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I'm trying to prove that the following polynomial

$x^7+3x^6+12x^5+6x^4+2x^3-4x^2+6x+2$

is irreducible in $\mathbb{Q}[x]$.

I began with using the reduction mod p (p a prime), using 5 as my prime, test to show that the polynomial is irreducible in $\mathbb{Z}_5[x]$, and hence irreducible in $\mathbb{Q}[x]$.

When I reduce mod 5, I get the following polynomial

$x^7+3x^6+2x^5+x^4+2x^3+x^2+x+2$

It is easily checked that this polynomial has no roots in $\mathbb{Z}_5[x]$. All that is left to show that our first polynomial is irreducible, is to show is that the second polynomial can't be reduced in $\mathbb{Z}_5[x]$. If it were to be reducible, then it would have to factor into one of the following:

$(x^5+ax^4+bx^3+cx^2+dx+e)(x^2+fx+g)$ $$or$$ $(x^4+ax^3+bx^2+cx+d)(x^3+ex^2+fx+g)$ $$or$$ $(x^3+ax^2+bx+c)(x^2+dx+e)(x^2+fx+g)$

Considering only the first case: I multiply the terms through, and group terms together, allowing for me to create a system between the coefficients on this polynomial and the polynomial $x^7+3x^6+2x^5+x^4+2x^3+x^2+x+2$.

I obtain the following system of equations:

$ \left\{ \begin{array}{c} f+a=3 \\ g+fa+b=2 \\ ag+bf+c=1 \\ bg+cf+d=2 \\ cg+df+e=1 \\ dg+ef=1 \\ eg=2 \\ \end{array} \right. $

however I don't see how to find a contradiction in the above system. Any help is greatly appreciated.

Mo Gainz
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  • Could you have done the reduction mod 2? – Parcly Taxel Dec 08 '17 at 02:28
  • unfortunately no as it was able to be reduced to $x^6(x+1)$ – Mo Gainz Dec 08 '17 at 02:30
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    How confident are you that it's irreducible? It looks so close to being an Eisenstein criterion question. – CJD Dec 08 '17 at 02:38
  • Fairly certain, but I'm open to evidence in favor of the contrary. – Mo Gainz Dec 08 '17 at 02:55
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    To me Z_2 seems more promising than Z_5. What about Z_4, does it factor over Z_4? I assume you also already factored it over Z_3? How does it factor there? I have a feeling that your x^6(x+1) factorization over Z_2 is a big constraint on how the polynomial can factor over Z. – CJD Dec 08 '17 at 02:59
  • I get $(x^2+1)(x^2+x+2)(x^3+2x^2+x+1)$ when I try reduction mod 3 – Mo Gainz Dec 08 '17 at 03:13
  • For whatever it's worth, I typed your polynomial into CoCalc and it says that it is irreducible over Z_5, so it sounds like your intuition is right! Is it from a class or do you have any reason to think it should have a nice solution? – CJD Dec 08 '17 at 03:22
  • Perhaps, for the cases with quadratic factors you can check the second polynomial's values on $a+b\sqrt{2}$, where $a=0,\pm1,\pm2$, and $b=1,2$. Note also that, mod 5, we have $2\sqrt{2}=\sqrt{-2}$. – Alexander Burstein Dec 08 '17 at 03:45
  • To prove irreducibility, it suffices to get incompatible partitions by factoring over different primes. See pages 10 and 11 in http://www.math.leidenuniv.nl/~hwl/papers/cheb.pdf – Will Jagy Dec 08 '17 at 03:50
  • How would that translate to showing irreducibility over $\mathbb{Q}$? Would it be sound to mod out the original polynomial by a prime (p) that I am confident will be irreducible over $\mathbb{Z}_p[x]$?Then use your above criteria to prove that that modded out polynomial is irreducible over $\mathbb{Z}_p[x]$, and hence the original is irreducible over $\mathbb{Q}$? Or is that not logically sound? – Mo Gainz Dec 08 '17 at 04:11
  • There are some other standard tricks, like Eisenstein. – Randall Dec 08 '17 at 12:44
  • Yes indeed, I don't see how Eisenstein could be applied here however. As the constant term is 2 and the second highest degree term has coefficient 3. – Mo Gainz Dec 08 '17 at 12:46
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    It's irreducible $\pmod 5$ but I don't see an easy way to confirm that (I just used wolfram alpha). – lulu Dec 08 '17 at 12:50
  • I agree, this polynomial does not seem to be an easy case. – Peter Dec 08 '17 at 12:56
  • One Eisenstein trick is to replace $x$ by $x+a$ for various integers $a$. Sometimes the resulting polynomial obeys Eisenstein's criterion. Having said that, I tried several values of $a$ and came up with nothing. – B. Goddard Dec 08 '17 at 13:05
  • Anyway, @CJD was right. Factorization modulo two does lead to a simpler solution. – Jyrki Lahtonen Dec 09 '17 at 06:58

1 Answers1

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The following trick seems to work.

Assume contariwise that your polynomial $f(x)$ is a product of two non-constant polynomials, $f(x)=p(x)q(x)$. The usual business with Gauss Lemma tells us that without loss of generality we can assume that $p(x),q(x)\in\Bbb{Z}[x]$.

Because $2=f(0)=p(0)q(0)$ and $p(0)$ and $q(0)$ are both integers we can without loss of generality (swap the roles of $p$ and $q$ if need be) assume that $|q(0)|=1$ and $|p(0)|=2$.

Consider your polynomial modulo two. We have $$ f(x)\equiv x^7+x^6=(x+1)x^6\pmod2.\tag{1} $$ On the other hand we also have (I use overline to indicate reduction modulo two) $$ \overline{f}(x)= \overline{q}(x)\overline{p}(x) $$ We can use uniqueness of factorization in the domain $\Bbb{Z}_2[x]$ as follows. Because $q(0)$ is odd, we know that $\overline{q}(x)$ has a non-zero constant term. But, from $(1)$ we see that the only divisor of $\overline{f}$ with this property is $(x+1)$.

This, in turn, implies that the putative factor $q(x)$ must be linear (it is monic, so $\deg q=\deg \overline{q}$). But the non-existence of a linear factor of $f(x)$ is easy to confirm by checking that it has no integer roots. The short list of candidates allowed by the rational root test is $\{\pm1,\pm2\}$. I hope you checked those out already :-)

Jyrki Lahtonen
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