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February 13th, 2009, 06:41 AM   #1
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Irreducible polynomial

Dear friends I've a little problem with this statement:

Prove that the trinomial is irreducible over GF(2) if and only if for some nonnegative integer k.

Any help will be highly appreciated.
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February 14th, 2009, 06:52 AM   #2
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Re: Irreducible polynomial

Consider this:

if n is not power of 3, i.e , then - reducible over GF(2)
If , then the only element that can be the root of and not of is primitive element of . Lets set q - is such element, then - has order 3.
So the last things to prove is that - is irreducible over GF(2), that is known fact.
zolden is offline  
February 14th, 2009, 07:29 AM   #3
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Re: Irreducible polynomial

Thanks for help!
Arczi1984 is offline  
February 14th, 2009, 08:44 PM   #4
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Re: Irreducible polynomial

Zolden, sorry I am rusty, but would you clarify the last step? We know x^2n + x^n + 1 factors over some extension as (x^n-alpha)*(x^n-beta), alpha, beta the roots of the irreducible x^2 + x + 1, but what fact do we need to assert that these terms don't admit grouping into a factorization over GF(2)?

Appreciate the pointer...

G
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February 16th, 2009, 04:13 PM   #5
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Re: Irreducible polynomial

Sorry for cumbersome explanation, did not elaborate for the last step.
Consider . All roots of are there, roots of as well.
When we divide first over second, the only will be left roots that are strictly belong to the first one. They are in the form , where is a primitive of our GF.

To see that such polynomial does not exist, consider following.
Lets see how first coefficient of the result polynomial is formed. It is sum of our roots. Sum of series belongs to GF(2), but if we remove any subset of it - it will not.
If is divisible on any polynomial of GF(2) - we could remove subset (those sum is in GF(2)) of roots and the left sum will still be in GF(2), is the only one like this.
Divisor degree has to divide and be less than it. Any polynomial with lesser degree cannot has roots from higher extension field. So all polynomials have roots only as linear combination of primitive elements from and its powers, so it will not remove any other roots of , other than those, that already removed by .

For the first step: set , then: nominator is , while denominator is . So it is reducible.

Hope I am clear this time.
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