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\begin{document}

\section{Gcd over multiple field
extensions}\label{gcd-over-multiple-field-extensions}\label{gcd-over-multiple-field-extensions}

\subsection{The problem}\label{the-problem}\label{the-problem}

Given $f_1 (x_1), ..., f_n (x_n), g (x_1, .., x_n, Y) \in \mathbb{Q} [x_1,
..x_n, Y]$, and $\alpha = (\alpha_1, .., \alpha_n) \in \mathbb{R}^n$, compute
the number of distinct roots of $g (x_1, .., x_n, Y)$.

We suppose that all the polynomial have a maximal degree in each variable
bounded by $d$ and that their coefficient's bitsizes are less than $\tau$.

\tmtextbf{Notation}. $d_i = \tmop{degree} (f_i)$, $d_y = \tmop{degree}_y (g)$.

\subsection{Computing with a unique field
extension}\label{computing-with-a-unique-field-extension}\label{computing-with-a-unique-field-extension}

In this part we focus on $f_1 (x_1) = 0, ..., f_n (x_n) = 0$ and denote by $I
= \langle f_1, .., f_n \rangle$ the ideal generated by $f_1 (x_1), ..., f_n
(x_n)$ in $\mathbb{Q} [x_1, .., x_n]$.

\tmtextbf{Remark [Radical]}: for a cost in $O (nd^2 \tau)$ bit operations, one
can assume that each $f_i$ is squarefree and thus that $I$ is radical.

\subsubsection{The quotient
algebra}\label{the-quotient-algebra}\label{the-quotient-algebra}

In this part, we study the quotient algebra $A_n = \frac{\mathbb{Q} [x_1, ..,
x_n]}{I}$ associated to $I$.

\tmtextbf{Notation}. Let $\bar{P}$ be the representant of the class of $P$ in
$A_n$ expressed in $B_n$.

\tmtextbf{Remark [Gr{\"o}bner basis]}: $\{f_1 (x_1), ..., f_n (x_n)\}$ is a
Gr{\"o}bner basis for any monomial ordering.

\tmtextbf{Notation}. A (monomial) basis of $A_n$ is $B_n = \{ \prod_{i = 1}^n
x_i^{e_i}, e_i < d\}$ and we will denote it by $B_n = \{w_1, .., w_{d^n} \}$
with $w_1 = 1$.

\tmtextbf{Remark [Muliplicative tensor]}: the multiplicative tensor $T_n = \{
\overline{w_i w_j}, 1 \leq i, j \leq d^n \}$ has cardinality $\sharp T_n < (2
d)^n$.

\tmtextbf{Lemma [Multiplication by one variable]}. Let $P = \sum_{i = 1}^{d^n}
a_i w_i \in A_n$. Then, $\forall i = 1.. n$, computing $\overline{x_i P}$ is
in $O (d^n)$ arithmetic operations in $\mathbb{Q}$.

\begin{quoteenv}
  {\tmem{Proof}}: Consider $I_1 = \{x_1^{d_1 - 1} m \in B_n \}$ and $I_2 = B_n
  \setminus I_1$. As $P = \sum_{\alpha \in I_1} a_{\alpha} x_1^{d_1 - 1}
  m_{\alpha} + \sum_{\beta \in I_2} a_{\beta} m_{\beta}$, then $x_1 P =
  \sum_{\alpha \in I_1} a_{\alpha} x_1^{d_1} m_{\alpha} + \sum_{\beta \in I_2}
  a_{\beta} x_1 m_{\beta}$. Remark that $x_1 m_{\beta} \in B_n$ so that $Q =
  \sum_{\beta \in I_2} a_{\beta} x_1 m_{\beta}$ is irreducible modulo $I$
  $\bar{Q} = Q$, and thus
  \[ \overline{x_1 P} = \sum_{\alpha \in I_1} a_{\alpha}  \frac{\tmop{tail}
     (f_1)}{\tmop{lc} (f_1)} m_{\alpha} + \bar{Q} \]
  where $\tmop{tail} (f_1) = f_1 - lc (f_1) x_1^{d_1}$. Remark also that the
  supports of the polynomials $a_{\alpha}  \frac{\tmop{tail} (f_1)}{\tmop{lc}
  (f_1)} m_{\alpha}$ are all disjoint so that computing $\sum_{\alpha \in I_1}
  a_{\alpha}  \frac{\tmop{tail} (f_1)}{\tmop{lc} (f_1)} m_{\alpha}$ is a
  simple concatenation that can be done in $O (d^n)$ arithmetic operations in
  $\mathbb{Q}$. The addition with $\bar{Q}$ can be done in $O (d^n)$
  arithmetic operations in $\mathbb{Q}$ which proves the lemma.
\end{quoteenv}

\tmtextbf{Lemma [Multiplicative tensor]}. $\overline{T_n} = \{ \bar{m}, m \in
T_n \}$ can be computed in $O (d^{2 n})$ arithmetic operations in
$\mathbb{Q}$.

\begin{quoteenv}
  {\tmem{Proof}}: As seen above (Remark [Muliplicative tensor]), $T_n$ has
  less than $d^n$ elements and each element $w_i w_j$ but $w_1 w_1$ can be
  rewritten $w_i w_j = x_k m$ for some $k$ and $m \in T_n$ so that $T_n$ can
  be computed in $O (d^{2 n})$ arithmetic operations in $\mathbb{Q}$.
\end{quoteenv}

\tmtextbf{Lemma [generalized Hermite's quadratic form]}. The first line
$\tmop{Tr}_1 = [\tmop{Trace} (w_1), \ldots, \tmop{Trace} (w_{d^n}))$] of the
generalized Hermite's quadratic form associated with $1$, $Q_1 = [\tmop{Trace}
(w_i w_j)]$ can be computed in $O (d^{2 n})$ arithmetic operations in
$\mathbb{Q}$.

