Is there a formula for quickly calculating the number of irreducible polynomials of degree $\displaystyle n$ over $\displaystyle Z_2$?

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- Sep 15th 2009, 02:25 PMjamixIrreducible polynomials
Is there a formula for quickly calculating the number of irreducible polynomials of degree $\displaystyle n$ over $\displaystyle Z_2$?

- Sep 15th 2009, 05:53 PMNonCommAlg
- Sep 15th 2009, 06:30 PMjamix
Ah right, I forgot out that.

I might have a couple more questions about irreducible polynomials, however I think I should be able to solve most of it myself (just wish I hadn't sold my textbook a couple years back (Doh)). - Sep 17th 2009, 01:07 PMjamixMore problems
I used this formula to calculate the number of irreducible polynomials for $\displaystyle n = 5$. Using it we conclude that there are 6 irreducible polynomials of this degree.

The problem I have though, is that when I tried calculating these irreducibles, I obtained the following 8 polynomials:

$\displaystyle x^5 + x + 1$

$\displaystyle x^5 + x^2 + 1$

$\displaystyle x^5 + x^3 + 1$

$\displaystyle x^5 + x^3 + x^2 + x + 1$

$\displaystyle x^5 + x^4 + 1$

$\displaystyle x^5 + x^4 + x^2 + x + 1$

$\displaystyle x^5 + x^4 + x^3 + x + 1$

$\displaystyle x^5 + x^4 + x^3 + x^2 + 1$

We know that if a degree 5 polynomial is reducible, then it must be divisible by one of $\displaystyle x,x+1,x^2 + x + 1$, yes? Taking the product of these yields the polynomial $\displaystyle x^4 + x$ over $\displaystyle Z_2$. Each of the above 8 polynomials $\displaystyle p(x)$ satisfies $\displaystyle gcd(x^4 + x,p(x)) = 1$, hence all 8 must be irreducible, right?

So whats going on (Headbang)? - Sep 17th 2009, 03:55 PMNonCommAlg