# Gcd over multiple field extensions

## 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 $\tau$.

**Notation**. $d_i=degree(f_i)$, $d_y=degree_y(g)$.

## 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 $\mathbb{Q}[x_1,..,x_n]$.

**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.

### The quotient algebra

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

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

**Remark [Gröbner basis]**: $\{f_1(x_1),...,f_n(x_n)\}$ is a Gröbner basis for any monomial ordering.

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

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

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

> *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 $x_1P=\sum_{\alpha\in I_1}a_{\alpha}x_1^{d_1}m_{\alpha}+\sum_{\beta\in I_2}a_{\beta}x_1m_{\beta}$.
> Remark that $x_1m_{\beta}\in B_n$ so that $Q=\sum_{\beta\in I_2}a_{\beta}x_1m_{\beta}$ is irreducible modulo $I$ $\overline{Q}=Q$, and thus 
> $$\overline{x_1P}=\sum_{\alpha\in I_1} a_{\alpha}\frac{tail(f_1)}{lc(f_1)} m_{\alpha}+\overline{Q}$$
> where $tail(f_1)=f_1-lc(f_1)x_1^{d_1}$.
> Remark also that the supports of the polynomials $a_{\alpha}\frac{tail(f_1)}{lc(f_1)} m_{\alpha}$ are all disjoint so that computing $\sum_{\alpha\in I_1} a_{\alpha}\frac{tail(f_1)}{lc(f_1)} m_{\alpha}$ is a simple concatenation that can be done in $O(d^n)$ arithmetic operations in $\mathbb{Q}$. The addition with $\overline{Q}$ can be done in $O(d^n)$ arithmetic operations in $\mathbb{Q}$ which proves the lemma.

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

> *Proof*: As seen above (Remark [Muliplicative tensor]),  $T_n$ has less than $d^n$ elements and each element $w_iw_j$ but $w_1w_1$ can be rewritten $w_iw_j=x_km$ for some $k$ and $m\in T_n$ so that $T_n$ can be computed in $O(d^{2n})$ arithmetic operations in $\mathbb{Q}$.
> 
**Lemma [generalized Hermite's quadratic form]**. The first line $Tr_1=[Trace(w_1),\ldots,Trace(w_{d^n}))]$ of the generalized Hermite's quadratic form associated with $1$, $Q_1=[Trace(w_iw_j)]$ can be computed in $O(d^{2n})$ arithmetic operations in $\mathbb{Q}$. 

**Remark**: $Trace(P)=\overrightarrow{P}\cdot Tr_1$ and thus $Trace(w_iw_j)=\overrightarrow{w_iw_j}\cdot Tr_1$, where $\overrightarrow{P}$ is the expression of $\overline{P}$ as a vector in the base $B_n$.

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

**Lemma [Newton sums of E_T]**

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

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

### 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 $T-a_2x_2+\cdots+a_nx_n$ 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 [^bs16] 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.

**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 $\tilde{O}(nlog(d))$.

### 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 $\prod_{i=1}^nlc(f_i)$, which is of bitsize $d^n$, one can compute $E_T$ in $\tilde{O}((\tau+nd)d^{3n-1})$ bit operations.

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

**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 $\tilde{O}((\tau+nd)d^{3n-1})$ bit operations in the worst case and $\tilde{O}((\tau+nd)d^{2n-1})$ axpected with a Las-Vegas algorithm. 

### 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 $\tilde{O}((\tau+nd)d^{3n-1})$ bit operations for $T$ is a linear form with coefficients' bitsizes in $\tilde{O}(nlog(d))$, on then can use Brand and Sagraloff Theorem 16 in [^bs16] with $\Pi=\tilde{O}((\tau+nd)d^{3n-1})$ to prove the following:

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

[^bs16]: Brand and Sagraloff *On the Complexity of Solving Zero-Dimensional Polynomial Systems via Projection*, 2016. [ArXiv version](https://arxiv.org/pdf/1604.08944.pdf)


### Compute a RUR

We have see how to compute a separating form $T$ and it characteristic polynomial $E_T$ using Traces.
$$E_{X_i}=\sum_{j=1}^{d^n}Trace(X_iT^j)H_{d^n-j-1}(T)$$
where $H_{d^n-j-1}$ is the $(d^n-j-1)$-th Horner polynomial associated to $E_T$.
As $Trace(X_iT^j)=\overrightarrow{X_iT^j}\cdot Tr_1$, it can be computed in $O(d^n)$ arithmetic operations in $\mathbb{Q}$ so that the entire set $\{Trace(X_iT^j),i=1\ldots n,j=1..d^n-1\}$ cn be computed in $\tilde{O}(nd^{2n-1})$ arithmetic operations in $\mathbb{Q}$ and thus the $E_{X_i}$ can be computed in $\tilde{O}(d^{2n-1})$ arithmetic operations in $\mathbb{Q}$.

### Bitsize of a RUR

Basically the u-resultant is the multivariate resultant of the system $f_1=\ldots f_n=0=u_0T+u_1x_1+..+u_nx_n$ and is a polynomial with rational coefficients.
In the present case, $R(u_0,..,u_n)=\prod_{a\in V}(u_0+a_{1}u_1+\ldots a_{n}u_n)$ (in particular $R$ is squarefree. Taking $T=t_1x_1+\ldots t_nx_n$ any linear form, then, in our situation, $R$ the characteristic polynomial of $T$/

On can notice that $R(u_0,-t_1,..,-t_n)\in \mathbb{Q}$ 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 $R_0=\frac{\partial R}{\partial u_0}=\sum_{b\in V}\prod_{a\neq b\in V}(u_0+a_{1}u_1+\ldots a_{n}u_n)$ 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 by some other determinant, the division being exact.