1
$\begingroup$

Let $f_1, f_2,\ldots,f_N$ be some affine polynomials. We consider the question if these polynomials have a common (affine) root. By homogenizing these polynomials, we can associate a projective resultant, in the literature, it is called Macaulay's or Dixon's resultant. But a problem arises: these $f_i$ may have a common root in the infinity (in the projective space), but no common root in the affine space.

My question: is there an analogue resultant for the case of affine polynomials?

$\endgroup$

1 Answer 1

5
$\begingroup$

The example of three linear equations in two variables is typical. The equations $ax+by+c=0$, $dx+ey+f=0$ and $gx+hy+i=0$ have a common root if

  • $\det \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} =0$ AND one of the following two conditions is met
  • EITHER at least one of $\det \begin{pmatrix} a & b \\ d & e \end{pmatrix}$, $\det \begin{pmatrix} a & b \\ g & h \end{pmatrix}$, $\det \begin{pmatrix} d & e \\ g & h\end{pmatrix}$ is nonzero OR
  • all nine $2 \times 2$ minors of $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ vanish.

Note that we cannot give a single polynomial $R(a,b,c,d,e,f,g,h,i)$ which vanishes if and only if the equations have a common root, because the property of these equations having a common root is not closed. For example, for all $\epsilon \neq 0$, the equations $y=0$, $y+\epsilon x=1$ and $y+2\epsilon x = 2$ have a common root at $(0, 1/\epsilon)$, but the equations $y=0$, $y=1$, $y=2$, obtained in the limit as $\epsilon \to 0$, have no common root. Note that this also shows that no list of equations could do the job.

In general, a constructible set is a subset of an algebraically closed field formed from finitely many polynomial equalities and the boolean connectors AND, OR and NOT. For example, the set $$\{ (a,b,c,d,e,f,g,h,i,x,y) : ax+by+c=dx+ey+f=gx+hy+i=0 \}.$$ A theorem of Chevalley says that the projection of a constructible set onto a subset of its coordinates is again constructible. So the projection of the above set onto the $(a,b,c,d,e,f,g,h,i)$ coordinates must be an constructible set, and I've given a boolean description above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.