**Proof.**
Set $J = (f_1, \ldots , f_ c)$. We prove the first assertion by induction on $c$. We have to show that given any relation $\sum _{|I| = n} a_ I f^ I \in J^{n + 1}$ with $a_ I \in R$ we actually have $a_ I \in J$ for all multi-indices $I$. Since any element of $J^{n + 1}$ is of the form $\sum _{|I| = n} b_ I f^ I$ with $b_ I \in J$ we may assume, after replacing $a_ I$ by $a_ I - b_ I$, the relation reads $\sum _{|I| = n} a_ I f^ I = 0$. We can rewrite this as

\[ \sum \nolimits _{e = 0}^ n \left( \sum \nolimits _{|I'| = n - e} a_{I', e} f^{I'} \right) f_ c^ e = 0 \]

Here and below the “primed” multi-indices $I'$ are required to be of the form $I' = (i_1, \ldots , i_{c - 1}, 0)$. We will show by descending induction on $l \in \{ 0, \ldots , n\} $ that if we have a relation

\[ \sum \nolimits _{e = 0}^ l \left( \sum \nolimits _{|I'| = n - e} a_{I', e} f^{I'} \right) f_ c^ e = 0 \]

then $a_{I', e} \in J$ for all $I', e$. Namely, set $J' = (f_1, \ldots , f_{c-1})$. Observe that $\sum \nolimits _{|I'| = n - l} a_{I', l} f^{I'}$ is mapped into $(J')^{n - l + 1}$ by $f_ c^{l}$. By induction hypothesis (for the induction on $c$) we see that $f_ c^ l a_{I', l} \in J'$. Because $f_ c$ is not a zerodivisor on $R/J'$ (as $f_1, \ldots , f_ c$ is a regular sequence) we conclude that $a_{I', l} \in J'$. This allows us to rewrite the term $(\sum \nolimits _{|I'| = n - l} a_{I', l} f^{I'})f_ c^ l$ in the form $(\sum \nolimits _{|I'| = n - l + 1} f_ c b_{I', l - 1} f^{I'})f_ c^{l-1}$. This gives a new relation of the form

\[ \left(\sum \nolimits _{|I'| = n - l + 1} (a_{I', l-1} + f_ c b_{I', l - 1}) f^{I'}\right)f_ c^{l-1} + \sum \nolimits _{e = 0}^{l - 2} \left( \sum \nolimits _{|I'| = n - e} a_{I', e} f^{I'} \right) f_ c^ e = 0 \]

Now by the induction hypothesis (on $l$ this time) we see that all $a_{I', l-1} + f_ c b_{I', l - 1} \in J$ and all $a_{I', e} \in J$ for $e \leq l - 2$. This, combined with $a_{I', l} \in J' \subset J$ seen above, finishes the proof of the induction step.

The second assertion means that given any formal expression $F = \sum _{|I| = n} m_ I X^ I$, $m_ I \in M$ with $\sum m_ I f^ I \in J^{n + 1}M$, then all the coefficients $m_ I$ are in $J$. This is proved in exactly the same way as we prove the corresponding result for the first assertion above.
$\square$

## Comments (2)

Comment #919 by JuanPablo on

Comment #923 by Johan on