# Reduction modulo: $\mathbb{Z}[x]\to\mathbb{Z}/p\mathbb{Z}[x]$

Let $$f(x)=\sum_{i}a_ix^i\in\mathbb{Z}[x],$$ and $$[f]_{p}(x)=\sum_{i}[a_i]_{p}x^i\in\mathbb{Z}/p\mathbb{Z}[x].$$ Is it true that: $$f(\xi)=0\Rightarrow [f]_{p}(\xi)=0,$$ where $\xi$ is some root of unity.

i.e. is it true proof: $$[f]_{p}(\xi)=\sum_{i}[a_i]_{p}{\xi}^i=\sum_{i}{a}_i{\xi}^i-\sum_{i}(a_i-[a_i]_{p}){\xi}^i=0-p\cdot\alpha=0-0\cdot\alpha=0.$$

Thanks.

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This question doesn't make sense to me. In what ring is addition and multiplication taking place when you evaluate $[f]_p(\xi)$? –  Alex B. Aug 18 '11 at 13:49
And, are you using the fact that $\xi$ is a root of unity anywhere? –  Álvaro Lozano-Robledo Aug 18 '11 at 13:53
@Alex: in $\mathbb{Z}/p\mathbb{Z}[\xi]=\{x_0+x_1{\xi}+\ldots+x_{n-2}{\xi}^{n-2}\mid x_i\in\mathbb{Z}/p\mathbb{Z}\}$.$\mathbb{Z}/p\mathbb{Z}[\xi]$ --- is extension of $\mathbb{Z}/p\mathbb{Z}$. –  Aspirin Aug 18 '11 at 13:57
@Álvaro: $\xi$ is a root of unity $\Rightarrow$ $\mathbb{Z}/p\mathbb{Z}[\xi]\cong\mathbb{F}_{p^{n-1}}$ $\Rightarrow$ $char(\mathbb{Z}/p\mathbb{Z})=p$ $\Rightarrow$ $\mathbb{Z}/p\mathbb{Z}\ni p = 0.$ –  Aspirin Aug 18 '11 at 14:02
@Alyushin: Careful there; if you are defining $\mathbb{Z}/p\mathbb{Z}[\xi]$ abstractly then the $\xi$ for $\mathbb{Z}[x]$ and the $\xi$ for $\mathbb{Z}/p\mathbb{Z}$ may be different animals. But it looks like you are trying to define $\mathbb{Z}/p\mathbb{Z}[\xi]$ as the group ring on $C_{p^{n-1}}$ over $\mathbb{Z}/p\mathbb{Z}$. Is that so? –  Arturo Magidin Aug 18 '11 at 15:39

I think that what you are trying to do is the following:

Pick an $f(x)\in\mathbb{Z}[x]$ which satisfies $f(\xi)=0$, where $\xi\in\mathbb{C}$ is a primitive $n$-th root of unity.

Then consider $(\mathbb{Z}/p\mathbb{Z})[\zeta]$, where $\zeta$ is a root of the reduction modulo $p$ of the $n$th cyclotomic polynomial $\Phi_n(x)$.

Is it the case that $[f]_p(\zeta)=0$ in $(\mathbb{Z}/p\mathbb{Z})[\zeta]$, where $[f]_p(x)$ is the reduction modulo $p$ of $f(x)$?

The answer is, I believe, "yes": since $\Phi_n(x)$ is the monic irreducible polynomial for $\xi$ over $\mathbb{Q}$, it follows that $f(\xi)=0$ if and only if $\Phi_n(x)$ divides $f(x)$ (in $\mathbb{Q}[x]$, but since both have integer coefficients, Gauss's Lemma guarantees the divisibility happens in $\mathbb{Z}[x]$). So we can write $f(x) = \Phi_n(x)g(x)$ for some $g(x)\in\mathbb{Z}[x]$.

Reducing modulo $p$, we get that $[f]_p(x) = [\Phi_n]_p(x)[g]_p(x) \in (\mathbb{Z}/p\mathbb{Z})[x]$, and evaluating at $\zeta$ we get $$[f]_p(\zeta) = [\Phi_n]_p(\zeta)[g]_p(\zeta) = 0,$$ since $\zeta$ satisfies $[\Phi_n]_p(x)$.

If you want to define $\zeta$ as satisfying $x^{n-1}+\cdots+x+1$, then the answer is "no". Take $n=4$; then you are defining $$(\mathbb{Z}/p\mathbb{Z})[\zeta]\cong \frac{(\mathbb{Z}/p\mathbb{Z})[x]}{(x^3+x^2+x+1)}.$$ But $f(x)=x^2+1$ satisfies $f(\xi)=0$ (since $\xi=i$), but $[f]_p(\zeta) = x^2+1\neq 0$.

Based on your comment above, it seems rather you are trying to use the "same" primitive root of unity for both $\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$. You have to be very careful here. It is of course possible to consider $\mathbb{Z}[\zeta]$, where $\zeta\in\mathbb{C}$ is a complex primitive $n$th root of unity. And similarly, you can consider the splitting field of $x^n-1$ over $\mathbb{Z}/p\mathbb{Z}$, and take a root $\xi$ which is (one hopes) a primitive element of the splitting field, so as to consider $\mathbb{Z}/p\mathbb{Z}[\xi]$, but $\xi\in\overline{\mathbb{F}_p}$, the algebraic closure of the field of $p$-elements. It really does not make sense to talk about $\mathbb{Z}[\xi]$: for this notation to make sense, you need an overring which contains both $\mathbb{Z}$ and $\xi$, and there is no obvious overring that does ($\xi$ "lives" in a field of characteristic $p$, $\mathbb{Z}$ does not).

