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Let $f$ be a non-constant polynomial over $\mathbb{Q}$ and $p$ be a positive prime number grater than $2\cdot \max^2$, where $\max$ is the maximum among the absolute values of all numerators and denominators of the coefficients of $f$.

For instance, if $f(x) = \frac{25}{7} x^3 + 11x - \frac{1}{90} x^3$, then $\max = 90$.

Consider $\psi: \mathbb{Q} \rightarrow \mathbb{Z}_p$ given by $\psi\left(\frac{a}{b}\right) = a\cdot b^{-1} \mod p$. Extend $\psi$ to polynomials by applying it to each coefficient.

The reason for that lower bound on $p$ is that it assures that $\psi$ has an inverse for this polynomial, the Rational reconstruction function.

That is, applying the rational reconstruction method to $\psi(f)$ gives us $f$ again.

So, the question is, when does it happen that the factors of $\psi(f)$ will not correspond to those of $f$?

More precisely, let $f_1, ..., f_n$ be the factors of $f$ over $\mathbb{Q}$, and $g_1, ..., g_m$ be the factors of $\psi(f)$ over $\mathbb{Z}_p$, then what are the conditions on $p$ and $f$ that imply

$n \not = m$

or

$n = m$ but $\{\psi(f_1), \dots, \psi(f_n)\} \not = \{g_1, \dots, g_n\}$?

Numerical example for case $n \not = m$:

Let $f(x) = x^4 - 7/20 x^3 + 13/180 x^2 - 1/900 x$.

Then, $\max = 900$ and the we can choose $p = 2430007$.

Thus, $\psi(f) = x^4 + 121500x^3 + 1363504x^2 + 1736105x$.

Using SAGE, I have checked that rational_reconstruction($\psi(f)$) is indeed equal to $f(x)$.

But the factors of $f$ over $\mathbb{Q}$ are $f_1(x) = x - 1/60, f_2(x) = x$, and $f_3(x) = x^2 - 1/3 x + 1/15$.

However, the factors of $\psi(f)$ over $\mathbb{Z}_p$ are $g_1(x) = x, g_2(x) = x + 1337712, g_3(x) = x + 1741505$, and $g_4(x) = x + 1902297$.

Numerical example for case where the factors match:

Let $f(x) = x^4 - 4/5 x^3 - 499/45 x^2 - 104/45 x - 11/9$.

Then $\max = 499$ and we pick $p = 747037$.

So, $\psi(f) = x^4 + 298814 x^3 + 365207 x^2 + 531224 x + 166007$.

Again, using SAGE, I have verified that rational_reconstruction of coefficients of $\psi(f)$ yields exactly the coefficients of $f$.

The factors of $f$ over $\mathbb{Q}$ are $f_1(x) = x^2 - x - 11$ and $f_2(x) = x^2 + 1/5 x + 1/9$.

The ones of $\psi(f)$ over $\mathbb{Z}_p$ are $g_1(x) =x^2 + 747036x + 747026$ and $g_2(x) = x^2 + 298815 x + 664033$.

This time, $\psi(f_1) = g_1$ and $\psi(f_2) = g_2$.

Hilder Vitor Lima Pereira
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    $x^2-2$ is irreducible over the rationals, but factors mod $p$ for any $p\equiv\pm1\bmod8$. Similar results apply for all irreducible quadratics. But I expect things are more complicated for higher degrees – I'd be surprised if there were a simple answer. – Gerry Myerson Oct 11 '18 at 09:31
  • I see. The point is that, for efficiency purposes, I would like to factor the polynomials mod some prime $p$ and reconstruct the factors back to $\mathbb{Q}$ instead of factoring over $\mathbb{Q}$, because over the rationals, the coefficients in the middle of the factoring process can get very big, even though the final coefficients in the factors are upper bounded by $\sqrt{p / 2}$. – Hilder Vitor Lima Pereira Oct 11 '18 at 09:37
  • There are even polynomials that are irreducible over the rationals but reducible over all finite fields: https://math.stackexchange.com/questions/77155/irreducible-polynomial-which-is-reducible-modulo-every-prime – darko Oct 11 '18 at 15:22

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