## 10.135 Local complete intersections

The property of being a local complete intersection is an intrinsic property of a Noetherian local ring. This will be discussed in Divided Power Algebra, Section 23.8. However, for the moment we just define this property for finite type algebras over a field.

Definition 10.135.1. Let $k$ be a field. Let $S$ be a finite type $k$-algebra.

We say that $S$ is a *global complete intersection over $k$* if there exists a presentation $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ such that $\dim (S) = n - c$.

We say that $S$ is a *local complete intersection over $k$* if there exists a covering $\mathop{\mathrm{Spec}}(S) = \bigcup D(g_ i)$ such that each of the rings $S_{g_ i}$ is a global complete intersection over $k$.

We will also use the convention that the zero ring is a global complete intersection over $k$.

Suppose $S$ is a global complete intersection $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ as in Definition 10.135.1. For a maximal ideal $\mathfrak m \subset k[x_1, \ldots , x_ n]$ we have $\dim (k[x_1, \ldots , x_ n]_\mathfrak m) = n$ (Lemma 10.114.1). If $(f_1, \ldots , f_ c) \subset \mathfrak m$, then we conclude that $\dim (S_\mathfrak m) \geq n - c$ by Lemma 10.60.13. Since $\dim (S) = n - c$ by Definition 10.135.1 we conclude that $\dim (S_\mathfrak m) = n - c$ for all maximal ideals of $S$ and that $\mathop{\mathrm{Spec}}(S)$ is equidimensional (Topology, Definition 5.10.5) of dimension $n - c$, see Lemma 10.114.5. We will often use this without further mention.

Lemma 10.135.2. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $g \in S$.

If $S$ is a global complete intersection so is $S_ g$.

If $S$ is a local complete intersection so is $S_ g$.

**Proof.**
The second statement follows immediately from the first. Proof of the first statement. If $S_ g$ is the zero ring, then it is true. Assume $S_ g$ is nonzero. Write $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $n - c = \dim (S)$ as in Definition 10.135.1. By the remarks following the definition $\dim (S_ g) = n - c$. Let $g' \in k[x_1, \ldots , x_ n]$ be an element whose residue class corresponds to $g$. Then $S_ g = k[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, x_{n + 1}g' - 1)$ as desired.
$\square$

Lemma 10.135.3. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. If $S$ is a local complete intersection, then $S$ is a Cohen-Macaulay ring.

**Proof.**
Choose a maximal prime $\mathfrak m$ of $S$. We have to show that $S_\mathfrak m$ is Cohen-Macaulay. By assumption we may assume $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $\dim (S) = n - c$. Let $\mathfrak m' \subset k[x_1, \ldots , x_ n]$ be the maximal ideal corresponding to $\mathfrak m$. According to Proposition 10.114.2 the local ring $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ is regular local of dimension $n$. In particular it is Cohen-Macaulay by Lemma 10.106.3. By Lemma 10.60.13 applied $c$ times the local ring $S_{\mathfrak m} = k[x_1, \ldots , x_ n]_{\mathfrak m'}/(f_1, \ldots , f_ c)$ has dimension $\geq n - c$. By assumption $\dim (S_{\mathfrak m}) \leq n - c$. Thus we get equality. This implies that $f_1, \ldots , f_ c$ is a regular sequence in $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ and that $S_{\mathfrak m}$ is Cohen-Macaulay, see Proposition 10.103.4.
$\square$

The following is the technical key to the rest of the material in this section. An important feature of this lemma is that we may choose any presentation for the ring $S$, but that condition (1) does not depend on this choice.

Lemma 10.135.4. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. Choose any presentation $S = k[x_1, \ldots , x_ n]/I$. Let $\mathfrak q'$ be the prime of $k[x_1, \ldots , x_ n]$ corresponding to $\mathfrak q$. Set $c = \text{height}(\mathfrak q') - \text{height}(\mathfrak q)$, in other words $\dim _{\mathfrak q}(S) = n - c$ (see Lemma 10.116.4). The following are equivalent

There exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is a global complete intersection over $k$.

