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How to prove that $f(x)=x^7+7x^2+2$is irreducible over $\mathbb{Q}$?

I have tried to use Eisenstein's Criterion. I pick $-2$,since $$(-2)^7+7(-2)^2+2\equiv 0~~ (\mod 7)$$ Let $y=x-2$,then

$$f(x)=g(y)=-98 + 420 x - 665 x^2 + 560 x^3 - 280 x^4 + 84 x^5 - 14 x^6 + x^7.$$

We see that $7\mid a_i$,$i=0,1,\cdots,6$,$7\nmid a_7=1$,but unfortunately $7^2\mid a_0=-98$,so $g(y)$ isn't an Eisenstein polynomial.

mbfkk
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  • Perhaps try $y=x+2$. Maybe it can work. – Anurag A Jun 19 '20 at 09:08
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    Be careful ending sentences with a number and then an exclamation mark. You do not mean that the constant term is $98!$, right? – Arthur Jun 19 '20 at 09:10
  • @Anurag A It turns out $158 + 476 x + 679 x^2 + 560 x^3 + 280 x^4 + 84 x^5 + 14 x^6 + x^7$ – mbfkk Jun 19 '20 at 09:12
  • @mbfkk which isn't an Eisenstein polynomial. – Angina Seng Jun 19 '20 at 09:13
  • @ Arthur Thanks for your suggestion. – mbfkk Jun 19 '20 at 09:20
  • There are clearly no linear factors, and I think (needs checking) that mod $3$ there are no quadratic factors either (there are only three quadratic irreducibles to check). So if it's not irreducible it's a product of a cubic irreducible and a degree 4 irreducible. I would check how it factorises mod 2 (and also perhaps mod 3) to see if this is plausible . – ancient mathematician Jun 19 '20 at 09:30
  • If my calculations are correct, you can Hensel lift the solution $x\equiv26\pmod{7^3}$ to a $7$-adic solution of $x^7+7x^2+2=0$, and so finding an Eisenstein criterion argument is hopeless. Maybe you can prove it's irreducible modulo some other prime? – Angina Seng Jun 19 '20 at 09:33
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    In fact , the polynomial is irreducible over $\mathbb Z_3$[$X$] – Peter Jun 19 '20 at 09:37
  • Excellent @Peter: did you do it by hand or machine? – ancient mathematician Jun 19 '20 at 09:54
  • To be honest , with PARI/GP , but there are not too many polynomials that have to be checked. – Peter Jun 19 '20 at 11:06

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