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Forum index » Science and Technology » Math » Undergraduate
Irreducibility of a Polynomial over Q
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Ohad Kammar
science forum beginner


Joined: 08 Jul 2006
Posts: 2

PostPosted: Sat Jul 08, 2006 5:57 pm    Post subject: Irreducibility of a Polynomial over Q Reply with 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^((p-1)p^k)) = (x^(p^(k+1)) -
1)/(x^(p^k) - 1)
Where p is prime and k is a non-negative integer.

Thanks in advance,
Ohad.
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G.E. Ivey
science forum Guru


Joined: 29 Apr 2005
Posts: 308

PostPosted: Sat Jul 08, 2006 10:23 pm    Post subject: Re: Irreducibility of a Polynomial over Q Reply with quote

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^((p-1)p^k)) = (x^(p^(k+1)) -
1)/(x^(p^k) - 1)
Where p is prime and k is a non-negative 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

PostPosted: Sat Jul 08, 2006 11:17 pm    Post subject: Re: Irreducibility of a Polynomial over Q Reply with quote

Quote:
1 + x^(p^k) + x^(2p^k) + x^(3p^k) + ... +
x^((p-1)p^k)) = (x^(p^(k+1)) - 1)/(x^(p^k) - 1)
Where p is prime and k is a non-negative 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, ..., p-1.

Ohad.
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