we have $i(c) = x$ for some $i{∈}G$. Thus $5 = \#(R) \mid \#(G)$. Since $5$ is prime, $G$ has a $5$-cycle (on $R$) by Cauchy's theorem. The swap and $5$-cycle imply that $G$ is isomorphic to $S_5$.

Is there any explicit unsolvable quintic with equally simple proof, and how small coefficients can you get? I know this is a bit subjective, but I'm simply asking for examples that don't need algebraic number theory.

Answer 1

Not a solution, just fleshing out my comments explaining, why I think this is difficult.

---

In the case of an irreducible quintic over $\Bbb{Q}$ we expect the Galois group $G$ to be $S_5$, but to prove it we need to rule out the possibilities that $G$ might be a subgroup of either $A_5$ or $F_{20}\simeq C_5\rtimes C_4$. The discriminant is the tailormade tool to cover the first alternative. If we are in the latter case, the sextic resolvent has a rational root, so that is the tool. As that was deemed unsatisfactory, this is a dead end.

---

For a different approach I considered the possibility that may be two quintics, $f(x)$ and $g(x)$ from $\Bbb{Q}[x]$, could share the same splitting field $L\subseteq \Bbb{C}$, hence also the Galois group, but still $f$ would have three real roots and $g$ only one. In such a case the simple argument using $f$ shows that $Gal(L/\Bbb{Q})\simeq S_5$, and $g$ would be an example fitting OP's criteria.

Unfortunately this is impossible. Basically because $G=S_5$ has a unique conjugacy class of subgroups of index five, namely the point stabilizers of $G$ as permutations of the set of roots $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5\}$ of $f$. By Galois correspondence the fields $K_i:=\Bbb{Q}(\alpha_i)$ are then the only intermediate fields of degree five over $\Bbb{Q}$.

If $\beta\in L$ is a real root of $g(x)$, then $[\Bbb{Q}(\beta):\Bbb{Q}]=5$, and our earlier observation implies that $$\Bbb{Q}(\beta)=K_j$$ for some $j\in\{1,2,3,4,5\}$ such that $\alpha_j\in L\cap \Bbb{R}$. It follows that $$\beta=r(\alpha_j)$$ for some at most quartic polynomial $r(x)\in\Bbb{Q}[x]$. Because all the automorphisms in $G$ fix the rationals, the five Galois conjugates of $\beta$ are $$ \beta_i=r(\alpha_i),\ i=1,2,3,4,5. $$ Alas, three of those are real, meaning that the polynomial $g(x)$ necessarily also has three real roots.

Answer 2

Another simple example is $f = x^5 + 2x + 2$ over $\Bbb Q$.
By Eisenstein, $f$ is irreducible over $\Bbb Q$. It has exactly one real root, namely $x \approx -0.817471019001$. Furthermore it has the Galois group $S_5$, which is not solvable.

The proof is as easy as it can be for this case. One can avoid the modulo $p$ reduction and use the resolvent instead, see the reference below.

Of course the case with exactly $2$ non-real roots is still the most basic one. It only requires Cauchy's theorem, and that the symmetric group $S_n$ can be generated by a transposition $(12)$ and an $n$-cycle $(123\cdots n)$.

Reference and proofs:

Is this Galois Group is isomorphic to $S_5$