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Question at hand is:

Is $x^4+3x^3-9x^2+7x+27$ irreducible in $\Bbb Q$ and/or $\Bbb Z$.

This is for an exam, reasoning is trivial, but no calculators in hand. Clearly, if there is a rational root, they are integers by Rational Root theorem and since $f$ is monic.

I am aware of

  1. Rational root theorem, which narrows down the options to $\pm1,\pm3,\pm9,\pm27$, and clearly, no roots.

  2. Eisenstein's Irreducibility Criteria, not helping here, thanks to $x$'s coefficient $7$

  3. Cohn's Irreducibility test: $12197$ is a prime, too large a number to prove that its a prime by hand.

  4. Descartes Rule of signs: at most 2 (or 0) positive/negative roots. Close enough.

None of which are helping me in any way since I can't use a calculator.

These are the solutions I tried:

  1. Alpha says all roots are complex. Made me search if there's some way to determine if all roots are complex, reaching nowhere.

  2. Check if there are any easy prime generation functions like Euler's, and if lucky 12197 falls in that list, the best I got is Euler's, $n^2+n+41, 1\le n<40$, and biggest such is $1601$, not helping.

Are there any better ways to determine if this polynomial is irreducible over $\Bbb Q$, without using calculators?

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    $\begingroup$ 12197 might be "large" but its square root is around 110, so you needn't check that many numbers to see if it is prime. $\endgroup$
    – Pedro
    Commented Dec 18, 2015 at 14:05
  • $\begingroup$ @PedroTamaroff, but without calculators, It's still a Himalayan task, isn't it? $\endgroup$ Commented Dec 18, 2015 at 14:06
  • $\begingroup$ If $x$ is a root in $\mathbb{Z}$, then $x + 3\mathbb{Z}$ is a root in $\mathbb{F}_3$. $\endgroup$
    – nombre
    Commented Dec 18, 2015 at 14:14
  • $\begingroup$ @PedroTamaroff, sorry for late ping, realised that Cohn's test is true only for $a_n\le9$,here $a_0=27$. Are there any other facts/theorems which are relevant which makes it right in this case? $\endgroup$ Commented Jun 18, 2016 at 6:41

5 Answers 5

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If you take the polynomial modulo 2 you get $x^4+x^3+x^2+x+1$. It doesn't have a root, so if it is decomposable, it is a product of two irreducible polynomials of degree 2. The only polynomial of degree 2 which doesn't have a root is $x^2+x+1$ and its square is $x^4+x^2+1$. It follows that your polynomial is irreducible modulo $2$ and therefore irreducible in over $\mathbb{Z}$.

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  • $\begingroup$ Accepting Jack's as answer since that was the first to answer the "concept" talked about here! Here's a +50 which will soon be a total of +150:-) $\endgroup$ Commented Dec 29, 2015 at 6:36
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Over $\mathbb{F}_3$ your polynomial splits as $x(x+1)^3$, no luck.

Over $\mathbb{F}_5$ your polynomial splits as $(x+1)(x^3+x^2+2)$. That is enough to state that your polynomial is irreducible over $\mathbb{Q}$, since we know in advance that it has no rational root.

You may also notice that $p(x)=x^4 + 3 x^3 - 9 x^2 + 7 x + 27$ takes prime values for any $x\in\{-20,-17,-14,-5,-2,1,4,10,13\}$, but no reducible polynomial over $\mathbb{Q}$ with degree $4$ can take nine prime values at nine different points.

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  • $\begingroup$ Can please tell me the technique used in detail/provide a link where it is explained? $\endgroup$ Commented Dec 18, 2015 at 14:20
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    $\begingroup$ @JessePFrancis: it is just a minor variation on the theme of "if a polynomial is irreducible over a finite field, it is irreducible over $\mathbb{Q}$, too". $\endgroup$ Commented Dec 18, 2015 at 14:25
  • $\begingroup$ So, if it is irreducible over any one finite field, then it is irreducible over $\Bbb Q$? $\endgroup$ Commented Dec 18, 2015 at 14:28
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    $\begingroup$ @JessePFrancis: exactly. $\endgroup$ Commented Dec 18, 2015 at 14:30
  • $\begingroup$ I had not heard of that! Could you provide some references? I'm interested. $\endgroup$ Commented Dec 21, 2015 at 5:21
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Echoing the answer by Prometheus note that modulo $2$ the reduced polynomial is $\Phi_5$, which is irreducible modulo $2$, since $2$ has order $4$ in $\Bbb Z_5^\times$: $2^2=4,2^3=8,2^4=16=1$. In general the $n$-th cyclotomic polynomial $\Phi_n$ is irreducible in $\Bbb F_q[X]$ iff $q$ has order $\varphi(n)$ in $\Bbb Z_n^\times$.

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  • $\begingroup$ ...and with these formulas I guess it makes things much simpler, for more complex cases! Thank you! $\endgroup$ Commented Dec 18, 2015 at 14:47
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Irreducibility over $\mathbb Q$ is the same as over $\mathbb Z$ (Gauss lemma). There is no root (real or complex) not exceeding $1$ in absolute value ($27>1+3+9+7$). Thus, if there is a factorization into two polynomials, the free terms must be $\pm 3$ and $\pm 9$ (otherwise, by Vieta, the polynomial with the free term $\pm 1$ has a root with absolute value $\le 1$). However then the coefficient at $x$ must be divisible by $3$ and it is not.

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Theorem. (Gauss). If $p(x)\in Z[x]$ is reducible over $Q$ then $p(x)$ is reducible over Z. And to show that $p(x)=x^4+3x^3-9x^2+7x+27$ is not the product of two quadratics in $Z[x]$, it suffices to show that if $p(x)=(x^2+A x+B)(x^2+C x+D)$ with four cases: $(B,D)\in \{(1,27),(-1,-27),(3,9),(-3,-9)\}$,.... then $A+B=3$ and $B+D+AC=-9$ and $ A D+B C=7$, which cannot be solved in integers $A,C$.

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  • $\begingroup$ "...which cannot be solved in integers A,C." How can one be so sure about it just by looking at it? Anyway, thank you, this surely contains useful information! $\endgroup$ Commented Jan 4, 2016 at 4:09

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