Answer to Question #134315 in Abstract Algebra for Aayushi Deshmukh

Question

Show that polynomial x⁵=9x+3 Is not solvable by radicals over.

Answer 1

$$Solution$$

We need to show that the Galois group is not solvable.

Firstly, we note that $f$ is irreducible by Eisenstein’s criterion with $p= 3$. Therefore, the zeros off are all conjugate, so the Galois group acts transitively on them. Next we observe that $f(x)$ has 3 real roots and two complex roots. We can see that $f(−∞) =-∞,f(0) = -3,f(1) =−11$ , and $f(∞) =∞$ , so by the intermediate value theorem, we see there are at least 3 real zeros. On the other hand, the roots can’t all be real, since the sum of their squares is 0.

Alternatively, we use some basic calculus to see that $f^′(x) = 5x^4−9$ , so the only turning points for fare zeros of $x^4−\frac95$ , which has only 2 real solutions.Therefore, $f$ has exactly 3 real solutions and two complex ones.

Therefore its Galois group is a transitive subgroup of $S_5$ , which contains a transposition, so it must be the whole of $S_5$, which is not solvable, so $f$ is not solvable by radicals over Q.