I'm having trouble with the following Lemma:

$\Phi_m$ is defined over the prime subfield of $K$ (that is, over $\mathbb Q$ or $\mathbb F_p$). When $\mathrm{char}K = 0$, $\Phi_m$ is defined over $\mathbb Z$.

The proof proceeds as follows:

The proof is by induction on $m$. The result is trivial if $m = 1$. If $m > 1 $ then $X^m - 1 = \Phi_m \displaystyle \prod _{d|m, d \neq m} \Phi_d = \Phi_m g$, where $g$ is monic and by the induction hypothesis is defined over the prime subfield of $K$ (and over $\mathbb Z$ if $\mathrm{char}K = 0$). By Gauss' Lemma, or by direct argument using the Remainder Theorem, $\Phi_m$ is also defined over the prime subfield (and over $\mathbb Z$ if $\mathrm{char}K = 0$).

I don't understand this at all. How is Gauss' Lemma being used? I'm also confused about the wording, in particular the characteristic $0$ bit. Any field with characteristic $0$ has $\mathbb Q$ as its prime subfield, right? So doesn't that make the $\mathbb Q$ bit redundant? Obviously if the polynomial is defined over $\mathbb Q$ then it's also defined over $\mathbb Z$, since the coefficients are all algebraic integers.


Let $R$ be the image of $\mathbb{Z}$ in $K$, so $R$ is the prime field if the characteristic of $K$ is positive and otherwise $R$ is isomorphic to $\mathbb{Z}$. Certainly $\Phi_m(X)=X-1\in R[X]$. Assume that $m>1$ and that $\Phi_d(X)\in R[X]$ for $d<m$. You have $X^m-1=\prod_{d\mid m}\Phi_d(X)=g(X)\Phi_m(X)$ with $g(X)$ monic in $R[X]$. By the division algorithm (here I use that $g(X)$ is monic) you can write $X^m-1=q(X)g(X)+r(X)$ for unique $q(X),r(X)\in R[X]$. But this equation also holds in $K[X]$, so by the uniqueness in the division algorithm, $r(X)=0$ and $\Phi_m(X)=q(X)\in R[X]$.

In general, if $S$ is a ring and $R$ a subring (commutative with identity and the identity of $R$ equals that of $S$), and if $p(X),g(X)\in R[X]$ with $g(X)$ monic are such that $g(X)\mid p(X)$ in $S[X]$, the same argument shows that $g(X)\mid p(X)$ in $R[X]$ (with the same quotient).

The Gauss' Lemma I remember is about irreducibility of a "primitive" polynomial over a UFD and its quotient field, but I know there is another result with the same name that says something like if a polynomial factors over the quotient field of a UFD $R$, then you can multiply the factors by some element of $R$ to get a factorization in $R$. Probably this in conjunction with the monicity of $g$ implies $\Phi_m(X)\in R[X]$. It's possible I'm not remembering the result correctly though. I think the division algorithm is easier.

  • $\begingroup$ Thanks for this response. Probably a stupid question - why do you use that $g$ is monic? $\endgroup$ – Matt Apr 10 '12 at 22:57
  • $\begingroup$ Because in the division algorithm for the ring $R[X]$ ($R$ any commutative ring with identity), the theorem states that if $p(X),g(X)\in R[X]$ and the leading coefficient of $g(X)$ is a unit, then there exist $q(X)$ and $r(X)$...If the leading coefficient of $g(X)$ is not a unit, we can't necessarily do division with remainder. If you look at the proof of that particular theorem, there is a point where you need to divide by the leading coefficient of $g(X)$. $\endgroup$ – Keenan Kidwell Apr 10 '12 at 23:05
  • $\begingroup$ Okay. But you could just take $R$ to be the prime subfield of $K$ also in the characteristic zero case (so $R$ is always a field), and then use the fact that the only algebraic integers in $\mathbb Q$ are $\mathbb Z$ at the end. Thanks a lot! $\endgroup$ – Matt Apr 10 '12 at 23:14

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