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Ohad Kammar science forum beginner
Joined: 08 Jul 2006
Posts: 2

Posted: Sat Jul 08, 2006 5:57 pm Post subject:
Irreducibility of a Polynomial over Q



Hello,
I was wondering whether the next polynomial is irreducible over the rational
field Q:
1 + x^(p^k) + x^(2p^k) + x^(3p^k) + ... + x^((p1)p^k)) = (x^(p^(k+1)) 
1)/(x^(p^k)  1)
Where p is prime and k is a nonnegative integer.
Thanks in advance,
Ohad. 

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G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308

Posted: Sat Jul 08, 2006 10:23 pm Post subject:
Re: Irreducibility of a Polynomial over Q



Quote:  Hello,
I was wondering whether the next polynomial is
irreducible over the rational
field Q:
1 + x^(p^k) + x^(2p^k) + x^(3p^k) + ... +
x^((p1)p^k)) = (x^(p^(k+1)) 
1)/(x^(p^k)  1)
Where p is prime and k is a nonnegative integer.
Thanks in advance,
Ohad.
That's not a polynomial it's an equation. Do you mean the polynomial you get on the left side of the equation if you multiply both sides by x^(p^k)1) and then subtract x^(p^(k+1))1 from both sides? 


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Ohad Kammar science forum beginner
Joined: 08 Jul 2006
Posts: 2

Posted: Sat Jul 08, 2006 11:17 pm Post subject:
Re: Irreducibility of a Polynomial over Q



Quote:  1 + x^(p^k) + x^(2p^k) + x^(3p^k) + ... +
x^((p1)p^k)) = (x^(p^(k+1))  1)/(x^(p^k)  1)
Where p is prime and k is a nonnegative integer.
That's not a polynomial it's an equation. Do you mean the polynomial
you get on the left side of the equation if you multiply both sides by
x^(p^k)1) and then subtract x^(p^(k+1))1 from both sides?

I meant the polynomial on the left side. The right side was written just to
clarify what I meant...
It can also be written as Sigma x^(jp^k) where j = 0, 1, 2, ..., p1.
Ohad. 

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