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Decide with a proof if $f(x)=x^4+x+1$ is irreducible in $\Bbb{Q}[x]$.

I was thinking of using DeMoivre's Theorem but I'm not sure how.

Thanks!

user26857
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jadealeska
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  • have you checked if there are any rational roots? –  Dec 05 '13 at 16:32
  • I am not very sure how do you feel Demoivre is useful in this case.. –  Dec 05 '13 at 16:32
  • $f(5)=631$ is a prime so f(x) is irreducible. See http://math.stackexchange.com/questions/568094/showing-the-polynomial-x4-x3-4x-1-is-irreducible-in-mathbbqx/568308#568308 – miracle173 Dec 05 '13 at 17:00

3 Answers3

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Much easier than that! Notice that $x^4+x+1 \in \mathbb{Z}[x]$ is primitive. So $x^4+x+1 \in \mathbb{Q}[x]$ is irreducible if and only if $x^4+x+1 \in \mathbb{Z}[x]$ is irreducible.

We can show that it is irreducible by looking at the $\mod 2$ reduction and showing that for $0,1$, $f(x)=x^4+x+1 \mod 2$ is not $0$. That shows that $f(x)$ has no linear factors.

Now we show it has no quadratic factors. Suppose it was reducible. Then it has to be the product of $2$ quadratic factors. But there are only $3$ reducible polynomials in $\mathbb{Z}_2[x]$ so only $1$ reducible one. That is $x^2+x+1$. But the square of this polynomial in $\mathbb{Z}_2[x]$ is not $f(x)$. So $f(x) \in \mathbb{Z}_2[x]$ is not reducible and is then irreducible in $\mathbb{Z}[x]$ and hence also $\mathbb{Q}[x]$.

mathematics2x2life
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  • I agree with you that it is only irreducible iff it is not irreducible in Z[x]. But when you use mod 2 to justify, isn't that being too specific, when I need to prove it in general? – jadealeska Dec 05 '13 at 16:45
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    @Jade No, if it was reducible then changing things through a $\mod 2$ does not change the fact that it would be reducible. In fact, this is a tool we use often to do these sort of things. Since it was irreducible in the $\mod 2$ case, it couldn't have been reducible in the original case. – mathematics2x2life Dec 05 '13 at 16:51
  • Can you please explain how there are only 3 reducible polynomials in $\mathbb{Z}/2\mathbb{Z}[x]$? The way I see it, $x^n+1$ is reducible for all integers $n>2$? That alone creates infinitely many reducible polynomials. – D777 Nov 16 '18 at 20:46
  • @D777 Infinitely many reducibles does not mean you need infinitely many irreducibles. You can just take products of larger powers of the irreducibles. For example, $x$ and $x+1$ are irreducible. From this, I can make infinitely many reducibles, e.g. $f_{n,m}(x)= x^n(x+1)^m$. I was meaning 3 irreducible polynomials of degree at most 3. – mathematics2x2life Nov 18 '18 at 00:42
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Let $f(x)=x^4+x+1$. Note that $f(x)$ has no rational root. Hence there will no linear factor of $f(x)$ and hence there will be no cubic factors also. So if $f(x)$ is reducible then there will only quadratic factor only. Clearly this will be of the form $(x^2+ax+1)(x^2+bx+1)$ which is as $x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1 $. Equating coefficients of similar terms we then get $a+b=0, ab=-2, a+b=1$. In otherwords, $0=1$, contradiction. Hence the proof

KON3
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  • I have a couple questions: Do I have to prove there are no rational roots or just state it? And how do you know a+b=0, ab=-2, a+b=1? I followed your thoughts up until there. – jadealeska Dec 05 '13 at 16:43
  • @Jade You'd really want to show this. But you can use the Rational Roots Theorem, http://en.wikipedia.org/wiki/Rational_root_theorem . We know $a+b=0$ because there is no $x^3$ term in $x^4+x+1$ and $ab=-2$ so $ab+2=0$ because there is no $x^2$ term in $x^4+x+1$. $a+b=1$ because the $x$ term in $x^4+x+1$ is $1$. Solving that system is inconsistent. Because $a+b=0$ and $a+b=1$ implying that $0=1$, contradiction. – mathematics2x2life Dec 05 '13 at 16:48
  • you should do this $\pmod 2$. Otherwise you have to investigate $(x^2+ax \pm 1)(x^2+bx \pm 1)$ – miracle173 Dec 05 '13 at 16:50
  • @Jade Indeed, miracle173 is correct. However, Anjan did give the correct approach for his case. See my answer for doing it $\mod 2$. – mathematics2x2life Dec 05 '13 at 16:52
  • Actuallu my intension was to proceed in the way miracle173 did. But looking at the question, I thought I should provide you such technique which you can apply easily (lengthy tough!) @Jade – KON3 Dec 05 '13 at 16:52
  • "Clearly" is only a guess. One can not assume that for two reasons: in $\mathbb Q$ we have $cd=1$ for numbers other than $c=d=1$, and whether for some reasons (which have to be explained!) we may assume $c,d\in\mathbb Z$, then we also have the case $c=d=-1$. – user26857 Sep 16 '20 at 05:59
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The $f(2)=19$ is a prime, and therefore $f(x)$ is irreducible in $\mathbb{Z}[x]$ by Cohn's irreducibility criterion, and especially it is irreducible in $\mathbb{Q}[x]$.

Sil
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