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Deformation Techniques for Efficient Polynomial Equation ... - RISC

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DEFORMATION TECHNIQUES<br />

85<br />

Denote by C[T] M and C[T] M [X](F 1 , ..., F n )=(C[T][X](F 1 , ..., F n )) M<br />

the localization by M of the corresponding C[T]-modules.<br />

Since the C-algebra C[X](F 1 (t, X), ..., F n (t, X)) is reduced we easily see<br />

that the Jacobian matrix<br />

F 1 (t, X) F<br />

}}}<br />

1 (t, X)<br />

\<br />

X 1 X n<br />

.<br />

b . . b<br />

+<br />

F n (t, X) F<br />

}}}<br />

n (t, X)<br />

X 1 X n<br />

is unimodular over this algebra (this means that the residue class of its<br />

determinant is a unit of the algebra).<br />

From this we deduce that the Jacobian matrix<br />

F 1 (T, X) F 1 (T, X)<br />

}}}<br />

X 1 X n<br />

.<br />

DF(X) :=DF(T, b<br />

X):=\<br />

. . b<br />

+<br />

(6)<br />

F n (T, X) F n (T, X)<br />

}}}<br />

X 1 X n<br />

is also unimodular over the localized C-algebra C[T] M [X](F 1 , ..., F n )<br />

and that this algebra is, there<strong>for</strong>e, reduced.<br />

This allows us to state the announced lifting procedure, which is in fact<br />

a variant of the NewtonHensel Lemma (see, e.g., [Ive73, Proposition 7.2]).<br />

Although this result is well known, we will give a self-contained and constructive<br />

proof in order to give a mathematical illustration of the particular<br />

algorithmic techniques we are using in this paper.<br />

Lemma 3. Let assumptions and notations be as be<strong>for</strong>e. Then, <strong>for</strong> any<br />

point !=(! 1 , ..., ! n )#V t , there exists a unique n-tuple R (!) =(R (!)<br />

1<br />

, ..., R(!) ) n<br />

# CT&t n of <strong>for</strong>mal power series such that the following two conditions are<br />

satisfied:<br />

v F 1 (T, R (!) )=0, ..., F n (T, R (!) )=0<br />

v R (!) (t) :=(R (!)<br />

1<br />

(t), ..., R(!) (t))=!.<br />

n<br />

Proof. Our arguments follow the lines of [Mat97]. They are based on<br />

the iterated application of the NewtonHensel operator we are going to<br />

explain now.

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