Let $A \to B$ be a local homomorphism of Noetherian complete local rings. A consequence of the Cohen structure theorem is that we can find a commutative diagram
of Noetherian complete local rings with $S \to B$ surjective, $A \to S$ flat, and $S/\mathfrak m_ A S$ a regular local ring. This follows from More on Algebra, Lemma 15.39.3. Let us (temporarily) say $A \to S \to B$ is a good factorization of $A \to B$ if $S$ is a Noetherian local ring, $A \to S \to B$ are local ring maps, $S \to B$ surjective, $A \to S$ flat, and $S/\mathfrak m_ AS$ regular. Let us say that $A \to B$ is a complete intersection homomorphism if there exists some good factorization $A \to S \to B$ such that the kernel of $S \to B$ is generated by a regular sequence. The following lemma shows this notion is independent of the choice of the diagram.
Lemma 23.9.1. Let $A \to B$ be a local homomorphism of Noetherian complete local rings. The following are equivalent
for some good factorization $A \to S \to B$ the kernel of $S \to B$ is generated by a regular sequence, and
for every good factorization $A \to S \to B$ the kernel of $S \to B$ is generated by a regular sequence.
Proof. Let $A \to S \to B$ be a good factorization. As $B$ is complete we obtain a factorization $A \to S^\wedge \to B$ where $S^\wedge $ is the completion of $S$. Note that this is also a good factorization: The ring map $S \to S^\wedge $ is flat (Algebra, Lemma 10.97.2), hence $A \to S^\wedge $ is flat. The ring $S^\wedge /\mathfrak m_ A S^\wedge = (S/\mathfrak m_ A S)^\wedge $ is regular since $S/\mathfrak m_ A S$ is regular (More on Algebra, Lemma 15.43.4). Let $f_1, \ldots , f_ r$ be a minimal sequence of generators of $\mathop{\mathrm{Ker}}(S \to B)$. 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 $S$ if and only if $f_1, \ldots , f_ r$ is a regular sequence in the completion $S^\wedge $ by Algebra, Lemma 10.68.5. Moreover, we have
because $B$ is $\mathfrak m_ B$-adically complete (first equality by Algebra, Lemma 10.97.1). Thus the kernel of $S \to B$ is generated by a regular sequence if and only if the kernel of $S^\wedge \to B$ is generated by a regular sequence. Hence it suffices to consider good factorizations where $S$ is complete.
Assume we have two factorizations $A \to S \to B$ and $A \to S' \to B$ with $S$ and $S'$ complete. By More on Algebra, Lemma 15.39.4 the ring $S \times _ B S'$ is a Noetherian complete local ring. Hence, using More on Algebra, Lemma 15.39.3 we can choose a good factorization $A \to S'' \to S \times _ B S'$ with $S''$ complete. Thus it suffices to show: If $A \to S' \to S \to B$ are comparable good factorizations, then $\mathop{\mathrm{Ker}}(S \to B)$ is generated by a regular sequence if and only if $\mathop{\mathrm{Ker}}(S' \to B)$ is generated by a regular sequence.
Let $A \to S' \to S \to B$ be comparable good factorizations. First, since $S'/\mathfrak m_ R S' \to S/\mathfrak m_ R S$ is a surjection of regular local rings, the kernel is generated by a regular sequence $\overline{x}_1, \ldots , \overline{x}_ c \in \mathfrak m_{S'}/\mathfrak m_ R S'$ which can be extended to a regular system of parameters for the regular local ring $S'/\mathfrak m_ R S'$, see (Algebra, Lemma 10.106.4). Set $I = \mathop{\mathrm{Ker}}(S' \to S)$. By flatness of $S$ over $R$ we have
Choose lifts $x_1, \ldots , x_ c \in I$. These lifts form a regular sequence generating $I$ as $S'$ is flat over $R$, see Algebra, Lemma 10.99.3.
We conclude that if also $\mathop{\mathrm{Ker}}(S \to B)$ is generated by a regular sequence, then so is $\mathop{\mathrm{Ker}}(S' \to B)$, see More on Algebra, Lemmas 15.30.13 and 15.30.7.
Conversely, assume that $J = \mathop{\mathrm{Ker}}(S' \to B)$ is generated by a regular sequence. Because the generators $x_1, \ldots , x_ c$ of $I$ map to linearly independent elements of $\mathfrak m_{S'}/\mathfrak m_{S'}^2$ we see that $I/\mathfrak m_{S'}I \to J/\mathfrak m_{S'}J$ is injective. Hence there exists a minimal system of generators $x_1, \ldots , x_ c, y_1, \ldots , y_ d$ for $J$. Then $x_1, \ldots , x_ c, y_1, \ldots , y_ d$ is a regular sequence and it follows that the images of $y_1, \ldots , y_ d$ in $S$ form a regular sequence generating $\mathop{\mathrm{Ker}}(S \to B)$. This finishes the proof of the lemma. $\square$
In the following proposition observe that the condition on vanishing of Tor's applies in particular if $B$ has finite tor dimension over $A$ and thus in particular if $B$ is flat over $A$.
