Let $R = \mathbb{C}[x_1,\ldots,x_n]$. Let $I$ be an ideal, and suppose we know a finite list of generators for $I$, say $I = \langle f_1,\ldots,f_k\rangle$.

Is this information enough to compute a new generating set $g_1,\ldots,g_\ell$ for $I$ such that $(g_1,\ldots,g_\ell)$ is a (strongly) regular sequence? If not, then what about if the $f_i$ are assumed to be irreducible and homogeneous?

Context: algebraic geometry, algebraic K-theory. I'm trying to use Koszul complexes to compute Grothendieck classes of some varieties, but I'm having trouble coming up with regular sequences (which I need for a Koszul resolution).

Specific case: $R = \mathbb{C}[x,y,z,r,s,t]$, $I = \langle f,g,h\rangle$ where $$ f = xr - yz, ~ g = xt - ys, ~ h = zt - rs. $$ These are the $2\times 2$ minors of the matrix $$ \begin{bmatrix} x & y\\ z & r\\ s & t \end{bmatrix}. $$ I'd like a new set of generators for $I$ that is a regular sequence, if one exists. The current set doesn't work because any two polynomials are a regular sequence, but the third is then a zero divisor in the quotient: e.g. $zg - sf = xh$.

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    $\begingroup$ First of all, I do not know what a strong regular sequence means. I do not believe you may choose $g_i$'s such that they form a regular sequence and generated the ideal $I$ unless you start with a complete intersection ideal. For a local ring, you may choose a generating sequence where the first grade $I$ form a regular sequence. $\endgroup$ – Youngsu Jun 17 '14 at 4:50
  • $\begingroup$ @Youngsu I think some authors exclude the condition that $R/\langle f\rangle \neq 0$ in their definition of regular. Strongly regular includes this condition. I'm not too familiar with the language of commutative algebra, so I've added the particular case I'm working on; can you tell me if it is possible to choose a regular sequence for this case? $\endgroup$ – Gyu Eun Lee Jun 17 '14 at 15:55
  • $\begingroup$ Thank you. I only knew the term "weakly regular" which includes $fR = R$. I think @zcn answered your question throughly. Good luck! $\endgroup$ – Youngsu Jun 19 '14 at 3:25

First, let me address the general question of ideals being generated by regular sequences: as Youngsu pointed out, such ideals are called complete intersections. In a Cohen-Macaulay ring (which the polynomial ring is), these are the same as ideals of the principal class - i.e. ideals which can be generated by $\DeclareMathOperator{\ht}{ht}$$\ht(I)$-many elements.

Now in your specific case, $I$ is the ideal of $2 \times 2$ minors of a generic $3 \times 2$ matrix. Such an ideal has height $2$ (e.g. by the Eagon-Northcott bound). As it happens, $I$ is Cohen-Macaulay (meaning $R/I$ is CM) but $I$ is not a complete intersection. One way to see this is via not the Koszul complex, but another famous complex, the Hilbert-Burch resolution:

$$0 \to R^2 \xrightarrow{\varphi} R^3 \xrightarrow{\psi} R \to R/I \to 0$$

where $\psi$ is the matrix $\begin{bmatrix} f & g & h \end{bmatrix}$ and $\varphi = \begin{bmatrix} s&t\\ -z&-r\\ x&y \end{bmatrix}$ is the matrix of relations among $f, g, h$ resulting from the equations $sf - zg + xh = 0 = tf - rg + yh$ - notice that the failure of $f, g, h$ to form a regular sequence gave rise exactly to the matrix $\varphi$. One immediately sees that this resolution is minimal (since all the entries of each map lies in the homogeneous maximal ideal), which implies that $I$ requires at least $3$ generators, and thus is not a complete intersection.

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    $\begingroup$ Just for the record: $t$-determinantal ideals of generic matrices (that is, matrices whose entries are indeterminates) are set-theoretic complete intersections iff $t=1$ or $m+n=2t$. $\endgroup$ – user26857 Jun 17 '14 at 20:24
  • $\begingroup$ Thanks! I'm not familiar with some of the terms in your answer, but I think I get the general picture of where I should be looking. $\endgroup$ – Gyu Eun Lee Jun 18 '14 at 0:58

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