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I'm asked to show that $27X^3 - 13X^2 + 180 \in \mathbb{Q}[x]$ is irreducible in $\mathbb{F}_{13}[X], \mathbb{Z}[X]$ and $\mathbb{Q}[X]$. I've managed to proof the first two. In $\mathbb{F}_{13}[X]$ I checked that it has no roots and therefore is irreducible. This also shows that is irreducible in $\mathbb{Z}[X]$. Normally I would now apply Gauss's lemma, which tells me that it is in fact irreducible in $\mathbb{Q}[X]$. The problem is, my text book says I can only use this result if the polynomial is monic, which is not the case here. Altough Wikipedia says a polynomial doesn't have to be monic, I'm only allowed to use my text book as reference.

I tried a general factorization: $$ 27X^3 - 13X^2 + 180 = (27X-a)\cdot (X^2 + bX + c), $$ but if I work this out, I get the original equation. Another way is to check every rational root, using the fact that if $\frac{b}{c}$ is a root (with $b$ and $c$ coprime), then $b$ divides $180$ and $c$ divides $27$. But these are just too many options to check by hand.

Is there any way I can show it is irreducible in $\mathbb{Q}[X]$ without the need to proof a generalisation of Gauss's lemma, or is that the only possibility?

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  • $\begingroup$ Does proving Gauss' lemma count? $\endgroup$ Jan 20 '15 at 18:43
  • $\begingroup$ Rational root theorem. $\endgroup$
    – Git Gud
    Jan 20 '15 at 18:43
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    $\begingroup$ @GitGud Did you even read the post? $\endgroup$ Jan 20 '15 at 18:44
  • $\begingroup$ You're right, forgot to add in it my post. I did do it right when trying to factorize, so it not the solution unfortunately. $\endgroup$
    – Pierre
    Jan 20 '15 at 18:44
  • $\begingroup$ @user2345215 No, I had not. $\endgroup$
    – Git Gud
    Jan 20 '15 at 18:48
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Hint $\ $ If it were reducible over $\Bbb Q$ it would have a rational root, which, by the Rational Root Test, has denominator $3^n$ dividing $27$, which would yield a root mod $13$, since $3^n$ is invertible mod $13$.

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  • $\begingroup$ How does modulo $13$ work if the root is not an integer? $\endgroup$
    – Pierre
    Jan 20 '15 at 18:59
  • $\begingroup$ If $\,13\nmid b$ then $a/b \equiv ab^{-1}\pmod{13}.\,$ As long as you restrict to fractions whose denominator is coprime to the modulus, then all the usual fractional arithmetic works fine. In particular it follows that $\,f(a/b) = 0\,$ in $\,\Bbb Q\,\Rightarrow\, f(ab^{-1})\equiv 0\pmod{13},\,$ so the root $\,a/b\,$ survives in $\,\Bbb F_{13}.\ $ $\endgroup$ Jan 20 '15 at 19:02
  • $\begingroup$ @Just You can find some examples of fractional modular arithmetic in these posts. $\endgroup$ Jan 20 '15 at 19:11

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