The ideal $I_{\mathfrak q'} \subset k[x_1, \ldots , x_ n]_{\mathfrak q'}$ can be generated by $c$ elements.

The conormal module $(I/I^2)_{\mathfrak q}$ can be generated by $c$ elements over $S_{\mathfrak q}$.

The conormal module $(I/I^2)_{\mathfrak q}$ is a free $S_{\mathfrak q}$-module of rank $c$.

The ideal $I_{\mathfrak q'}$ can be generated by a regular sequence in the regular local ring $k[x_1, \ldots , x_ n]_{\mathfrak q'}$.

In this case any $c$ elements of $I_{\mathfrak q'}$ which generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$ form a regular sequence in the local ring $k[x_1, \ldots , x_ n]_{\mathfrak q'}$.

**Proof.**
Set $R = k[x_1, \ldots , x_ n]_{\mathfrak q'}$. This is a Cohen-Macaulay local ring of dimension $\text{height}(\mathfrak q')$, see for example Lemma 10.135.3. Moreover, $\overline{R} = R/IR = R/I_{\mathfrak q'} = S_{\mathfrak q}$ is a quotient of dimension $\text{height}(\mathfrak q)$. Let $f_1, \ldots , f_ c \in I_{\mathfrak q'}$ be elements which generate $(I/I^2)_{\mathfrak q}$. By Lemma 10.20.1 we see that $f_1, \ldots , f_ c$ generate $I_{\mathfrak q'}$. Since the dimensions work out, we conclude by Proposition 10.103.4 that $f_1, \ldots , f_ c$ is a regular sequence in $R$. By Lemma 10.69.2 we see that $(I/I^2)_{\mathfrak q}$ is free. These arguments show that (2), (3), (4) are equivalent and that they imply the last statement of the lemma, and therefore they imply (5).

If (5) holds, say $I_{\mathfrak q'}$ is generated by a regular sequence of length $e$, then $\text{height}(\mathfrak q) = \dim (S_{\mathfrak q}) = \dim (k[x_1, \ldots , x_ n]_{\mathfrak q'}) - e = \text{height}(\mathfrak q') - e$ by dimension theory, see Section 10.60. We conclude that $e = c$. Thus (5) implies (2).

We continue with the notation introduced in the first paragraph. For each $f_ i$ we may find $d_ i \in k[x_1, \ldots , x_ n]$, $d_ i \not\in \mathfrak q'$ such that $f_ i' = d_ i f_ i \in k[x_1, \ldots , x_ n]$. Then it is still true that $I_{\mathfrak q'} = (f_1', \ldots , f_ c')R$. Hence there exists a $g' \in k[x_1, \ldots , x_ n]$, $g' \not\in \mathfrak q'$ such that $I_{g'} = (f_1', \ldots , f_ c')$. Moreover, pick $g'' \in k[x_1, \ldots , x_ n]$, $g'' \not\in \mathfrak q'$ such that $\dim (S_{g''}) = \dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S)$. By Lemma 10.116.4 this dimension is equal to $n - c$. Finally, set $g$ equal to the image of $g'g''$ in $S$. Then we see that

\[ S_ g \cong k[x_1, \ldots , x_ n, x_{n + 1}] / (f_1', \ldots , f_ c', x_{n + 1}g'g'' - 1) \]

and by our choice of $g''$ this ring has dimension $n - c$. Therefore it is a global complete intersection. Thus each of (2), (3), and (4) implies (1).

