I'm working on an exercise in Janusz's Algebraic Number Fields. (I simplified it.)

Let $\Phi(x)$ be the minimal polynomial of the primitive $p$-th root of unity. ($p$ is an odd prime.) Let $q\ne p$ be a prime and consider the reduced polynomial $\bar \Phi(x)$ modulo $q$. Show that the splitting field of $\bar \Phi(x)$ over $GF(q)$ is the field $GF(q^m)$, where $m$ is the order of $q$ in the multiplicative group $\mathbb{F}_p^\times$. Conclude that every prime factor of $\bar \Phi(x)$ over $GF(q)$ has degree $m$.

I proved that the splitting field is $GF(q^m)$ for such $m$, but I don't know how to conclude the final statement. I expect that $\bar \Phi(x)$ splits as a product of linear polynomials over $GF(q^m)$ and $m$ linear polynomials make a irreducible factor of $\bar \Phi(x)$ over $GF(q)$ in some sense. I also thought that if we can show first that every irreducible factor of $\bar \Phi(x)$ over $GF(q)$ has the same degree then it should be $m$, since the degree of the splitting field over $GF(q)$ is equal to the lcm of all degrees of irreducible factors. But I cannot finish both approaches.


Note that if you add one single primitive $p$-th root of unity to any field, you add ALL $p$-th roots of unity, since you can obtain them just by taking powers of that one single primitive $p$-th root of unity.

Hence for any irreducible factor $f$ of $\Phi$, the field $\operatorname{GF}(q)[x]/(f)$ is a splitting field of $\Phi$, hence equal to $\operatorname{GF}(q^m)$. Thus $\deg f = m$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.