The Stacks project

Proof. Let $k$ be the residue field of $A$. If the characteristic of $k$ is $p > 0$, then we denote $\Lambda $ a Cohen ring (Algebra, Definition 10.160.5) with residue field $k$ (Algebra, Lemma 10.160.6). If the characteristic of $k$ is $0$ we set $\Lambda = k$. Recall that $\Lambda [[x_1, \ldots , x_ n]]$ for any $n$ is formally smooth over $\mathbf{Z}$, resp. $\mathbf{Q}$ in the $\mathfrak m$-adic topology, see More on Algebra, Lemma 15.39.1. Fix a surjection $\Lambda [[x_1, \ldots , x_ n]] \to A$ as in the Cohen structure theorem (Algebra, Theorem 10.160.8).

Let $R \to A$ be a surjection from a regular local ring $R$. Let $f_1, \ldots , f_ r$ be a minimal sequence of generators of $\mathop{\mathrm{Ker}}(R \to A)$. We will use without further mention that an ideal in a Noetherian local ring is generated by a regular sequence if and only if any minimal set of generators is a regular sequence. Observe that $f_1, \ldots , f_ r$ is a regular sequence in $R$ if and only if $f_1, \ldots , f_ r$ is a regular sequence in the completion $R^\wedge $ by Algebra, Lemmas 10.68.5 and 10.97.2. Moreover, we have

because $A$ is $\mathfrak m_ A$-adically complete (first equality by Algebra, Lemma 10.97.1). Finally, the ring $R^\wedge $ is regular since $R$ is regular (More on Algebra, Lemma 15.43.4). Hence we may assume $R$ is complete.

If $R$ is complete we can choose a map $\Lambda [[x_1, \ldots , x_ n]] \to R$ lifting the given map $\Lambda [[x_1, \ldots , x_ n]] \to A$, see More on Algebra, Lemma 15.37.5. By adding some more variables $y_1, \ldots , y_ m$ mapping to generators of the kernel of $R \to A$ we may assume that $\Lambda [[x_1, \ldots , x_ n, y_1, \ldots , y_ m]] \to R$ is surjective (some details omitted). Then we can consider the commutative diagram

By Algebra, Lemma 10.135.6 we see that the condition for $R \to A$ is equivalent to the condition for the fixed chosen map $\Lambda [[x_1, \ldots , x_ n]] \to A$. This finishes the proof of the lemma. $\square$

Proof. The completion of $A$ is equal to $A^\wedge = R_\mathfrak p^\wedge /(f_1, \ldots , f_ r)R_\mathfrak p^\wedge $ because completion for finite modules over the Noetherian ring $R_\mathfrak p$ is exact (Algebra, Lemma 10.97.1). The image of the sequence $f_1, \ldots , f_ r$ in $R_\mathfrak p$ is a regular sequence by Algebra, Lemmas 10.97.2 and 10.68.5. Moreover, $R_\mathfrak p^\wedge $ is a regular local ring by More on Algebra, Lemma 15.43.4. Hence the result holds by our definition of complete intersection for complete local rings. $\square$

Proof. We may and do replace $R$ by its localization at $\mathfrak p$. Then $\mathfrak p = \mathfrak m$ is the maximal ideal of $R$ and $A = R/I$. Let $f_1, \ldots , f_ r \in I$ be a minimal sequence of generators. The completion of $A$ is equal to $A^\wedge = R^\wedge /(f_1, \ldots , f_ r)R^\wedge $ because completion for finite modules over the Noetherian ring $R_\mathfrak p$ is exact (Algebra, Lemma 10.97.1).

If (1) holds, then the image of the sequence $f_1, \ldots , f_ r$ in $R^\wedge $ is a regular sequence by assumption. Hence it is a regular sequence in $R$ by Algebra, Lemmas 10.97.2 and 10.68.5. Thus (1) implies (2).

Assume (3) holds. Set $c = \dim (R) - \dim (A)$ and let $f_1, \ldots , f_ c \in I$ map to generators of $I/I^2$. by Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $I = (f_1, \ldots , f_ c)$. Since $R$ is regular and hence Cohen-Macaulay (Algebra, Proposition 10.103.4) we see that $f_1, \ldots , f_ c$ is a regular sequence by Algebra, Proposition 10.103.4. Thus (3) implies (2). Finally, (2) implies (1) by Lemma 23.8.2. $\square$

Proof. Consider the diagram

Since the horizontal maps are faithfully flat (Algebra, Lemma 10.97.3) we conclude that the right vertical arrow is flat (for example by Algebra, Lemma 10.99.15). Moreover, we have $(B/\mathfrak m_ A B)^\wedge = B^\wedge /\mathfrak m_{A^\wedge } B^\wedge $ by Algebra, Lemma 10.97.1. Thus we may assume $A$ and $B$ are complete local Noetherian rings.