Assume (1). Let $S_ g \cong k[y_1, \ldots , y_ m]/(f_1, \ldots , f_ t)$ be a presentation of $S_ g$ as a global complete intersection. Write $J = (f_1, \ldots , f_ t)$. Let $\mathfrak q'' \subset k[y_1, \ldots , y_ m]$ be the prime corresponding to $\mathfrak qS_ g$. Note that $t = m - \dim (S_ g) = \text{height}(\mathfrak q'') - \text{height}(\mathfrak q)$, see Lemma 10.116.4 for the last equality. As seen in the proof of Lemma 10.135.3 (and also above) the elements $f_1, \ldots , f_ t$ form a regular sequence in the local ring $k[y_1, \ldots , y_ m]_{\mathfrak q''}$. By Lemma 10.69.2 we see that $(J/J^2)_{\mathfrak q}$ is free of rank $t$. By Lemma 10.134.16 we have

\[ J/J^2 \oplus S_ g^ n \cong (I/I^2)_ g \oplus S_ g^ m \]

Thus $(I/I^2)_{\mathfrak q}$ is free of rank $t + n - m = m - \dim (S_ g) + n - m = n - \dim (S_ g) = \text{height}(\mathfrak q') - \text{height}(\mathfrak q) = c$. Thus we obtain (4).
$\square$

The result of Lemma 10.135.4 suggests the following definition.

Definition 10.135.5. Let $k$ be a field. Let $S$ be a local $k$-algebra essentially of finite type over $k$. We say $S$ is a *complete intersection (over $k$)* if there exists a local $k$-algebra $R$ and elements $f_1, \ldots , f_ c \in \mathfrak m_ R$ such that

$R$ is essentially of finite type over $k$,

$R$ is a regular local ring,

$f_1, \ldots , f_ c$ form a regular sequence in $R$, and

$S \cong R/(f_1, \ldots , f_ c)$ as $k$-algebras.

By the Cohen structure theorem (see Theorem 10.160.8) any complete Noetherian local ring may be written as the quotient of some regular complete local ring. Hence we may use the definition above to define the notion of a complete intersection ring for any complete Noetherian local ring. We will discuss this in Divided Power Algebra, Section 23.8. In the meantime the following lemma shows that such a definition makes sense.

Lemma 10.135.6. Let $A \to B \to C$ be surjective local ring homomorphisms. Assume $A$ and $B$ are regular local rings. The following are equivalent

$\mathop{\mathrm{Ker}}(A \to C)$ is generated by a regular sequence,

$\mathop{\mathrm{Ker}}(A \to C)$ is generated by $\dim (A) - \dim (C)$ elements,

$\mathop{\mathrm{Ker}}(B \to C)$ is generated by a regular sequence, and

$\mathop{\mathrm{Ker}}(B \to C)$ is generated by $\dim (B) - \dim (C)$ elements.

**Proof.**
A regular local ring is Cohen-Macaulay, see Lemma 10.106.3. Hence the equivalences (1) $\Leftrightarrow $ (2) and (3) $\Leftrightarrow $ (4), see Proposition 10.103.4. By Lemma 10.106.4 the ideal $\mathop{\mathrm{Ker}}(A \to B)$ can be generated by $\dim (A) - \dim (B)$ elements. Hence we see that (4) implies (2).

It remains to show that (1) implies (4). We do this by induction on $\dim (A) - \dim (B)$. The case $\dim (A) - \dim (B) = 0$ is trivial. Assume $\dim (A) > \dim (B)$. Write $I = \mathop{\mathrm{Ker}}(A \to C)$ and $J = \mathop{\mathrm{Ker}}(A \to B)$. Note that $J \subset I$. Our assumption is that the minimal number of generators of $I$ is $\dim (A) - \dim (C)$. Let $\mathfrak m \subset A$ be the maximal ideal. Consider the maps

\[ J/ \mathfrak m J \to I / \mathfrak m I \to \mathfrak m /\mathfrak m^2 \]

By Lemma 10.106.4 and its proof the composition is injective. Take any element $x \in J$ which is not zero in $J /\mathfrak mJ$. By the above and Nakayama's lemma $x$ is an element of a minimal set of generators of $I$. Hence we may replace $A$ by $A/xA$ and $I$ by $I/xA$ which decreases both $\dim (A)$ and the minimal number of generators of $I$ by $1$. Thus we win.
$\square$

Lemma 10.135.7. Let $k$ be a field. Let $S$ be a local $k$-algebra essentially of finite type over $k$. The following are equivalent:

$S$ is a complete intersection over $k$,

for any surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by a regular sequence,

for some surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by $\dim (R) - \dim (S)$ elements,

there exists a global complete intersection $A$ over $k$ and a prime $\mathfrak a$ of $A$ such that $S \cong A_{\mathfrak a}$, and

there exists a local complete intersection $A$ over $k$ and a prime $\mathfrak a$ of $A$ such that $S \cong A_{\mathfrak a}$.

