I'm working through the following question:

Question Reference: Oxford Part I Paper B2 2003

Find the monic polynomial $f \in \mathbb{Z}[X]$ whose roots are the complex primitive $12^{\text{th}}$ roots of unity. Let $p$ be a prime and $\mathbb{F}_p$ be the field with $p$ elements. Let $f_p$ be the image of $f$ under the (unique) ring homomorphism $\mathbb{Z} \rightarrow \mathbb{F}_p$.

Claim 1: $f_p$ is reducible

Claim 2: $f$ is irreducible in $\mathbb{Q}[X]$


So, I've found $f=x^4-x^2+1$ and so its image under the ring homomorphism is $\overline{1}x^4+\overline{(p-1)}x^2+\overline{1}$. Now, to show it's reducible, I imagine it's easiest to show that, if it has no root, it cannot factor as quadratics.

Am I missing something here? Are their any results that govern the properties of a polynomial under this map?

So assume $f_p(\alpha)$ is non-zero for all $\alpha \in \mathbb{F}_p$. Now the field only contains $p$ elements, and so in the case $p=2$ for example it's easy to see what the required factorisation is. However, I cannot then generalise this to the arbitrary $p$ case.

Can someone help? It would be useful to see the result for the '12 case', but I'm also interested in more general results which can be used.

Regarding the second claim, I can prove this anyway, without using the above result, since all cyclotomic polynomials are irreducible over $\mathbb{Q}$.

Is there any way I can apply the first claim to give rise to the second more immediately?


One can first consider the primes $p=2,3$ that divide $12$ separately. For $p=2$ the reduction is $\overline f=X^4+X^2+1=(X^2+X+1)^2$, and for $p=3$ it is $\overline f=X^4+2X^2+1=(X^2+1)^2$, so both are reducible. For a prime$~p$ that does not divide$~12$, the polynomial $x^{12}-1$ is separable over $\def\Fp{\Bbb F_p}\Fp$ (its roots in some extension field are all distinct), and so is therefore $\overline f$ which divides it. If $\alpha$ is a root of$~\overline f$, then other roots in $\Bbb F_p(\alpha)$ can be obtained from it by iterating the Frobenius automorphism $x\mapsto x^p$ of that field, and taking the product of factors $X-\beta$ where $\beta$ runs through the orbit of $\alpha$, one obtains a polynomial $\mu\in\Fp[X]$ (the minimal polynomial of$~\alpha$) that divides$~\overline f$. The fact that this is a proper divisor come from the circumstance that in fact $\alpha^{p^2}=\alpha$, so that the size of the orbit and the degree of$~\mu$ are$~2$. This is because $\alpha^{12}=1$ and every invertible element$~x$ of $\Bbb Z/12\Bbb Z$ satisfies $x^2=1$ as is easily checked; the class of$~p$ being such an element one therefore has $p^2\equiv1\pmod{12}$ and $\alpha^{p^2}=\alpha^1$.

This gives your first claim. The second is a special case of the known fact that all cyclotomic polynomials are irreducible over$~\Bbb Q$, as is proved for instance here; this in fact uses reduction modulo$~p$ for many primes$~p$, and arguments similar but not identical to the ones above.

  • $\begingroup$ This is very very helpful. The argument is clear - thanks! $\endgroup$ – Mathmo Mar 21 '14 at 17:00

You certainly cannot deduce the second claim from the first, since if some $f_p$ was irreducible, then $f$ would have to be irreducible, too. You cannot hope to have an argument that goes like $A\Rightarrow B$ and (not $A)\Rightarrow B$, unless you think that $B$ is always true.

Regarding claim 1, what do you know about the Galois group of $\mathbb{Q}(e^{2\pi i/12})/\mathbb{Q}$ and about Galois groups of irreducible polynomials over finite fields?

  • $\begingroup$ I'm not too sure what I know about Galois Groups of irreducible polynomials over finite fields that will help here... If $f$ were irreducible in $\mathbb{F}_p$, then $\mathbb{F}_p(\alpha)$ would be a splitting field for $f$ over $\mathbb{F}_p$ for any root $\alpha$ of $f$. Perhaps this helps, but I'm not too sure. Also, can we say that the Galois Group you've mentioned is cyclic? $\endgroup$ – Mathmo Mar 21 '14 at 14:56
  • $\begingroup$ Galois groups of irreducible polynomials over finite fields are finite, generated by Frobenius (this is also covered in the Oxford course the exam for which you are looking at). But is the Galois group of $f$ over $\mathbb{Q}$ cyclic? $\endgroup$ – Alex B. Mar 21 '14 at 17:58

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