Lemma 23.11.1. Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal of finite projective dimension over $R$. If $F \subset I/I^2$ is a direct summand isomorphic to $R/I$, then there exists a nonzerodivisor $x \in I$ such that the image of $x$ in $I/I^2$ generates $F$.

## 23.11 Freeness of the conormal module

Tate resolutions and derivations on them can be used to prove (stronger) versions of the results in this section, see [Iyengar]. Two more elementary references are [Vasconcelos] and [Ferrand-lci].

**Proof.**
By assumption we may choose a finite free resolution

Then $\varphi _1 : R^{\oplus n_1} \to R$ has rank $1$ and we see that $I$ contains a nonzerodivisor $y$ by Algebra, Proposition 10.102.9. Let $\mathfrak p_1, \ldots , \mathfrak p_ n$ be the associated primes of $R$, see Algebra, Lemma 10.63.5. Let $I^2 \subset J \subset I$ be an ideal such that $J/I^2 = F$. Then $J \not\subset \mathfrak p_ i$ for all $i$ as $y^2 \in J$ and $y^2 \not\in \mathfrak p_ i$, see Algebra, Lemma 10.63.9. By Nakayama's lemma (Algebra, Lemma 10.20.1) we have $J \not\subset \mathfrak m J + I^2$. By Algebra, Lemma 10.15.2 we can pick $x \in J$, $x \not\in \mathfrak m J + I^2$ and $x \not\in \mathfrak p_ i$ for $i = 1, \ldots , n$. Then $x$ is a nonzerodivisor and the image of $x$ in $I/I^2$ generates (by Nakayama's lemma) the summand $J/I^2 \cong R/I$. $\square$

Lemma 23.11.2. Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal of finite projective dimension over $R$. If $F \subset I/I^2$ is a direct summand free of rank $r$, then there exists a regular sequence $x_1, \ldots , x_ r \in I$ such that $x_1 \bmod I^2, \ldots , x_ r \bmod I^2$ generate $F$.

**Proof.**
If $r = 0$ there is nothing to prove. Assume $r > 0$. We may pick $x \in I$ such that $x$ is a nonzerodivisor and $x \bmod I^2$ generates a summand of $F$ isomorphic to $R/I$, see Lemma 23.11.1. Consider the ring $R' = R/(x)$ and the ideal $I' = I/(x)$. Of course $R'/I' = R/I$. The short exact sequence

splits because the map $I/xI \to I/I^2$ sends $xR/xI$ to a direct summand. Now $I/xI = I \otimes _ R^\mathbf {L} R'$ has finite projective dimension over $R'$, see More on Algebra, Lemmas 15.74.3 and 15.74.9. Hence the summand $I'$ has finite projective dimension over $R'$. On the other hand, we have the short exact sequence $0 \to xR/xI \to I/I^2 \to I'/(I')^2 \to 0$ and we conclude $I'/(I')^2$ has the free direct summand $F' = F/(R/I \cdot x)$ of rank $r - 1$. By induction on $r$ we may we pick a regular sequence $x'_2, \ldots , x'_ r \in I'$ such that there congruence classes freely generate $F'$. If $x_1 = x$ and $x_2, \ldots , x_ r$ are any elements lifting $x'_1, \ldots , x'_ r$ in $R$, then we see that the lemma holds. $\square$

Proposition 23.11.3. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal which has finite projective dimension and such that $I/I^2$ is finite locally free over $R/I$. Then $I$ is a regular ideal (More on Algebra, Definition 15.32.1).

**Proof.**
By Algebra, Lemma 10.68.6 it suffices to show that $I_\mathfrak p \subset R_\mathfrak p$ is generated by a regular sequence for every $\mathfrak p \supset I$. Thus we may assume $R$ is local. If $I/I^2$ has rank $r$, then by Lemma 23.11.2 we find a regular sequence $x_1, \ldots , x_ r \in I$ generating $I/I^2$. By Nakayama (Algebra, Lemma 10.20.1) we conclude that $I$ is generated by $x_1, \ldots , x_ r$.
$\square$

For any local complete intersection homomorphism $A \to B$ of rings, the naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ is perfect of tor-amplitude in $[-1, 0]$, see More on Algebra, Lemma 15.85.4. Using the above, we can show that this sometimes characterizes local complete intersection homomorphisms.

Lemma 23.11.4. Let $A \to B$ be a perfect (More on Algebra, Definition 15.82.1) ring homomorphism of Noetherian rings. Then the following are equivalent

$\mathop{N\! L}\nolimits _{B/A}$ has tor-amplitude in $[-1, 0]$,

$\mathop{N\! L}\nolimits _{B/A}$ is a perfect object of $D(B)$ with tor-amplitude in $[-1, 0]$, and

$A \to B$ is a local complete intersection (More on Algebra, Definition 15.33.2).

**Proof.**
Write $B = A[x_1, \ldots , x_ n]/I$. Then $\mathop{N\! L}\nolimits _{B/A}$ is represented by the complex

of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is finite free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see More on Algebra, Lemma 15.66.2. Since $I/I^2$ is a finite $B$-module and $B$ is Noetherian, this is true if and only if $I/I^2$ is a finite locally free $B$-module (Algebra, Lemma 10.78.2). Thus the equivalence of (1) and (2) is clear. Moreover, the equivalence of (1) and (3) also follows if we apply Proposition 23.11.3 (and the observation that a regular ideal is a Koszul regular ideal as well as a quasi-regular ideal, see More on Algebra, Section 15.32). $\square$

Lemma 23.11.5. Let $A \to B$ be a flat ring map of finite presentation. Then the following are equivalent

$\mathop{N\! L}\nolimits _{B/A}$ has tor-amplitude in $[-1, 0]$,

$\mathop{N\! L}\nolimits _{B/A}$ is a perfect object of $D(B)$ with tor-amplitude in $[-1, 0]$,

$A \to B$ is syntomic (Algebra, Definition 10.136.1), and

$A \to B$ is a local complete intersection (More on Algebra, Definition 15.33.2).

**Proof.**
The equivalence of (3) and (4) is More on Algebra, Lemma 15.33.5.

If $A \to B$ is syntomic, then we can find a cocartesian diagram

such that $A_0 \to B_0$ is syntomic and $A_0$ is Noetherian, see Algebra, Lemmas 10.127.18 and 10.168.9. By Lemma 23.11.4 we see that $\mathop{N\! L}\nolimits _{B_0/A_0}$ is perfect of tor-amplitude in $[-1, 0]$. By More on Algebra, Lemma 15.85.3 we conclude the same thing is true for $\mathop{N\! L}\nolimits _{B/A} = \mathop{N\! L}\nolimits _{B_0/A_0} \otimes _{B_0}^\mathbf {L} B$ (see also More on Algebra, Lemmas 15.66.13 and 15.74.9). This proves that (3) implies (2).

Assume (1). By More on Algebra, Lemma 15.85.3 for every ring map $A \to k$ where $k$ is a field, we see that $\mathop{N\! L}\nolimits _{B \otimes _ A k/k}$ has tor-amplitude in $[-1, 0]$ (see More on Algebra, Lemma 15.66.13). Hence by Lemma 23.11.4 we see that $k \to B \otimes _ A k$ is a local complete intersection homomorphism. Thus $A \to B$ is syntomic by definition. This proves (1) implies (3) and finishes the proof. $\square$

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