Deducing factorization over $\mathbb{F}_p$ from factorization over $\mathbb{Q}$

Question

I want to show rigorously that factorization over algebraic extensions of $\mathbb{Q}$ automatically yields a corresponding factorization over $\mathbb{F}_p.$

Consider for example the polynomial $x^4-1.$ If we view it initially over the field $\mathbb{Q}(i),$ it factors as $(x-1)(x-i)(x+1)(x+i).$ But if we view it now as a polynomial over $\mathbb{F}_5,$ then since $\mathbb{F}_5$ already has a square root of $-1$ (namely $2$) we expect that our polynomial should factor as $x^4-1=(x-1)(x-2)(x+1)(x+2).$ It's easy to check that this works.

More generally, let $p(x) \in Q[x]$ and suppose $p(x)=(x-r_1)\dots(x-r_n)$ over $\mathbb{C}.$ For each $r_i,$ let $m_{r_i, \mathbb{Q}}(x)$ be its minimal polynomial over $Q.$

Let us now view $m_{r_i, \mathbb{Q}}(x)$ as a polynomial over $F_p[x]$ and let $s_i$ be a root of this polynomial in an algebraic closure of $F_p.$

Then my claim/hope is that $p(x)=(x-s_1)\dots(x-s_n),$ where these polynomials are now viewed in an algebraic closure of $F_p.$

Answer 1

You can always do the following. It requires a tiny bit of algebraic number theory.

Assume that the polynomial has integer coefficients, so $$ p(x)=(x-r_1)\cdots(x-r_n)\in\Bbb{Z}[x],\qquad(*) $$ where $r_i\in\Bbb{C}$ are the zeros. Let $K=\Bbb{Q}(r_1,r_2,\ldots,r_n)$ be the splitting field of $p(x)$. Then $[K:\Bbb{Q}]<\infty$. Let $\mathcal{O}_K$ be the ring of algebraic integers of $K$. That is, elements of $\mathcal{O}_K$ are exactly those elements of $K$ whose minimal polynomials have integer coefficients. A fact from basic algebraic number theory is that $\mathcal{O}_K$ has a prime ideal $\mathfrak{p}$ with the following properties:

• $\mathfrak{p}\cap\Bbb{Q}=p\Bbb{Z}$,
• $[\mathcal{O}_K:\mathfrak{p}]$ is finite.

Let's study the quotient ring $E=\mathcal{O}_K/\mathfrak{p}$. Because $\mathfrak{p}$ is a prime ideal, $E$ is an integral domain. All finite integeral domains are fields, so by the second bullet we know that $E$ is a finite field. By the first bullet $p\cdot 1_E=0_E$, so $E$ has characteristic $p$. Therefore $E$ is an extension field of $\Bbb{F}_p$.

If $p(x)$ has integer coefficients, then all its zeros $r_i\in\mathcal{O}_K$. Therefore the factorization $(*)$ of $p(x)$ into linear factors actually takes place in the ring $\mathcal{O}_K[x]$.

Reduction modulo $\mathfrak{p}$ gives a homomorphism of rings (= a natural projection) $\pi:\mathcal{O}_K\to E$, and then, by applying $\pi$ that to coefficients, a homomorphism of rings from $\mathcal{O}_K[x]$ to $E[x]$. This is very much analogous to the "reduce the coefficients modulo $p$" mapping from $\Bbb{Z}[x]$ to $\Bbb{F}_p[x]$.

Applying this reduction modulo $\mathfrak{p}$ mapping to $(*)$ gives what you want. We get $$ p(x)=(x-s_1)(x-s_2)\cdots(x-s_n),\qquad(**) $$ where $s_i=r_i+\mathfrak{p}$ is the coset of the complex root $r_i$ modulo the chosen prime ideal $\mathfrak{p}$. Because $r_i\in\mathcal{O}_K$ we get that $s_i\in E$.

Because we can identify a (unique) copy of $E$ inside $\overline{\Bbb{F}_p}$, $(**)$ is a factorization of $p(x)$ over the algebraic closure of $\Bbb{F}_p$.

To make this a bit more concrete let's redo your example of $p(x)=x^4-1$. Now the splitting field is $K=\Bbb{Q}[i]$. Here the ring $\mathcal{O}_K$ is the ring of Gaussian integers $$\mathcal{O}_K=\Bbb{Z}[i]=\{a+bi\mid a,b\in\Bbb{Z}\}.$$ You were interested in factorization over an algebraic closure of $\Bbb{F}_5$, so $p=5$. Finding a suitable prime ideal $\mathfrak{p}$ is easy, because in the ring $\Bbb{Z}[i]$ we have $$ 5=(2+i)(2-i), $$ and it is easy to show (assuming basic properties of the ring of Gaussian integers) that both $2+i$ and $2-i$ generate a (principal) prime ideal of $\Bbb{Z}[i]$. This time it is easy to show that the quotient rings $\Bbb{Z}[i]/\langle 2+i\rangle$ and $\Bbb{Z}[i]/\langle 2-i\rangle$ both have five elements, and are thus both isomorphic to $\Bbb{F}_5$.

So if we select $\mathfrak{p}=\langle 2+i\rangle$, then $E\cong\Bbb{F}_5$. We also see that $i$ and $-2\equiv3\pmod5$ are both in the same coset of $\mathfrak{p}$. Basically this is because $2+i=0$ in the field $E$, so we get $i=-2=3$. Of course, then $-i=2$ in $E$. We have just seen that "reduction modulo $\mathfrak{p}$" maps the zeros as $1\mapsto 1$, $-1\mapsto -1=4$, $i\mapsto -2=3$ and $-i\mapsto 2$.

Note that if we, instead, use $\mathfrak{p}=\langle 2-i\rangle$, then the scenery is superficially a bit different because this time $i+\mathfrak{p}=2+\mathfrak{p}$. But we see that the overall picture does not change at all in the sense that now $-i\mapsto 3$. This always happens for polynomials with coefficients in $\Bbb{Z}$. The explanations comes from Galois theory, but may be I won't get into that here.

Answer 2

Let's tweak your example above slightly and take $f(x)=x^2-2$. Then over $\mathbb{C}$ we have $f=(x-\sqrt{2})(x+\sqrt{2})$, each factor having minimal polynomial $f$.

Over $\mathbb{F}_5$, $f$ is still irreducible, however from what you're saying is that we can take any root in each case so we shall choose the same root $\alpha$ twice.

Then $(x-\alpha)^2=x^2-2\alpha x-2$ which is not a polynomial over $\mathbb{F}_5$ so certianly does not equal the reduction of $f$ to $\mathbb{F}_5$.

To avoid this problem, you would need to state that you acquire all roots of the irreducible factors with equal multiplicities in order for your end polynomial to be defined over $\mathbb{F}_p$ but this should then suffice.