**Proof.**
It is clear that (2) implies (1) and (1) implies (3). It is also clear that (4) implies (5). Let us show that (3) implies (4). Thus we assume there exists a surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ such that the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by $\dim (R) - \dim (S)$ elements. We may write $R = (k[x_1, \ldots , x_ n]/J)_{\mathfrak q}$ for some $J \subset k[x_1, \ldots , x_ n]$ and some prime $\mathfrak q \subset k[x_1, \ldots , x_ n]$ with $J \subset \mathfrak q$. Let $I \subset k[x_1, \ldots , x_ n]$ be the kernel of the map $k[x_1, \ldots , x_ n] \to S$ so that $S \cong (k[x_1, \ldots , x_ n]/I)_{\mathfrak q}$. By assumption $(I/J)_{\mathfrak q}$ is generated by $\dim (R) - \dim (S)$ elements. We conclude that $I_{\mathfrak q}$ can be generated by $\dim (k[x_1, \ldots , x_ n]_{\mathfrak q}) - \dim (S)$ elements by Lemma 10.135.6. From Lemma 10.135.4 we see that for some $g \in k[x_1, \ldots , x_ n]$, $g \not\in \mathfrak q$ the algebra $(k[x_1, \ldots , x_ n]/I)_ g$ is a global complete intersection and $S$ is isomorphic to a local ring of it.

To finish the proof of the lemma we have to show that (5) implies (2). Assume (5) and let $\pi : R \to S$ be a surjection with $R$ a regular local $k$-algebra essentially of finite type over $k$. By assumption we have $S = A_{\mathfrak a}$ for some local complete intersection $A$ over $k$. Choose a presentation $R = (k[y_1, \ldots , y_ m]/J)_{\mathfrak q}$ with $J \subset \mathfrak q \subset k[y_1, \ldots , y_ m]$. We may and do assume that $J$ is the kernel of the map $k[y_1, \ldots , y_ m] \to R$. Let $I \subset k[y_1, \ldots , y_ m]$ be the kernel of the map $k[y_1, \ldots , y_ m] \to S = A_{\mathfrak a}$. Then $J \subset I$ and $(I/J)_{\mathfrak q}$ is the kernel of the surjection $\pi : R \to S$. So $S = (k[y_1, \ldots , y_ m]/I)_{\mathfrak q}$.

By Lemma 10.126.7 we see that there exist $g \in A$, $g \not\in \mathfrak a$ and $g' \in k[y_1, \ldots , y_ m]$, $g' \not\in \mathfrak q$ such that $A_ g \cong (k[y_1, \ldots , y_ m]/I)_{g'}$. After replacing $A$ by $A_ g$ and $k[y_1, \ldots , y_ m]$ by $k[y_1, \ldots , y_{m + 1}]$ we may assume that $A \cong k[y_1, \ldots , y_ m]/I$. Consider the surjective maps of local rings

\[ k[y_1, \ldots , y_ m]_{\mathfrak q} \to R \to S. \]

We have to show that the kernel of $R \to S$ is generated by a regular sequence. By Lemma 10.135.4 we know that $k[y_1, \ldots , y_ m]_{\mathfrak q} \to A_{\mathfrak a} = S$ has this property (as $A$ is a local complete intersection over $k$). We win by Lemma 10.135.6.
$\square$

Lemma 10.135.8. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. The following are equivalent:

The local ring $S_{\mathfrak q}$ is a complete intersection ring (Definition 10.135.5).

There exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is a local complete intersection over $k$.

There exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is a global complete intersection over $k$.

