It is a theorem that any finite separable extension is simple. We also know that any finite Galois extension is normal and separable. Then the conclusion is that each Galois extension must be simple. Is this reasoning correct?

Now it is easy to show that $f(x) = x^5-8x+2$ is irreducible over $\mathbb Q$ by Eisenstein. Then 5 must divide the Galois group $G$ which is a subgroup of $S_5$. That is all I can think of as of now.


The polynomial $x^5 - 8x+2$ is irreducible, and its derivative is $5x^4 - 8$ which has roots as the fourth roots of $1.6$. But none of these satisfy $x^4(x-8)+2$ because otherwise $x- 8 = 1.25$ so $x = 9.25$, contradiction. Therefore the minimal polynomial has distinct roots.

How many real roots does it have? Note that the derivative $5x^4 - 8$ has exactly two real roots, namely $\pm \sqrt[4]{1.6}$. Therefore, the poynomial $x^5 - 8x+2$ has at most three real roots, and therefore at least two complex roots.

However, we explicitly find intervals for two real roots, since if $f(x) = x^5 -8x+2$ then $f(0) = 2$ and $f(1) = -5$ and $f(2) = 14$ so there are roots lying in $(0,1)$ and $(1,2)$. However, this implies there cannot be more than three complex roots, and since the number of complex roots are even, this implies there are at most two complex roots.

In conclusion, there are exactly three real roots and two complex roots of $f(x)$.

Also, for any root $\alpha$ of $f(x)$, if $L$ denotes the splitting field of $f(x)$ then $[L:\mathbb Q]$ is a multiples of $[\mathbb Q(\alpha) : \mathbb Q]$ which is five, so the Galois group is contained in $S_5$. Now, we need to identify the kinds of elements which are in this group.

For example, complex conjugation is an automorphism of $L$ sending each real root to itself and the complex roots to each other. Thus the Galois group of $L/F$ contains a transposition i.e. an element of order $2$.

However, note that the Galois group also contains an element of order $5$ since it has order divisible by $5$.

From the above two facts, you can anyway see that the Galois group must have order a multiple of $10$. I leave you to show the following : if a subgroup of $S_5$ contains an element of order two and an element of order $5$ it is the whole group. The idea is to use these two elements to construct an element of order $3$. Once you do this, the order of the group must be divisible by $30$ and hence must be $60$ or $120$ (noting that there is no subgroup of order $30$, which can be shown by assuming there is one, and studying its intersection with $A_5$ as a subgroup of $A_5$), but then the unique group of order $60$ is $A_5$, the collection of all even permutations, while an element of order $2$ is not an even permutation. Therefore you can conclude.

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  • $\begingroup$ Thank you. To show it has three distinct roots, I guess it is easier by using the derivative test. $\endgroup$ – Amanda Mar 20 '19 at 4:48
  • $\begingroup$ (+1). You mean, though, that "the unique group of order $\color{red}{60}$ is $A_5$. $\endgroup$ – Travis Willse Mar 20 '19 at 5:08
  • $\begingroup$ Ah, yes,I will make a correction. $\endgroup$ – Teresa Lisbon Mar 20 '19 at 5:44
  • $\begingroup$ How do you conclude from that the derivative has only two roots the fact that the original function has at most three real roots? $\endgroup$ – Amanda Mar 20 '19 at 20:01
  • $\begingroup$ Because of Rolle's theorem which states that between any two real roots of the function there is a root of the derivative. In particular if a function has $n$ distinct real roots then it's derivative will have $n-1$ distinct real roots. But then the derivative has only two real roots which I have shown, so my claim follows $\endgroup$ – Teresa Lisbon Mar 21 '19 at 3:33

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