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Is there any way to determine the Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$ not using the discriminant? Thanks!

Ergin Süer
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    Are you unsatisfied with the discriminant proof or you don't want to go through it? I mean, determining a Galois group of a polynomial of degree $5$ is hard enough as it is... (P.S. Your polynomial has no rational roots.) – Patrick Da Silva Aug 30 '14 at 14:29
  • P.S. 2 : Wolfram Alpha says the discriminant is $7494933 = 3 \times 991 \times 2521$... I haven't done this kind of Galois group computation before, but I have a hunch that this big number won't mean much to any of us. – Patrick Da Silva Aug 30 '14 at 14:32
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    Yes. This was a question in an exam. It was not allowed to use calculator and I don't know how to determine whether the discriminant is square or not. – Ergin Süer Aug 30 '14 at 14:40
  • If this was in an exam, I think you just cried... at least I would've. – Patrick Da Silva Aug 30 '14 at 14:55

4 Answers4

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Following Nishant's suggestion the calculation might go as follows.

Modulo $p=2$ our polynomial $f(x)=x^5-2x+7$ factors as $$ f(x)\equiv x^5+1=(x+1)(x^4-x^3+x^2-x+1). $$ This tells us many things. For example the only possible non-trivial factorization over $\Bbb{Q}$ must have a linear factor. But the rational root test quickly tells us that those don't exist. Thus $p(x)$ is irreducible in $\Bbb{Q}[x]$, and the Galois group $G\le S_5$ thus acts transitively on the set of five roots of $f(x)$.

By Dedekind's theorem we also see that $G$ contains a 4-cycle, so we can tell that $G$ is doubly transitive, and that $|G|\ge 5\cdot4=20$. Note that the modulo $2$ factors above were distinct. This was crucial, as otherwise we could not have made this deduction. This also implies that $p=2$ does not divide the discriminant.

What other primes to try? Somehow $p=7$ suggests itself. We immediately get the factorization $$ f(x)=x^5-2x=x(x^4-2). $$ Here we get a bit of luck as modulo $7$ we have $2\equiv9=3^2$. Therefore we can factor further: $$ f(x)=x(x^4-9)=x(x^2-3)(x^2+3)=x(x^2-3)(x^2-4)=x(x^2-3)(x-2)(x+2). $$ It is, indeed, our lucky day, as $3$ is a quadratic non-residue modulo $7$, so we have a single irreducible quadratic factor and three distinct linear factors. Dedekind's theorem thus tells us that $G$ contains a 2-cycle.

It is a known fact that when a subgroup of $S_5$ contains a 5-cycle and a 2-cycle, it must be all of $S_5$. Alternatively we could deduce that $G$ must have at least $40$ elements, and, because it contains a 2-cycle, is not a subgroup of $A_5$.

Anyway: $G=S_5$ is the only remaining option.

Community
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Jyrki Lahtonen
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    For example one might take advantage of the known list of transitive subgroups of $S_5$. They have orders $5$ (cyclic), $10$ (dihedral), $20$ ($C_5\rtimes C_4$), $60$ ($A_5$) and $120$ ($S_5$). Out of these $S_5$ is the only one that contains a $2$-cycle. – Jyrki Lahtonen Aug 30 '14 at 17:14
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Well, you can factor it modulo a bunch of primes to see what cycle types of elements of $S_5$ are in the Galois group, but of course, you can't use primes that divide the discriminant.

Nishant
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for $f(x)=x^5-2x+7$ what is $f'(x)$

how many real roots do you think this would have??

  • What has the derivative got to do with this? – Ali Caglayan Aug 30 '14 at 14:51
  • it does tell you how many real roots do you have for this.... which would help to find galois group... –  Aug 30 '14 at 14:52
  • Please could you expand on how the number of real roots can help determine the Galois group? – Ali Caglayan Aug 30 '14 at 14:54
  • @Alizter : suppose we can see that it has exactly three real roots then it follows that galois group is $S_5$ but in this case you ave only one real root.. This procedure (unchanged) may not be of any use.. let me try for some other way... –  Aug 30 '14 at 15:03
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    I would be interested to see if you can flesh this argument out in your answer :) – Ali Caglayan Aug 30 '14 at 15:04
  • @PraphullaKoushik I don't follow. $f'$ has two real roots. – zibadawa timmy Aug 30 '14 at 15:22
  • A good idea, but Mathematica suggests that it only has a single real root $\in(-2,-1)$. The local minimum at $\root4\of{2/5}$ is positive, so that's all. – Jyrki Lahtonen Aug 30 '14 at 17:05
  • @zibadawatimmy : Yes... that is what i said, as $f'$ has two real roots i can not decide anything... Though this does not answer this particular question i thought it is kind of relevant so i did not delete it... –  Aug 30 '14 at 18:49
  • @JyrkiLahtonen : Yes Yes... with a wrong calculation i thought this would give the answer... i was wrong.. –  Aug 30 '14 at 18:49
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Using Dummit & Foote's Abstract Algebra book Page 639 Exercise 21, I can tell you that the Galois group is not solvable. This because they know that the Galois group of a quintic in the form $$ f(x) = x^5 + Ax + B $$ is solvable over $\mathbb Q$ (but do not prove it there, they just give a reference ; note that your polynomial has this form) if and only if the following associated polynomial of degree $6$ has a rational root : $$ g(x) = x^6 + 8Ax^5 + 40A^2 x^4 + 160 A^3 x^3 + 400 A^4 x^2 + (512 A^5 - 3125 B^4)x - 9375 AB^4 + 256 A^6. $$ So in this case you can use the rational root theorem to compute the solvability of your polynomial. I did this for you ; in our case $-9375 AB^4 + 256 A^6 = 45035134 = 2 \times 821 \times 27427$ a product of three primes, so there were only $16$ evaluations of $g(x)$ to do, which I did with Wolfram Alpha ; your polynomial is not solvable over $\mathbb Q$.

Hope that helps,

Patrick Da Silva
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