For any presentation $S = k[x_1, \ldots , x_ n]/I$ with $\mathfrak q' \subset k[x_1, \ldots , x_ n]$ corresponding to $\mathfrak q$ any of the equivalent conditions (1) – (5) of Lemma 10.135.4 hold.

**Proof.**
This is a combination of Lemmas 10.135.4 and 10.135.7 and the definitions.
$\square$

Lemma 10.135.9. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. The following are equivalent:

The ring $S$ is a local complete intersection over $k$.

All local rings of $S$ are complete intersection rings over $k$.

All localizations of $S$ at maximal ideals are complete intersection rings over $k$.

**Proof.**
This follows from Lemma 10.135.8, the fact that $\mathop{\mathrm{Spec}}(S)$ is quasi-compact and the definitions.
$\square$

The following lemma says that being a complete intersection is preserved under change of base field (in a strong sense).

Lemma 10.135.10. Let $K/k$ be a field extension. Let $S$ be a finite type algebra over $k$. Let $\mathfrak q_ K$ be a prime of $S_ K = K \otimes _ k S$ and let $\mathfrak q$ be the corresponding prime of $S$. Then $S_{\mathfrak q}$ is a complete intersection over $k$ (Definition 10.135.5) if and only if $(S_ K)_{\mathfrak q_ K}$ is a complete intersection over $K$.

**Proof.**
Choose a presentation $S = k[x_1, \ldots , x_ n]/I$. This gives a presentation $S_ K = K[x_1, \ldots , x_ n]/I_ K$ where $I_ K = K \otimes _ k I$. Let $\mathfrak q_ K' \subset K[x_1, \ldots , x_ n]$, resp. $\mathfrak q' \subset k[x_1, \ldots , x_ n]$ be the corresponding prime. We will show that the equivalent conditions of Lemma 10.135.4 hold for the pair $(S = k[x_1, \ldots , x_ n]/I, \mathfrak q)$ if and only if they hold for the pair $(S_ K = K[x_1, \ldots , x_ n]/I_ K, \mathfrak q_ K)$. The lemma will follow from this (see Lemma 10.135.8).

By Lemma 10.116.6 we have $\dim _{\mathfrak q} S = \dim _{\mathfrak q_ K} S_ K$. Hence the integer $c$ occurring in Lemma 10.135.4 is the same for the pair $(S = k[x_1, \ldots , x_ n]/I, \mathfrak q)$ as for the pair $(S_ K = K[x_1, \ldots , x_ n]/I_ K, \mathfrak q_ K)$. On the other hand we have

\begin{eqnarray*} I \otimes _{k[x_1, \ldots , x_ n]} \kappa (\mathfrak q') \otimes _{\kappa (\mathfrak q')} \kappa (\mathfrak q_ K') & = & I \otimes _{k[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K') \\ & = & I \otimes _{k[x_1, \ldots , x_ n]} K[x_1, \ldots , x_ n] \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K') \\ & = & (K \otimes _ k I) \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K') \\ & = & I_ K \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q'_ K). \end{eqnarray*}

Therefore, $\dim _{\kappa (\mathfrak q')} I \otimes _{k[x_1, \ldots , x_ n]} \kappa (\mathfrak q') = \dim _{\kappa (\mathfrak q'_ K)} I_ K \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K')$. Thus it follows from Nakayama's Lemma 10.20.1 that the minimal number of generators of $I_{\mathfrak q'}$ is the same as the minimal number of generators of $(I_ K)_{\mathfrak q'_ K}$. Thus the lemma follows from characterization (2) of Lemma 10.135.4.
$\square$

Lemma 10.135.11. Let $k \to K$ be a field extension. Let $S$ be a finite type $k$-algebra. Then $S$ is a local complete intersection over $k$ if and only if $S \otimes _ k K$ is a local complete intersection over $K$.

**Proof.**
This follows from a combination of Lemmas 10.135.9 and 10.135.10. But we also give a different proof here (based on the same principles).