Proposition 23.9.2. Let $A \to B$ be a local homomorphism of Noetherian local rings. Then the following are equivalent
$B$ is a complete intersection and $\text{Tor}^ A_ p(B, A/\mathfrak m_ A)$ is nonzero for only finitely many $p$,
$A$ is a complete intersection and $A^\wedge \to B^\wedge $ is a complete intersection homomorphism in the sense defined above.
Proof. Let $F_\bullet \to A/\mathfrak m_ A$ be a resolution by finite free $A$-modules. Observe that $\text{Tor}^ A_ p(B, A/\mathfrak m_ A)$ is the $p$th homology of the complex $F_\bullet \otimes _ A B$. Let $F_\bullet ^\wedge = F_\bullet \otimes _ A A^\wedge $ be the completion. Then $F_\bullet ^\wedge $ is a resolution of $A^\wedge /\mathfrak m_{A^\wedge }$ by finite free $A^\wedge $-modules (as $A \to A^\wedge $ is flat and completion on finite modules is exact, see Algebra, Lemmas 10.97.1 and 10.97.2). It follows that
By flatness of $B \to B^\wedge $ we conclude that
In this way we see that the condition in (1) on the local ring map $A \to B$ is equivalent to the same condition for the local ring map $A^\wedge \to B^\wedge $. Thus we may assume $A$ and $B$ are complete local Noetherian rings (since the other conditions are formulated in terms of the completions in any case).
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 $J = \mathop{\mathrm{Ker}}(S \to B)$. The proposition now follows from Lemma 23.7.6. $\square$
Remark 23.9.3. It appears difficult to define an good notion of “local complete intersection homomorphisms” for maps between general Noetherian rings. The reason is that, for a local Noetherian ring $A$, the fibres of $A \to A^\wedge $ are not local complete intersection rings. Thus, if $A \to B$ is a local homomorphism of local Noetherian rings, and the map of completions $A^\wedge \to B^\wedge $ is a complete intersection homomorphism in the sense defined above, then $(A_\mathfrak p)^\wedge \to (B_\mathfrak q)^\wedge $ is in general not a complete intersection homomorphism in the sense defined above. A solution can be had by working exclusively with excellent Noetherian rings. More generally, one could work with those Noetherian rings whose formal fibres are complete intersections, see [Rodicio-ci]. We will develop this theory in Dualizing Complexes, Section 47.23.
To finish of this section we compare the notion defined above with the notion introduced in More on Algebra, Section 23.8.
Lemma 23.9.4. Consider a commutative diagram of Noetherian local rings with $S \to B$ surjective, $A \to S$ flat, and $S/\mathfrak m_ A S$ a regular local ring. The following are equivalent
$\mathop{\mathrm{Ker}}(S \to B)$ is generated by a regular sequence, and
$A^\wedge \to B^\wedge $ is a complete intersection homomorphism as defined above.
\[ \xymatrix{ S \ar[r] & B \\ & A \ar[lu] \ar[u] } \]
Proof. Omitted. Hint: the proof is identical to the argument given in the first paragraph of the proof of Lemma 23.9.1. $\square$
Lemma 23.9.5. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. The following are equivalent
$A \to B$ is a local complete intersection in the sense of More on Algebra, Definition 15.33.2,
for every prime $\mathfrak q \subset B$ and with $\mathfrak p = A \cap \mathfrak q$ the ring map $(A_\mathfrak p)^\wedge \to (B_\mathfrak q)^\wedge $ is a complete intersection homomorphism in the sense defined above.
Proof. Choose a surjection $R = A[x_1, \ldots , x_ n] \to B$. Observe that $A \to R$ is flat with regular fibres. Let $I$ be the kernel of $R \to B$. Assume (2). Then we see that $I$ is locally generated by a regular sequence by Lemma 23.9.4 and Algebra, Lemma 10.68.6. In other words, (1) holds. Conversely, assume (1). Then after localizing on $R$ and $B$ we can assume that $I$ is generated by a Koszul regular sequence. By More on Algebra, Lemma 15.30.7 we find that $I$ is locally generated by a regular sequence. Hence (2) hold by Lemma 23.9.4. Some details omitted. $\square$
Lemma 23.9.6. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map such that the image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ contains all closed points of $\mathop{\mathrm{Spec}}(A)$. Then the following are equivalent
$B$ is a complete intersection and $A \to B$ has finite tor dimension,
$A$ is a complete intersection and $A \to B$ is a local complete intersection in the sense of More on Algebra, Definition 15.33.2.
Proof. This is a reformulation of Proposition 23.9.2 via Lemma 23.9.5. We omit the details. $\square$
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