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I denote by $m_n(q)$ the number of irreducible monic polynomials with degree $n$ over the finite field of order $q$. So the number of monic polynomials with degree is just $q^n$.

From this, how does it follow that $ \prod_{n\geq 1}\frac{1}{(1-x^n)^{m_n(q)}}=\frac{1}{1-qx}. $ I know $ \frac{1}{1-qx}=1+qx+q^2x^2+\cdots $ but I don't see an obvious relation between the coefficients of one to the other.

Thanks,

2nd try: Based on the help, I try to write this out in form $ (1+x+x^2+\cdots)^{m_1(q)}(1+x^2+x^4+c\dots)^{m_2(q)}\cdots=\sum_{f\text{ monic}}x^{\deg f}. $

Due to unique factorization, distinct products of monic irreducibles of determine unique monic polynomials. So the coefficient of $x^{\deg f}$ is the number of monic polynomials of degree $\deg f$, But I'm having a hard time picturing how the expanded left hand side would count them for the identity to hold. Would someone be willing to explain this subtlety?

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    $T$hanks @AndréNicolas. I tried to get a little further based on your hint. If you find the time, do you mind explaining how your method works? So far I'm not able to imagine how the expanded version multiply out correctly to get the identity.2012-01-29

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$q^n$ is the number of monic polynomials of degree $n$ over $\mathbb{F}_q$, so we can write $\frac{1}{1 - qx}$ as $\sum_{f \text{ a monic polynomial}} x^{\deg f}.$

Now, monic polynomials can be uniquely factored into a product of irreducible polynomials, and $\deg$ satisfies $\deg fg = \deg f + \deg g$. What does that tell you about how the above generating function can be factored? (This is closely analogous to how we derive the Euler product for the Riemann zeta function.)

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    @Hailie: yes, if I understand what you're saying correctly. Each factor $(1 + x^n + x^{2n} + ...)$ corresponds to one of the possible irreducible factors of $f$, where the different terms describe the multiplicity of the factor.2012-01-30