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In my math course, it is written that since over $\mathbb{F}_5[X]$, $P$ has no root in $\mathbb{F}_{25}$, it is irreducible over $\mathbb{Q}[X]$ by Gauss Lemma.

I don't really understand…

If we prove that $P$ is irreducible over $\mathbb{Z}[X]$, then Gauss proves it is over $\mathbb{Q}[X]$, but I don't understand what $\mathbb{F}_{25}$ is doing there…

The exact question was :" Find a prime $p$ such that the reduction of $P(X)$ over $\mathbb{F}_p[X]$ has no root in $\mathbb{F}_{p^2}$, then show that $P$ is irreducible"

Jack D'Aurizio
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Labo
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2 Answers2

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$x^5-6x+3$ is irreducible over $\mathbb{F}_5$ because it is an Artin–Schreier polynomial, of the form $x^p-x+\alpha$ for $p=5$: since such polynomial is irreducible over $\mathbb{F}_5$, it is irreducible over $\mathbb{Q}$, too.

$x^5-x+3$ has no root in $\mathbb{F}_5$ or $\mathbb{F}_{25}$ because the $\gcd$ between $x^{5^2}-x$ and $x^5-x+3$ is $1$: by assuming $x^5=x-3$ it follows that $x^{5^2}=(x-3)^{5}=x^5-3=x-6$ hence $x^{5^2}-x = -1$.
It follows that $x^5-x+3$ factors over $\mathbb{F}_5$ as the product of irreducible polynomials with degree $\geq 3$ and the conclusion is the same as before: $x^5-x+3$ is irreducible over $\mathbb{F}_5$, hence it is irreducible over $\mathbb{Q}$.

Jack D'Aurizio
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  • Sure, but I don't think we are supposed to know that… The previous question was: "Find a prime $p$ such that the reduction of $P(X)$ over $F_p[X]$ has no root in $F_{p^2}$" – Labo Mar 23 '17 at 17:03
  • @Labo: the splitting field of $x^5-x+3$ over $\mathbb{F}5$ is $\mathbb{F}{5^5}$: you may easily prove that (that is the same as proving the mentioned lemma: if someone does not teach you something, it does not imply you cannot learn it on your own) and deduce that such polynomial has no roots in $\mathbb{F}_{5^2}$. – Jack D'Aurizio Mar 23 '17 at 17:05
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    @Labo: anyway, I have modified my answer including the approach it was probably meant by the problem poser. – Jack D'Aurizio Mar 23 '17 at 17:15
  • Thanks! But why "It follows that $x^5-x+3$ factors over $\mathbb{F}_5$ as the product of irreducible polynomials with degree $\geq 3$"? Does the 3 come from the 2 of $p^2$? – Labo Mar 23 '17 at 17:20
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    @Labo: there cannot be irreducible factors with degree $1$ or $2$ because $x^5-x+3$ has no roots in $\mathbb{F}5$ or $\mathbb{F}{5^2}$, hence the degree of every irreducible factor is $\geq 3$, i.e. there is a single irreducible factor with degree $5$. – Jack D'Aurizio Mar 23 '17 at 17:27
  • With a view of making the site as self-contained as possible I will comment that we have collected many proofs for the irreducibility of Artin-Schreier polynomials in the case of a prime field. A nice observation that it applies here. – Jyrki Lahtonen Mar 24 '17 at 06:21
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So Gauss' lemma states that irreducibility over $\mathbb{Z}$ implies irreducibility over $\mathbb{Q}$; so it suffices to show that $f = x^5 -6x + 3$ is irreducible over $\mathbb{Z}$. Now irreducibility in $\mathbb{F}_{p}$ (provided that $p$ does not divide the leading coefficient of $f$) implies irreducibility over $\mathbb{Z}$. So if $f$ was reducible, what are the possible factorizations? Well since $\mathbb{F}_{5}[x]$ admits unique factorization, there are the following options:

  1. $f$ is the product of 5 linear factors.
  2. $f$ is the product of 3 linear factors and an irreducible quadratic
  3. $f$ is the product of 2 linear factors and an irreducible cubic.
  4. $f$ is the product of a single linear factor and an irreducible quartic.
  5. $f$ is the product of a single linear factor and two irreducible quadratics.
  6. $f$ is the product of a single irreducible quadratic and a single irreducible cubic.

Checking that $f$ has no root in $\mathbb{F}_{5}$ rules out options 1-5. Now since $\mathbb{F}_{25}$ is the splitting field extension for all quadratics over $\mathbb{F}_{5}$, if $f$ possessed an irreducible quadratic in its factorization then $f$ would have a root in $\mathbb{F}_{25}$.

Oiler
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  • Ok, I understand better now, but why "$\mathbb{F}{25}$ is the splitting field extension for all quadratics over $\mathbb{F}{5}$"? – Labo Mar 23 '17 at 17:41
  • Ahah I think I understood. If there is an irreductible quadradic, then its splitting field has $5^2=25$ elements, hence is $\mathbb{F}_{25}$! – Labo Mar 23 '17 at 17:47
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    Looks like you got it. – Oiler Mar 23 '17 at 18:04