Is this 7-th degree polynomial solvable?

Question

Unluckily I don't have a solid background in Galois theory; I ran into this polynomial equation: \begin{equation} x^7-\frac{1}{2}x^6-\frac{3}{2}=0 \end{equation} I know it has only one positive real root, so the other $6$ must be complex conjugates. I also know that every complex root has nonzero real part. My question is: is this equation solvable, and how?

Answer 1

Since you mention Galois theory in the question, I presume you are about whether $$ f(x) = x^7 - \frac{1}{2} x^6 - \frac{3}{2} $$ is solvable in radicals.

If we factor the polynomial modulo various primes (not dividing the discriminant), and look at the degrees of the resulting irreducible factors we determine which cycle types must appear in the Galois group over $ \mathbb{Q} $. This is a theorem of Dedekind. More formally

Let $ f(x) \in \mathbb{Q}[x] $ be an irreducible polynomial of degree $ n = \deg(f) $. Suppose $ p $ is a prime not dividing $ \operatorname{disc}(f) $, and that modulo $ p $ the polynomial $ f(x) $ factors as $$ f(x) \equiv h_1(x) \cdots h_k(x) \pmod{p} $$ where the $ h_i(x) $ are monic irreducibles modulo $ p $, with degrees $ d_i = \deg(h_i) $. Then the is an element in the Galois group $ G < S_n $ of $ f $ over $ \mathbb{Q} $, which has cycle type $ (d_1 \, \cdots \, d_k) $.

Conrad has a short note about this, and using it to recognise Galois groups $ S_n $ and $ A_n $.

For this example, if we reduce modulo $ p = 1019 $, the polynomial $ f(x) = x^7 - \frac{1}{2} x^6 - \frac{3}{2} $ factors as $$ f(x) \equiv (53 + x) (348 + x) (542 + x) (961 + x) (1014 + x) (20 + 648 x + x^2) \pmod{1019} \, . $$ This means the Galois group of $ f(x) $ over $ \mathbb{Q} $ must contain a cycle of type $ (1,1,1,1,1,2) $, i.e. a transposition. From user21820's answer, we also know that it must contain a 7-cycle (or get this by factoring modulo 5).

[Edit: Actually, rather than reducing mod $ p = 1019 $, a better approach is to reduce modulo $ p = 23 $. We find that $$ f(x) \equiv (3 + x) (18 + x) (12 + 22 x + x^2) (14 + 20 x + 14 x^2 + x^3) \pmod{23} $$ This shows that the Galois group of $ f(x) $ over $ \mathbb{Q} $ contains a cycle of type $ (1,1,2,3) $. But taking the third power of this gives a cycle of type $ (1,1,1,1,1,2) $, so we can continue as we would above.]

By Theorem 2.1 in Conrad's note, a 7-cycle and a transposition generate all of $ S_7 $. So we conclude the Galois group of $ f(x) $ over $ \mathbb{Q} $ is $ S_7 $. And since $ S_7 $ is not a solvable group, the polynomial $ f(x) $ is not solvable in radicals.

Answer 2

Let $p$ be the polynomial $( x \mapsto 2 x^7 - x^6 - 3 )$. First check that $p$ is irreducible. Note that $p$ is irreducible iff $q = ( y \mapsto p(y+3) )$ is irreducible, and $q = 2 y (y+3)^6 - 3$ is irreducible by Eisenstein's criterion applied to prime $3$.

Then we have to figure out the automorphism group $G$ that permutes the roots of the polynomial while fixing the rationals. We know that it acts on the roots transitively since $p$ is irreducible, so $G$ has order divisible by $7$ and hence by Sylow's theorem $G$ has a subgroup of size $7$, which must have be generated by a $7$-cycle $a = (1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7)$. We label the roots from $1$ to $7$ appropriately. Also, complex conjugation is an automorphism in $G$ and by what you've noted it does three 2-cycles on the roots.

If $b = (1 \ 2)(3 \ 4)(5 \ 6)$, we have $c = a^{-2}ba^2b = (7 \ 6 \ 5)$, and using $a,c$ we can obtain:

• $(7 \ 6 \ 4) = (7 \ 6 \ 5) \ (4 \ 5 \ 6)$
• $(7 \ 6 \ 3) = (7 \ 6 \ 5) \ (5 \ 4 \ 3) \ (4 \ 5 \ 6)$
• $(7 \ 6 \ 2)$ and $(7 \ 6 \ 1)$ by symmetry

And from there we can construct any 3-cycle and hence the whole alternating group $A_7$, which is not solvable.

Now there are so many other cases for $b$, but I think all will turn out to be similar, and that $a,b$ will always generate $A_7$. If so, then we are done, and the roots of $p$ cannot be expressed using radicals over the rationals.

I really have not much idea how to tackle all the different cases at one go.