Set $S' = S \otimes _ k K$. Let $\alpha : k[x_1, \ldots , x_ n] \to S$ be a presentation with kernel $I$. Let $\alpha ' : K[x_1, \ldots , x_ n] \to S'$ be the induced presentation with kernel $I'$.

Suppose that $S$ is a local complete intersection. Pick a prime $\mathfrak q \subset S'$. Denote $\mathfrak q'$ the corresponding prime of $K[x_1, \ldots , x_ n]$, $\mathfrak p$ the corresponding prime of $S$, and $\mathfrak p'$ the corresponding prime of $k[x_1, \ldots , x_ n]$. Consider the following diagram of Noetherian local rings

\[ \xymatrix{ S'_{\mathfrak q} & K[x_1, \ldots , x_ n]_{\mathfrak q'} \ar[l] \\ S_{\mathfrak p}\ar[u] & k[x_1, \ldots , x_ n]_{\mathfrak p'} \ar[u] \ar[l] } \]

By Lemma 10.135.4 we know that $S_{\mathfrak p}$ is cut out by some regular sequence $f_1, \ldots , f_ c$ in $k[x_1, \ldots , x_ n]_{\mathfrak p'}$. Since the right vertical arrow is flat we see that the images of $f_1, \ldots , f_ c$ form a regular sequence in $K[x_1, \ldots , x_ n]_{\mathfrak q'}$. Because tensoring with $K$ over $k$ is an exact functor we have $S'_{\mathfrak q} = K[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ c)$. Hence by Lemma 10.135.4 again we see that $S'$ is a local complete intersection in a neighbourhood of $\mathfrak q$. Since $\mathfrak q$ was arbitrary we see that $S'$ is a local complete intersection over $K$.

Suppose that $S'$ is a local complete intersection. Pick a maximal ideal $\mathfrak m$ of $S$. Let $\mathfrak m'$ denote the corresponding maximal ideal of $k[x_1, \ldots , x_ n]$. Denote $\kappa = \kappa (\mathfrak m)$ the residue field. By Remark 10.17.8 the primes of $S'$ lying over $\mathfrak m$ correspond to primes in $K \otimes _ k \kappa $. By the Hilbert-Nullstellensatz Theorem 10.34.1 we have $[\kappa : k] < \infty $. Hence $K \otimes _ k \kappa $ is finite nonzero over $K$. Hence $K \otimes _ k \kappa $ has a finite number $> 0$ of primes which are all maximal, each of which has a residue field finite over $K$ (see Section 10.53). Hence there are finitely many $> 0$ prime ideals $\mathfrak n \subset S'$ lying over $\mathfrak m$, each of which is maximal and has a residue field which is finite over $K$. Pick one, say $\mathfrak n \subset S'$, and let $\mathfrak n' \subset K[x_1, \ldots , x_ n]$ denote the corresponding prime ideal of $K[x_1, \ldots , x_ n]$. Note that since $V(\mathfrak mS')$ is finite, we see that $\mathfrak n$ is an isolated closed point of it, and we deduce that $\mathfrak mS'_{\mathfrak n}$ is an ideal of definition of $S'_{\mathfrak n}$. This implies that $\dim (S_{\mathfrak m}) = \dim (S'_{\mathfrak n})$ for example by Lemma 10.112.7. (This can also be seen using Lemma 10.116.6.) Consider the corresponding diagram of Noetherian local rings

\[ \xymatrix{ S'_{\mathfrak n} & K[x_1, \ldots , x_ n]_{\mathfrak n'} \ar[l] \\ S_{\mathfrak m}\ar[u] & k[x_1, \ldots , x_ n]_{\mathfrak m'} \ar[u] \ar[l] } \]

