7
$\begingroup$

I was illustrating the theorem on solvability by radicals through some examples of degree $5$ polynomials. One I chose was $x^5-5x+10$. I was (perhaps wrongly) going to prove that the Galos group is $S_5$, with expectation that it has exactly three real roots.

However, it has exactly one real root (which I saw by looking graph of this polynomial in $\mathbb{R}^2$ using some online software). Then I get stucked, and couldn't completely find the Galois group of this polynomial.

How do we proceed to determine the Galois group of this polynomial over $\mathbb{Q}$?

$\endgroup$
5
$\begingroup$

Observing discriminant of polynomial is very helpful to find its Galois group. It is known that if $D$ is discriminant of polynomial $f(x)$, then Galois group $G=G_{f}$ of $f(x)$ is contained in $A_{5}$ iff $D\in \mathbb{Q}^{2}$. (This holds for general fields with $char\neq 2$, not only $\mathbb{Q}$. Proof is not hard and you can find proof in "Abstract Algebra", Chap 14, Proposition 33 of Dummit-Foote.) As Wolfram says, its discriminant is 30450000 which is not a square. So $G$ is not contained in $A_{5}$.

Also, by Eisenstein criterion with $p=5$, $f(x)$ is irreducible and $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$ for any $f(\alpha)=0$. So order of $|G|$ has to be divided by $5$.

Then if you see here, the only candidates are $S_{5}$ and general Affine group $GA(1,5)$. If we reduce the polynomial modulo $3$, then $\overline{f}(x)=x^{5}+x+1=(x-1)^{2}(x^{3}-x^{2}+1)$ in $\mathbb{F}_{3}[x]$. Since $x^{3}-x^{2}+1$ is irreducible over $\mathbb{F}_{3}$, order of $G_{\overline{f}}$ has to be divided by $3$. Since $G_{\overline{f}}\leq G_{f}$ (this is non-trivial result, and see here for the Tate's proof), $|G_{f}|$ is also divided by $3$ and contains a $3$-cycle. We can show that $5$-cycle and $3$-cycle in $S_{5}$ generates $A_{5}$, so the answer is $G_{f}=S_{5}$.

I think it will be possible to find a prime $p$ s.t. $\overline{f}(x)\in\mathbb{F}_{p}[x]$ has an irreducible factor of degree $2$, and then you don't have to calculate discriminant of $f$.

$\endgroup$
  • $\begingroup$ Useful points. Just to make sure I will remark that finding a prime such tha $\overline{f}$ has an irreducible quadratic factor does not quite suffice. If $\overline{f}$ splits as a product of a linear factor and two distinct quadratics that only guarantees the existence of a permutation of cycle type $(2,2,1)$ in $G_f$. But such a permutation is still even. You need cycle types $(2,1,1,1)$ or $(4,1)$ or $(3,2)$ to be sure of finding odd permutations. Also, if $\overline{f}$ has a repeated factor, then you cannot use that prime. For example, your modulo $3$ factorization doesn't help... $\endgroup$ – Jyrki Lahtonen Jul 4 '16 at 9:06
  • $\begingroup$ (cont'd) because the factor $(x-1)$ is repeated. Theorem A in your link does require that $\overline{f}$ should not have roots of multiplicity $>1$. So, modulo $13$ factorization gives a product of two disjoint 2-cycles, modulo $17$ gives a 3-cycle, and modulo $41$ a 4-cycle. At that point we can conclude that $G_f=S_5$. $\endgroup$ – Jyrki Lahtonen Jul 4 '16 at 9:11
  • $\begingroup$ @JyrkiLahtonen Thank you for your comment! Actually I find the theorem $G_{\overline{f}}\leq G_{f}$ right before answer the question, so I missed about some important details. I will modify this answer later. $\endgroup$ – Seewoo Lee Jul 6 '16 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.