Here is one thing you can do: keep them straight and proceed more or less along the lines I described above.

First, fix an integer $n$. Let $\Phi_n(x)\in\mathbb{Z}[x]$ be the $n$th cyclotomic polynomial over $\mathbb{Z}$, so that the roots of $\Phi_n(x)$ are precisely the complex numbers that are primitive $n$th roots of unity. Let $\zeta$ be such a root, and will consider $\mathbb{Z}[\zeta]$.

By the usual theory, for all $f(x)\in\mathbb{Z}[x]$, $f(\zeta)=0$ if and only if $\Phi_n(x)$ divides $f(x)$ over $\mathbb{Z}[x]$.

Similarly, let $\Psi_{n,p}(x)\in\mathbb{Z}/p\mathbb{Z}[x]$ be the polynomial $$\Psi_{n,p}(x) = \prod(x-\omega)$$ where $\omega$ runs over all elements of (multiplicative) order $n$ in $\overline{\mathbb{F}_p}$. Note that $\Psi_{n,p}(x)$ need not be equal to the reduction modulo $p$ of $\Phi_n(x)$. For example, $\Phi_p(x)=x^{p-1}+\cdots+1$, but in characteristic $p$ we have $x^p-1 = (x-1)^p$, so the only root of $x^p-1$ in $\overline{\mathbb{F}_p}$ is $1$, which is not of multiplicative order $p$; so $\Psi_{p,p}(x) = 1$, since the product is empty. In fact,t his holds whenever $n$ is a power of $p$; more generally, if $n$ is divisible by $p$, you are going to run into problems when you work in characteristic $p$.

However, if $n$ is not divisible by $p$, then I believe the proof for $\Phi_n(x)$ will go through to prove that $\Psi_{n,p}(x)$ will be irreducible, and you can consider $\xi$ to be a root in the algebraic closure. Then $\overline{g}(x)\in\mathbb{Z}/p\mathbb{Z}[x]$ is such that $\overline{g}(\xi)=0$ if and only if $\Psi_{n,p}(x)$ divides $\overline{g}(x)$. In particular, by reducing $\Phi_n(x)$ modulo $p$ you get that $\Psi_{n,p}(x)$ divides $[\Phi_n]_p(x)$.

Now, suppose you have $f(x)\in\mathbb{Z}[x]$ such that $f(x)$ vanishes at a complex primitive $n$th root of unity, $f(\zeta)=0$. Then $\Phi_n(x)$ divides $f(x)$, hence $[\Phi_n]_p(x)$ divides $[f]_p(x)$, hence $\Psi_{n,p}(x)$ divides $[f]_p(x)$, hence $[f]_p(x)$ vanishes at the $\overline{\mathbb{F}_p}[x]$ primitive $n$th root of unity, $[f]_p(\xi)=0$, which proves your claim at least in the case where $p$ does not divide $n$.

What if $p$ divides $n$? Consider $n=p$ for simplicity. $\Phi_p(x) = x^{p-1}+\cdots+x+1$. You can reduce modulo $p$, but since $$(x-1)^p = x^p-1 = (x-1)(x^{p-1}+\cdots+x+1),$$ it follows that $x^{p-1}+\cdots+x+1 = (x-1)^{p-1}$ in $\mathbb{Z}/p\mathbb{Z}[x]$. The only root is $1$, and there are no primitive $p$th roots of unity, so it makes no sense to talk about $\xi$, a primitive $p$th root of unity in $\overline{\mathbb{F}_p}$. You can ask whether, if $f(\zeta)=0$ for $f(x)\in\mathbb{Z}[x]$, does it follow that $1$ is a root of $[f]_p(x)$ with multiplicity at least $p-1$ in $\mathbb{Z}/p\mathbb{Z}[x]$, and the answer is "yes": If $f(x)\in\mathbb{Z}[x]$ is a multiple of $\Phi_p(x)$, then reducing modulo $p$ you get that $[f]_p(x)$ is a multiple of $(x-1)^{p-1}$, and therefore $[f]_p(1)=0$ and $1$ is a root with multiplicity at least $p-1$.

But again: you have to be careful because you are really working with two different "primitive roots of unity" (and in characteristic dividing $n$, there may not even be such a thing).

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If $\xi$ is any algebraic integer, then there is a ring homomorphism $\mathbb{Z}[\xi] \rightarrow \bar{\mathbb{F}}_p$ (actually taking values in $\mathbb{F}_{p^n}$ for some $n$). Just take a maximal ideal $\mathfrak{m}$ of $\mathbb{Z}[\xi]$ containing $p$ (there is at least one thanks to Zorn's lemma because $p$ is not invertible), then $\mathbb{Z}[\xi]/\mathfrak{m} \simeq \mathbb{F}_{p^n}$ for some $n$. This ring homomorphism lifts the one you described ($\mathbb{Z} \rightarrow \mathbb{F}_p$), and allows to give a meaning to your computation. Note that there can be more than one such lift (i.e. more than one maximal ideal containing $p$).

In the special case that $\xi$ is a $p^{\text{something}}$-th or $k$-th ($k$ dividing $p-1$) root of unity, $n$ is equal to $1$. In the first case, there is a unique lift, but in the second case there are several if $k>1$.

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