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My question regards a very specific part of an example in Chapter 14.8 of Dummit and Foote (page 641, image shown below) in which the authors are computing the Galois group of $f(x) = x^5 - x - 1$ over $\mathbb Q$. They want to prove $f(x)$ is irreducible modulo 3 in order to show the Galois group over $\mathbb Q$ contains a $5$-cycle. To rule out a factorization consisting of an irreducible quadratic and a cubic, they provide the following justification:

[If $f(x)$ had an irreducible quadratic factor modulo 3 then] it would have a factor in common with $x^9 - x$ (which is the product of all irreducible polynomials of degrees 1 and 2 over $\mathbb F_3$), hence a factor in common with either $x^4 - 1$ or $x^4 + 1$, hence a factor in common with either $x^5 - x$ or $x^5 + x$, hence a factor in common with either $-1$ or $2x + 1$.

I understand this reasoning up until the final phrase. Why does sharing a factor with $x^5 - x$ or $x^5 + x$ modulo $3$ imply sharing a factor with $-1$ or $2x - 1$?

A more general question regarding this example was asked here, but it was never answered and is less specific.

A screenshot of the example in question

LéKitty
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    Answer: because $f(x) = x^5 - x - 1$. – Anne Bauval Jan 05 '24 at 07:23
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    Why is $\gcd(a,b)=\gcd(a-kb,b)$? – Jyrki Lahtonen Jan 05 '24 at 07:41
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    $,g\mid x^4!-!1,\ x^5!-!x!-!1,\Rightarrow, g\mid x(x^4!-!1)-(x^5!-!x!-!1) = 1,,$ i.e. $,g\mid f,f'\Rightarrow g\mid f'\bmod f,,$ the reduction step of the euclidean algorithm. Generally $,g\mid f,f'\iff g\mid\gcd(f,f').,$ It's easier to work $!\bmod g,,$ i.e. in $,\Bbb F_9,$ as in the linked dupe. $\ \ $ – Bill Dubuque Jan 05 '24 at 08:36
  • Ahh, one should work mod $g$. Makes complete sense. Thanks for clearing it up and for answering the question a second time @Bill Dubuque! – LéKitty Jan 05 '24 at 22:59

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