Assume $A$ and $B$ are complete local Noetherian rings. Choose a diagram

as in More on Algebra, Lemma 15.39.3. Let $I = \mathop{\mathrm{Ker}}(R \to A)$ and $J = \mathop{\mathrm{Ker}}(S \to B)$. Note that since $R/I = A \to B = S/J$ is flat the map $J/IS \otimes _ R R/\mathfrak m_ R \to J/J \cap \mathfrak m_ R S$ is an isomorphism. Hence a minimal system of generators of $J/IS$ maps to a minimal system of generators of $\mathop{\mathrm{Ker}}(S/\mathfrak m_ R S \to B/\mathfrak m_ A B)$. Finally, $S/\mathfrak m_ R S$ is a regular local ring.

Assume (1) holds, i.e., $J$ is generated by a regular sequence. Since $A = R/I \to B = S/J$ is flat we see Lemma 23.7.6 applies and we deduce that $I$ and $J/IS$ are generated by regular sequences. We have $\dim (B) = \dim (A) + \dim (B/\mathfrak m_ A B)$ and $\dim (S/IS) = \dim (A) + \dim (S/\mathfrak m_ R S)$ (Algebra, Lemma 10.112.7). Thus $J/IS$ is generated by

elements (Algebra, Lemma 10.60.13). It follows that $\mathop{\mathrm{Ker}}(S/\mathfrak m_ R S \to B/\mathfrak m_ A B)$ is generated by the same number of elements (see above). Hence $\mathop{\mathrm{Ker}}(S/\mathfrak m_ R S \to B/\mathfrak m_ A B)$ is generated by a regular sequence, see for example Lemma 23.8.3. In this way we see that (2) holds.

If (2) holds, then $I$ and $J/J \cap \mathfrak m_ RS$ are generated by regular sequences. Lifting these generators (see above), using flatness of $R/I \to S/IS$, and using Grothendieck's lemma (Algebra, Lemma 10.99.3) we find that $J/IS$ is generated by a regular sequence in $S/IS$. Thus Lemma 23.7.6 tells us that $J$ is generated by a regular sequence, whence (1) holds. $\square$

Proof. Choose a prime $\mathfrak q$ of $A^\wedge $ lying over $\mathfrak p$ (this is possible as $A \to A^\wedge $ is faithfully flat by Algebra, Lemma 10.97.3). Then $A_\mathfrak p \to (A^\wedge )_\mathfrak q$ is a flat local ring homomorphism. Thus by Proposition 23.8.4 we see that $A_\mathfrak p$ is a complete intersection if and only if $(A^\wedge )_\mathfrak q$ is a complete intersection. Thus it suffices to prove the lemma in case $A$ is complete (this is the key step of the proof).

Assume $A$ is complete. By definition we may write $A = R/(f_1, \ldots , f_ r)$ for some regular sequence $f_1, \ldots , f_ r$ in a regular local ring $R$. Let $\mathfrak q \subset R$ be the prime corresponding to $\mathfrak p$. Observe that $f_1, \ldots , f_ r \in \mathfrak q$ and that $A_\mathfrak p = R_\mathfrak q/(f_1, \ldots , f_ r)R_\mathfrak q$. Hence $A_\mathfrak p$ is a complete intersection by Lemma 23.8.2. $\square$

Proof. This follows immediately from Lemma 23.8.6 and the definitions. $\square$

Proof. Proof of (1). Let $k[x_1, \ldots , x_ n] \to S$ be a surjection. Let $\mathfrak p \subset k[x_1, \ldots , x_ n]$ be the prime ideal corresponding to $\mathfrak q$. Let $I \subset k[x_1, \ldots , x_ n]$ be the kernel of our surjection. Note that $k[x_1, \ldots , x_ n]_\mathfrak p \to S_\mathfrak q$ is surjective with kernel $I_\mathfrak p$. Observe that $k[x_1, \ldots , x_ n]$ is a regular ring by Algebra, Proposition 10.114.2. Hence the equivalence of the two notions in (1) follows by combining Lemma 23.8.3 with Algebra, Lemma 10.135.7.

Having proved (1) the equivalence in (2) follows from the definition and Algebra, Lemma 10.135.9. $\square$

Proof. Now that the definition makes sense this is a trivial reformulation of the (nontrivial) Proposition 23.8.4. $\square$