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In the finite field of $q^n$ elements, the product of all monic irreducible polynomials with degree dividing $n$ is known to simply be $X^{q^n}-X$. Why is this?

I understand that $q^n=\sum_{d\mid n}dm_d(q)$, where $m_d(q)$ is the number of irreducible monic polynomials with degree $d$, and that each element of the field satisfies $X^{q^n}-X$.

How does the conclusion follow from this? I tried substituting the exponent to $X^{\sum_{d\mid n}dm_d(q)}-X$ but this doesn't seem to do much good.

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    Your first $q^n$ should be $q$. The product of all monic irreducible polynomials of degree dividing $n$ over the field of $q^n$ elements is $X^{(q^n)^n}-X$: there are at least $q^n$ of degree $1$, and you also have irreducibles of degree $n$, among others.2012-02-07

3 Answers 3