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Suppose you have a (smooth) affine algebraic variety $X$ (over an algebraically closed field $k$) with ring of regular functions $A$, of dimension $n$. Let $Y$ a complete intersection in $X$, of codimension $c$. Take two minimal sets $f_1, \dots, f_c$ and $g_1, \dots, g_c$ of generators for $I_Y$. Set $f:=(f_1, \dots, f_c) \in A^c$ and $g = (g_1, \dots, g_c) \in A^c$.

Question: Does there exist a matrix $P \in M_{c \times c}(A)$ such that $Pf=g$ and that $\det P$ never vanishes over $X$?

Comment. Of course there are always (a lot of) matrices $P$ such that $Pf=g$. Moreover, it is easy to prove that any matrix $P$ such that $Pf=g$ has determinant which does not vanish over $Y$, and hence over an affine open subset $V$ of $Y$ inside $X$. (see for example, Matsumura, Commutative Ring Theory, thm 2.3).

On the other hand, it is also easy to find examples where a matrix $P$ such that $Pf=g$, has determinant vanishing somewhere on $X \setminus Y$. Take for example $X=\mathbb{C}^2$ and $f=g=(x,y)$ and $$A = \left( \begin{array}{cc}1-y & x \\ -y^2 & 1+xy \end{array} \right) \;.$$

However, Arrondo, in https://arxiv.org/pdf/math/0610015.pdf, beginning of paragraph 3, suggests that my question might have a positive answer.

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This is in general not true. Consider the ideal (complete intersection) $I=(x,y,zt-1)\subset k[x,y,z,t]$ for any field $k$. Then $I$ can also be generated by $(x, yz, zt-1)$. It can be shown that there are no non-singular $3\times 3$ matrix as you seek in this case. For details, see my paper (N. Mohan Kumar, A note on Cancellation of Reflexive modules, J. Ramanujan Society, 2002).

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  • $\begingroup$ Dear @Mohan, thanks so much!. It seems to me that your counterexample can be easily adapted to give counterexamples when codimension $Y =c \geq 3$. What happens if codimension of $Y$ is 2? $\endgroup$
    – skywalker
    Apr 29, 2018 at 12:42
  • $\begingroup$ @skywalker The same argument works even for $I=(x,yz-1)=(xy, yz-1)$, codimension two in $k[x,y,z]$. I did not want to rewrite the arguments from my paper. $\endgroup$
    – Mohan
    Apr 29, 2018 at 13:36
  • $\begingroup$ @ Mohan, Thank you very much! $\endgroup$
    – skywalker
    Apr 29, 2018 at 13:48

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