Let $I \subset A$ be generated by a regular sequence $x_1, \dots, x_r$. Then, $I/I^2$ is a free $A/I$ module of rank $r$.

My question is regarding the proof of Proposition 12.2.16 in Vakil's FOAG notes.

Precisely, the proposition says (once reduced to the local case) that if $A$ is a noetherian local ring, $I = (x_1, \dots, x_r)$ is generated by a regular sequence, then the map $\gamma: (A/I)^r \to I/I^2$ by $$\gamma(a_1, \dots, a_r) = a_1x_1 + \cdots +a_rx_r$$ is an isomorphism.

Surjectivity is clear. If $(\overline{a_1}, \dots, \overline{a_r}) \in \ker \gamma$, we show $a_i \in I$ to conclude that $\overline{a_i} = 0$. Now, it is claimed that $a_rx_r \equiv 0 \operatorname{mod} \;(x_1, \dots, x_{r - 1})$. I don't see why this is the case.

The rest of the proof is clear: since $x_i$ form a regular sequence, this implies $a_r \in (x_1, \dots, x_{r - 1}) \subseteq I$. The result follows from the fact that the we can switch $a_i$ and $a_r$ for any $i$ since $A$ is local(the $x_i$ can be reordered).

Why is $a_rx_r \equiv 0 \operatorname{mod}\; (x_1, \dots, x_{r-1})$? Do we take the $x_1, \cdots x_{r - 1}$ as an $I/I^2$-regular sequence somehow? This seems to use the fact that $\sum_i a_ix_i \equiv 0 \operatorname{mod} I^2$ since $(\overline{a_1}, \dots, \overline{a_r}) \in \ker \gamma$. I am not sure how to apply this though.

Can someone help me fill this seeming gap?

Thank you!

Solution 1:

The Proposition in Ravi Vakil's notes is a special case of the well-known fact that regular sequences are quasi-regular. You might have a look at [1,2] for more details.

What you can do is replace $a_1,\dots,a_r$ with elements $a_1',\dots,a_r'$ such that $$\sum a_i'x_i=0. $$ In fact, since $(\overline{a_1},\dots,\overline{a_r})\in \ker\gamma$, we have $\sum a_ix_i\in I^2$. A typical element of $I^2$ can be written as $\sum_{ij}b_{ij}x_ix_j$. Hence, there exist $c_1,\dots,c_r\in I$ such that $$\sum a_ix_i = \sum c_ix_i$$ We let $a_i' = a_i-c_i$. Then, in $A/I$, we have $\overline{a_i}=\overline{a_i'}$. Meanwhile, we can run the argument in Vakil's notes for $a_i'$ (using $\sum a_i'x_i=0$) to get $\overline{a_i'} = 0$ in $A/I$, as intended.

1. Bruns, W. and Herzog, J. Cohen-Macaulay Rings. Cambridge University Press. 1993. p. 6

2. Matsumura, H. Commutative Ring Theory. Cambridge University Press. 2006. Section 16.