Irreductible Polynomial

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Irreductible Polynomial

November 2, 2009 beni22sof Leave a comment Go to comments

Prove that the polynomial $\displaystyle f(x)=\frac{x^n+x^m-2}{x^{\gcd (m,n)}-1}$ is irreductible over $\mathbb Q$ for all integers $n>m>0$ .
Miklos Schweitzer 2009 Problem 4
A generalization of this problem was proposed by myself in the Romanian TST 2010, the solution being similar to the one below. The generalization said to prove that if $p$ is a prime and $k_1,...,k_p$ are positive integers with $d=\gcd (k_1,...,k_p)$ then $\displaystyle f(x)=\frac{x^{k_1}+...+x^{k_n}-p}{x^d-1}$ is irreductible over $\Bbb{Q}$ .

Solution: Say $d=\gcd(n,m),\ da=m,\ db=n$ . It’s easy to see that $f(x)=x^{(b-1)d}+...+x^{ad}+2x^{(a-1)d}+...+2x^d+2$ . Therefore, if we consider $g(x)=x^{b-1}+...+x^{a}+2x^{a-1}+...+2x+2$ , then this polynomial satisfies the hypothesis of the following Problem so all the roots of $g$ have modulus strictly greater than 1, or $g$ has a root $z$ which is the root of unity of some order $k$ different of 1. Then, since the minimal polynomial of $z$ is $1+x+...+x^{k-1}$ and $z$ is a root of $g$ which has integer coefficients, it follows that $1+...+x^{k-1}$ divides $g$ . Suppose $r,q$ are the remainders of $a,b$ modulo $k$ , and suppose they are non zero. Then $1+...+x^{k-1} | 1+...+x^{r-1} +1+...+x^{q-1}$ which shows that at least one of $r,q$ is greater or equal to $k$ , which is a contradiction, because $r,q <k$ . Therefore, one of $a,b$ is divisible by $k$ . But we know that $k | a+b$ , so $k$ divides the other one too. Therefore $k | \gcd(a,b)=1$ , which is a contradiction, because $k$ is different from 1 since 1 is not a root for $g$ . Therefore $g$ has only roots which have modulus greater than 1.
Take $z$ a root of $f$ . Then $z^d$ is a root for $g$ , so $|z^d|=|z|^d >1$ . This implies $|z|>1$ , so $f$ has all roots of modulus strictly greater than 1.
Suppose $f=gh,\ g,h \in \mathbb{Z}[X]$ . Then $g(0)h(0)=2$ . This implies that one of $|g(0)|,|h(0)|$ is 1. Suppose $|g(0)|=1$ . But $|g(0)|=\displaystyle \prod_{g(z)=0} |z|>1$ . This is a contradiction proving that our assumption was false. Therefore $f$ is irreductible. $\Box$

Categories: Problem Solving Tags: irreductible, miklos schweitzer, polynomial

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Blord

December 30, 2010 at 5:40 pm

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Hi,

I don’t understand a step in your solution: why is the minimal polynomial of a root of unity is 1+x+x^2+..+x^(k-1)? The minimal polynomial of a root of unity is a cyclotomic polynomial which is not necessarily in the form you wrote. Please help me to clarify what is the problem. :S

And could you mention a book which contains the lemma that helped to solve this problem? I don’t see any other way to find a solution, and I never heard about this lemma before.

Thank you for your help, and keep up the excellent work!
- beni22sof
  
  January 8, 2011 at 6:39 pm
  
  Reply
  
  Hello.
  
  I see what you mean. I must modify the solution, using other polynomial as the minimal one.
  What lemma are you referring to?
  - Blord
    
    January 9, 2011 at 12:11 am
    
    I mean this: https://mathproblems123.wordpress.com/2009/11/09/position-of-roots/
    
    Thanks!
beni22sof

January 9, 2011 at 11:58 pm

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I found the property as a lemma in the solving of the following Romanian National Math Olympiad Test in 1990:
Prove that if $a \in \Bbb{Z}^*$ and $n \geq 2$ then the polynomial
$f(x)=x^n+ax^{n-1}+ax^{n-2}+...+ax-1$ is irreductible in $\Bbb{Z}[X]$ .
The lemma was stated without the fact that the polynomial can have a root of unity of some order (namely, if the polynomial has coefficients in descending order, and they are not all equal, then all the roots are OUTSIDE the closed unit disk), which I found curious, since the polynomial $x^{n-1}+...+x^{m}+2x^{m-1}+2x^{m-2}+...+2x+2$ satisfies the lemma, and for $m |n$ any root of unity of order $m$ is a root for the polynomial above.
I don’t know any book which contains the result. I found some posts on mathlinks which use this lemma, but in the “old” form.

Hope this helps.

Beni