Can I use Krull dimension to test if a sequence of polynomials is regular?

A sequence $(f_1, \ldots, f_n)$ of elements of a commutative ring $R$ is said to be regular if for each $i$, $f_i$ is not a zero divisor in $R/(f_1, \ldots, f_{i-1})$. Call a sequence dimension dropping$^1$ if for each $i$, we have $\dim R/(f_1,\ldots,f_i) = \dim R/(f_1,\ldots,f_{i-1}) - 1$ (where $\dim$ denotes the Krull dimension of a ring).

For the polynomial ring $R=\mathbb{F}_2[x_1, \ldots, x_m]$ is it true that a sequence of elements is regular if and only if it is dimension dropping?

I think one can deduce that for this ring, regular implies dimension dropping by Krull's principal ideal theorem. And I also know that dimension dropping does not imply regular for general rings: for example, if $A = \mathbb{F}_2[x,y]/(x^2,xy)$, then $y$ is a zero divisor in $A$, but $\dim A = 1 \neq 0 =\dim A/(y)$.

Note that the sequence $(x^2,xy,y)$ in $\mathbb{F}_2[x,y]$ is not a counterexample to dimension dropping implying regular for the polynomial ring: the initial portion $(x^2,xy)$ is neither regular or dimension dropping.

$^1$ This is not standard terminology; is there a standard name for this?

• I believe the terminology is a system of parameters in case $R$ is local. I believe the answer to your question is positive if $f_i$ are homogeneous or you are in the local setup. In a Cohen-Macaulay (local) ring, every system of parameters forms a regular sequence. – Youngsu May 16 '16 at 1:52
• Thanks @Youngsu! That's excellent news. If you put that down as answer I'd be happy to accept it. (And if you added a reference I'd be even more grateful.) I didn't mention it because I didn't know it would be relevant, but I only needed this for homogeneous elements. – Omar Antolín-Camarena May 16 '16 at 18:11
• Homogeneous is necessary as far as my knowledge goes since all these statements are local (or graded). A general reference like Matsumura's book or Eisenbud's has the material. – Youngsu May 17 '16 at 4:47

What you ask is true without any restrictions for nice rings like polynomial rings. So, assume you have a sequence $f_1,\ldots, f_{k-1}$ and $f_k$ is dimension dropping in your terminology. Then $\dim R/(f_1,\ldots, f_{k-1})=\dim R-k+1$ and this quotient ring is also equidimensional Cohen-Macaulay. Thus, all its associated primes are minimal and since $f_k$ is dimension dropping, it can not belong to any of the minimal primes. Thus it is a non-zero divisor.