According to Lemma 10.134.8 we have $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' = \mathop{N\! L}\nolimits (\alpha ')$, in particular $I'/(I')^2 = I/I^2 \otimes _ S S'$. Thus $(I/I^2)_{\mathfrak m} \otimes _{S_{\mathfrak m}} \kappa $ and $(I'/(I')^2)_{\mathfrak n} \otimes _{S'_{\mathfrak n}} \kappa (\mathfrak n)$ have the same dimension. Since $(I'/(I')^2)_{\mathfrak n}$ is free of rank $n - \dim S'_{\mathfrak n}$ we deduce that $(I/I^2)_{\mathfrak m}$ can be generated by $n - \dim S'_{\mathfrak n} = n - \dim S_{\mathfrak m}$ elements. By Lemma 10.135.4 we see that $S$ is a local complete intersection in a neighbourhood of $\mathfrak m$. Since $\mathfrak m$ was any maximal ideal we conclude that $S$ is a local complete intersection.
$\square$

We end with a lemma which we will later use to prove that given ring maps $T \to A \to B$ where $B$ is syntomic over $T$, and $B$ is syntomic over $A$, then $A$ is syntomic over $T$.

Lemma 10.135.12. Let

\[ \xymatrix{ B & S \ar[l] \\ A \ar[u] & R \ar[l] \ar[u] } \]

be a commutative square of local rings. Assume

$R$ and $\overline{S} = S/\mathfrak m_ R S$ are regular local rings,

$A = R/I$ and $B = S/J$ for some ideals $I$, $J$,

$J \subset S$ and $\overline{J} = J/\mathfrak m_ R \cap J \subset \overline{S}$ are generated by regular sequences, and

$A \to B$ and $R \to S$ are flat.

Then $I$ is generated by a regular sequence.

**Proof.**
Set $\overline{B} = B/\mathfrak m_ RB = B/\mathfrak m_ AB$ so that $\overline{B} = \overline{S}/\overline{J}$. Let $f_1, \ldots , f_{\overline{c}} \in J$ be elements such that $\overline{f}_1, \ldots , \overline{f}_{\overline{c}} \in \overline{J}$ form a regular sequence generating $\overline{J}$. Note that $\overline{c} = \dim (\overline{S}) - \dim (\overline{B})$, see Lemma 10.135.6. By Lemma 10.99.3 the ring $S/(f_1, \ldots , f_{\overline{c}})$ is flat over $R$. Hence $S/(f_1, \ldots , f_{\overline{c}}) + IS$ is flat over $A$. The map $S/(f_1, \ldots , f_{\overline{c}}) + IS \to B$ is therefore a surjection of finite $S/IS$-modules flat over $A$ which is an isomorphism modulo $\mathfrak m_ A$, and hence an isomorphism by Lemma 10.99.1. In other words, $J = (f_1, \ldots , f_{\overline{c}}) + IS$.

By Lemma 10.135.6 again the ideal $J$ is generated by a regular sequence of $c = \dim (S) - \dim (B)$ elements. Hence $J/\mathfrak m_ SJ$ is a vector space of dimension $c$. By the description of $J$ above there exist $g_1, \ldots , g_{c - \overline{c}} \in I$ such that $J$ is generated by $f_1, \ldots , f_{\overline{c}}, g_1, \ldots , g_{c - \overline{c}}$ (use Nakayama's Lemma 10.20.1). Consider the ring $A' = R/(g_1, \ldots , g_{c - \overline{c}})$ and the surjection $A' \to A$. We see from the above that $B = S/(f_1, \ldots , f_{\overline{c}}, g_1, \ldots , g_{c - \overline{c}})$ is flat over $A'$ (as $S/(f_1, \ldots , f_{\overline{c}})$ is flat over $R$). Hence $A' \to B$ is injective (as it is faithfully flat, see Lemma 10.39.17). Since this map factors through $A$ we get $A' = A$. Note that $\dim (B) = \dim (A) + \dim (\overline{B})$, and $\dim (S) = \dim (R) + \dim (\overline{S})$, see Lemma 10.112.7. Hence $c - \overline{c} = \dim (R) -\dim (A)$ by elementary algebra. Thus $I = (g_1, \ldots , g_{c - \overline{c}})$ is generated by a regular sequence according to Lemma 10.135.6.
$\square$

## Comments (2)

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