\tmtextbf{Remark}: $\tmop{Trace} (P) = \overrightarrow{P} \cdot \tmop{Tr}_1$
and thus $\tmop{Trace} (w_i w_j) = \overrightarrow{w_i w_j} \cdot
\tmop{Tr}_1$, where $\overrightarrow{P}$ is the expression of $\bar{P}$ as a
vector in the base $B_n$.

\tmtextbf{Notation [Characteristic polynomial]} Given $T = \sum_{i = 1}^n a_i
x_i \in \mathbb{Q} [x_1, .., x_n]$, we denote by $E_T$ the characteristic
polynomial of the multiplication by $T$ in $A_n$.

\tmtextbf{Lemma [Newton sums of E\_T]}

\begin{quoteenv}
  {\tmem{Proof}}. Compute $1, T, T^{d^n}$ in $O (nd^{2 n})$ arithmetic
  operations in $\mathbb{Q}$ using Lemma [Multiplication by one variable] and
  then $\tmop{Trace} (1), \tmop{Trace} (T), .., \tmop{Trace} (T^{d^n})$ in $O
  (d^{2 n})$ arithmetic operations in $\mathbb{Q}$.
  
  Now, it suffice to remember that $\tmop{Trace} (T^i)$ is the i-th Newton sum
  of $E_T$ to prove the following proposition:
\end{quoteenv}

\tmtextbf{Proposition [Computation of $E_T$]} Given $T = \sum_{i = 1}^n a_i
x_i \in \mathbb{Q} [x_1, .., x_n]$, $E_T$ can be computed in $O (nd^{2 n})$
arithmetic operations in $\mathbb{Q}$.

\subsubsection{Bitsizes of elimination
polynomials}\label{bitsizes-of-elimination-polynomials}\label{bitsizes-of-elimination-polynomials}

$E_T$ can be viewed as a factor determinant of the Macaulay matrix in the
hidden-variable strategy for computing the multivariate resultant after a
change of variable. Replacing $x_1$ by $T - a_2 x_2 + \cdots + a_n x_n$
remains to consider coefficients of bitsize $\tau + d \cdot s$, where $s$ is
the bitsize of the coefficients of the linear form. Using Hadamard's bound on
the Macaulay matrix, which has dimension $< (nd)^{n - 1}$ one can then bound
this resultant and its coefficients by a rational of bitsize $\tilde{O} ((\tau
+ ds) d^{n - 1})$. Accordingly to Brand and Sagraloff \footnote{Brand and
Sagraloff {\tmem{On the Complexity of Solving Zero-Dimensional Polynomial
Systems via Projection}}, 2016.
\href{https://arxiv.org/pdf/1604.08944.pdf}{ArXiv version}} result (Theorem
16), one can take $s = O (nlog (d))$, so that the bitsize of the coefficients
of $E_T$ are in $sO ((\tau + nd) d^{n - 1})$ where $\tilde{O} (.)$ forgets the
logarithmic terms.

\tmtextbf{Proposition [bitsize of $E_T$]}. $E_T$ has degree $d^n$ and
coefficients of bitsize in $\tilde{O} ((\tau + nd) d^{n - 1})$ if $T$ is a
linear form with coefficients' bitsizes in $\tilde{O} (nlog (d))$.

\subsubsection{Multi-modular
computation}\label{multi-modular-computation}\label{multi-modular-computation}

In general, $E_T$ is a polynomial with rational coefficients. It is easy to
see that all the operations performed for computing it commute with the images
modulo a prime number that does not divide any of the leading terms of the
$f_i$. More precisely, the denominator appear in the reduction of the
multiplication by one variable which is the only operation involving a
division by some of the above leading coefficients. All the other operations
are linear combinations.

Also, according to the previous section and choosing prime numbers that do not
divide $\prod_{i = 1}^n lc (f_i)$, which is of bitsize $d^n$, one can compute
$E_T$ in $\tilde{O} ((\tau + nd) d^{3 n - 1})$ bit operations.

\tmtextbf{Proposition [computation of $E_T$]}. $E_T$ can be computed in
$\tilde{O} ((\tau + nd) d^{3 n - 1})$ if $T$ is a linear form with
coefficients' bitsizes in $\tilde{O} (n \log (d))$.

