The Stacks project

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. $\square$

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. $\square$

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

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

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). $\square$

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

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. $\square$

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

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. $\square$

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

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

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

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. $\square$

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

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 10.18.5 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

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. $\square$

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. $\square$