**Proof.**
It is clear that (1) implies (2). We prove (2) implies (1) by induction on $r$. The case $r = 1$ is trivial. The case $r = 2$ says that if $a, b \in R$ are a regular sequence and $b$ is a nonzerodivisor, then $b, a$ is a regular sequence. This is clear because the kernel of $a : R/(b) \to R/(b)$ is isomorphic to the kernel of $b : R/(a) \to R/(a)$ if both $a$ and $b$ are nonzerodivisors. The case $r > 2$. Assume (2) holds and say we want to prove $f_{\sigma (1)}, \ldots , f_{\sigma (r)}$ is a regular sequence for some permutation $\sigma $. We already know that $f_{\sigma (1)}, \ldots , f_{\sigma (r - 1)}$ is a regular sequence by induction. Hence it suffices to show that $f_ s$ where $s = \sigma (r)$ is a nonzerodivisor modulo $f_1, \ldots , \hat f_ s, \ldots , f_ r$. If $s = r$ we are done. If $s < r$, then note that $f_ s$ and $f_ r$ are both nonzerodivisors in the ring $R/(f_1, \ldots , \hat f_ s, \ldots , f_{r - 1})$ (by induction hypothesis again). Since we know $f_ s, f_ r$ is a regular sequence in that ring we conclude by the case of sequence of length $2$ that $f_ r, f_ s$ is too.

Note that $R[x_1, \ldots , x_ r]/(f_1x_1, \ldots , f_ ix_ i)$ as an $R$-module is a direct sum of the modules

\[ R/I_ E \cdot x_1^{e_1} \ldots x_ r^{e_ r} \]

indexed by multi-indices $E = (e_1, \ldots , e_ r)$ where $I_ E$ is the ideal generated by $f_ j$ for $1 \leq j \leq i$ with $e_ j > 0$. Hence $f_{i + 1}x_ i$ is a nonzerodivisor on this if and only if $f_{i + 1}$ is a nonzerodivisor on $R/I_ E$ for all $E$. Taking $E$ with all positive entries, we see that $f_{i + 1}$ is a nonzerodivisor on $R/(f_1, \ldots , f_ i)$. Thus (3) implies (2). Conversely, if (2) holds, then any subsequence of $f_1, \ldots , f_ i, f_{i + 1}$ is a regular sequence in particular $f_{i + 1}$ is a nonzerodivisor on all $R/I_ E$. In this way we see that (2) implies (3).
$\square$

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