\tmtextbf{Corollary [computation of $\overline{E_T}$]} The result has degree
$d^n$ and coefficients of bitsize in $sO ((\tau + nd) d^{n - 1})$ and thus its
squarefree part can be computed in $\tilde{O} ((\tau + nd) d^{3 n - 1})$ bit
operations in the worst case and $\tilde{O} ((\tau + nd) d^{2 n - 1})$
axpected with a Las-Vegas algorithm.

\subsubsection{Separating linear forms and their characteristic
polynomial}\label{separating-linear-forms-and-their-characteristic-polynomial}\label{separating-linear-forms-and-their-characteristic-polynomial}

One key problem is to be able to compute a separating linear form for $f_1
(x_1) = 0, ..., f_n (x_n) = 0$. As we have seen, one can suppose this ideal to
be radical so that testing is an arbitrary chosen linear for is separating
remains to testing if $E_T$ is squarefree.

As computing $E_T$ can be done in $\tilde{O} ((\tau + nd) d^{3 n - 1})$ bit
operations for $T$ is a linear form with coefficients' bitsizes in $\tilde{O}
(n \log (d))$, on then can use Brand and Sagraloff Theorem 16 in
\footnote{Brand and Sagraloff {\tmem{On the Complexity of Solving
Zero-Dimensional Polynomial Systems via Projection}}, 2016.
\href{https://arxiv.org/pdf/1604.08944.pdf}{ArXiv version}} with $\Pi =
\tilde{O} ((\tau + nd) d^{3 n - 1})$ to prove the following:

\tmtextbf{Proposition [$E_T$ with $T$ separant]} One can compute a separating
linear form with coefficients in $\tilde{O} (n \log (d))$ and its
characteristic polynomial $E_T$ in $\tilde{O} (n (\tau + nd) d^{3 n - 1})$ bit
operations. $E_T$ is of degree $d^n$ with coefficients of bitsize in
$\tilde{O} ((\tau + nd) d^{n - 1})$

\subsubsection{Compute a RUR}\label{compute-a-rur}\label{compute-a-rur}

We have see how to compute a separating form $T$ and it characteristic
polynomial $E_T$ using Traces.
\[ E_{X_i} = \sum_{j = 1}^{d^n} \tmop{Trace} (X_i T^j) H_{d^n - j - 1} (T) \]
where $H_{d^n - j - 1}$ is the $(d^n - j - 1)$-th Horner polynomial associated
to $E_T$. As $\tmop{Trace} (X_i T^j) = \overrightarrow{X_i T^j} \cdot
\tmop{Tr}_1$, it can be computed in $O (d^n)$ arithmetic operations in
$\mathbb{Q}$ so that the entire set $\{\tmop{Trace} (X_i T^j), i = 1 \ldots n,
j = 1.. d^n - 1\}$ cn be computed in $\tilde{O} (nd^{2 n - 1})$ arithmetic
operations in $\mathbb{Q}$ and thus the $E_{X_i}$ can be computed in
$\tilde{O} (d^{2 n - 1})$ arithmetic operations in $\mathbb{Q}$.

\subsubsection{Bitsize of a
RUR}\label{bitsize-of-a-rur}\label{bitsize-of-a-rur}

Basically the u-resultant is the multivariate resultant of the system $f_1 =
\ldots f_n = 0 = f_0$ with $f_0 = x_0 + u_1 x_1 + .. + u_n x_n$ and is a
polynomial with rational coefficients. In the present case, $R (x_0, .., u_n)
= \prod_{a \in V} (x_0 + a_1 u_1 + \ldots a_n u_n)$ (in particular $R$ is
squarefree. Taking $T = - (t_1 x_1 + \ldots t_n x_n)$ any linear form in
$\mathbb{Q} [x_1, \ldots, x_n]$, then, in our situation, $R (x_0, t_1, \ldots,
t_n)$ the characteristic polynomial of $T$.

On can notice that $R (x_0, t_1, .., t_n) \in \mathbb{Q}$ cancels at all the
$T (a), a \in V$ and thus is proportional $E_T (u_0) \in \mathbb{Q}$.

In the same spirit, let define $R_i = \frac{\partial R}{\partial u_i} =
\sum_{b \in V} \prod_{a \neq b \in V} (u_0 + a_1 u_1 + \ldots a_n u_n)$ and
$R_0 = \frac{\partial R}{\partial u_0} = \sum_{b \in V} \prod_{a \neq b \in V}
(x_0 + a_1 u_1 + \ldots a_n u_n)$ and then $E_{T, i} (u_0) = R_i (u_0, - t_1,
.., - t_n)$.

Then, if $\beta = T (\alpha)$ is a root of $E_T$, $\alpha_i = \frac{E_{T, i}
(\beta)}{E_{T, 0} (\beta)}$. In short, $E_T, E_{T, 1}, \ldots, E_{T, n}$ is
the RUR of $f_1, .., f_n$ associated to $T$.

$R$ is a quotient of the determinant of the Macaulay matrix associated to the
system $\{ f_1 (x_1), ..., f_n (x_n), \}$ by some other determinant, the
division being exact. Accordingly to Mignotte's bound on the factors of a
polynomial, a bound on the determinant of the Macaulay matrix will give a good
bound.

\end{document}