Lemma 115.4.1. Let $M$ be an $R$-module of finite presentation. For any surjection $\alpha : R^{\oplus n} \to M$ the kernel of $\alpha $ is a finite $R$-module.

## 115.4 Obsolete algebra lemmas

**Proof.**
This is a special case of Algebra, Lemma 10.5.3.
$\square$

Lemma 115.4.2. Let $\varphi : R \to S$ be a ring map. If

for any $x \in S$ there exists $n > 0$ such that $x^ n$ is in the image of $\varphi $, and

for any $x \in \mathop{\mathrm{Ker}}(\varphi )$ there exists $n > 0$ such that $x^ n = 0$,

then $\varphi $ induces a homeomorphism on spectra. Given a prime number $p$ such that

$S$ is generated as an $R$-algebra by elements $x$ such that there exists an $n > 0$ with $x^{p^ n} \in \varphi (R)$ and $p^ nx \in \varphi (R)$, and

the kernel of $\varphi $ is generated by nilpotent elements,

then (1) and (2) hold, and for any ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ also satisfies (a), (b), (1), and (2) and in particular induces a homeomorphism on spectra.

**Proof.**
This is a combination of Algebra, Lemmas 10.46.3 and 10.46.7.
$\square$

The following technical lemma says that you can lift any sequence of relations from a fibre to the whole space of a ring map which is essentially of finite type, in a suitable sense.

Lemma 115.4.3. Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$. Assume $S_{\mathfrak q}$ is essentially of finite type over $R_\mathfrak p$. Assume given

an integer $n \geq 0$,

a prime $\mathfrak a \subset \kappa (\mathfrak p)[x_1, \ldots , x_ n]$,

a surjective $\kappa (\mathfrak p)$-homomorphism

\[ \psi : (\kappa (\mathfrak p)[x_1, \ldots , x_ n])_{\mathfrak a} \longrightarrow S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}, \]and

elements $\overline{f}_1, \ldots , \overline{f}_ e$ in $\mathop{\mathrm{Ker}}(\psi )$.

Then there exist

an integer $m \geq 0$,

and element $g \in S$, $g \not\in \mathfrak q$,

a map

\[ \Psi : R[x_1, \ldots , x_ n, x_{n + 1}, \ldots , x_{n + m}] \longrightarrow S_ g, \]and

elements $f_1, \ldots , f_ e, f_{e + 1}, \ldots , f_{e + m}$ of $\mathop{\mathrm{Ker}}(\Psi )$

such that

the following diagram commutes

\[ \xymatrix{ R[x_1, \ldots , x_{n + m}] \ar[d]_\Psi \ar[rr]_-{x_{n + j} \mapsto 0} & & (\kappa (\mathfrak p)[x_1, \ldots , x_ n])_{\mathfrak a} \ar[d]^\psi \\ S_ g \ar[rr] & & S_{\mathfrak q}/\mathfrak p S_{\mathfrak q} }, \]the element $f_ i$, $i \leq n$ maps to a unit times $\overline{f}_ i$ in the local ring

\[ (\kappa (\mathfrak p)[x_1, \ldots , x_{n + m}])_{ (\mathfrak a, x_{n + 1}, \ldots , x_{n + m})}, \]the element $f_{e + j}$ maps to a unit times $x_{n + j}$ in the same local ring, and

the induced map $R[x_1, \ldots , x_{n + m}]_{\mathfrak b} \to S_{\mathfrak q}$ is surjective, where $\mathfrak b = \Psi ^{-1}(\mathfrak qS_ g)$.

**Proof.**
We claim that it suffices to prove the lemma in case $R$ and $S$ are local with maximal ideals $\mathfrak p$ and $\mathfrak q$. Namely, suppose we have constructed

and $f_1', \ldots , f_{e + m}' \in R_{\mathfrak p}[x_1, \ldots , x_{n + m}]$ with all the required properties. Then there exists an element $f \in R$, $f \not\in \mathfrak p$ such that each $ff_ k'$ comes from an element $f_ k \in R[x_1, \ldots , x_{n + m}]$. Moreover, for a suitable $g \in S$, $g \not\in \mathfrak q$ the elements $\Psi '(x_ i)$ are the image of elements $y_ i \in S_ g$. Let $\Psi $ be the $R$-algebra map defined by the rule $\Psi (x_ i) = y_ i$. Since $\Psi (f_ i)$ is zero in the localization $S_{\mathfrak q}$ we may after possibly replacing $g$ assume that $\Psi (f_ i) = 0$. This proves the claim.

Thus we may assume $R$ and $S$ are local with maximal ideals $\mathfrak p$ and $\mathfrak q$. Pick $y_1, \ldots , y_ n \in S$ such that $y_ i \bmod \mathfrak pS = \psi (x_ i)$. Let $y_{n + 1}, \ldots , y_{n + m} \in S$ be elements which generate an $R$-subalgebra of which $S$ is the localization. These exist by the assumption that $S$ is essentially of finite type over $R$. Since $\psi $ is surjective we may write $y_{n + j} \bmod \mathfrak pS = \psi (h_ j)$ for some $h_ j \in \kappa (\mathfrak p)[x_1, \ldots , x_ n]_{\mathfrak a}$. Write $h_ j = g_ j/d$, $g_ j \in \kappa (\mathfrak p)[x_1, \ldots , x_ n]$ for some common denominator $d \in \kappa (\mathfrak p)[x_1, \ldots , x_ n]$, $d \not\in \mathfrak a$. Choose lifts $G_ j, D \in R[x_1, \ldots , x_ n]$ of $g_ j$ and $d$. Set $y_{n + j}' = D(y_1, \ldots , y_ n) y_{n + j} - G_ j(y_1, \ldots , y_ n)$. By construction $y_{n + j}' \in \mathfrak p S$. It is clear that $y_1, \ldots , y_ n, y_ n', \ldots , y_{n + m}'$ generate an $R$-subalgebra of $S$ whose localization is $S$. We define

to be the map that sends $x_ i$ to $y_ i$ for $i = 1, \ldots , n$ and $x_{n + j}$ to $y'_{n + j}$ for $j = 1, \ldots , m$. Properties (1) and (4) are clear by construction. Moreover the ideal $\mathfrak b$ maps onto the ideal $(\mathfrak a, x_{n + 1}, \ldots , x_{n + m})$ in the polynomial ring $\kappa (\mathfrak p)[x_1, \ldots , x_{n + m}]$.

Denote $J = \mathop{\mathrm{Ker}}(\Psi )$. We have a short exact sequence

The surjectivity comes from our choice of $y_1, \ldots , y_ n, y_ n', \ldots , y_{n + m}'$ above. This implies that

is exact. By construction $x_ i$ maps to $\psi (x_ i)$ and $x_{n + j}$ maps to zero under the last map. Thus it is easy to choose $f_ i$ as in (2) and (3) of the lemma. $\square$

Remark 115.4.4 (Projective resolutions). Let $R$ be a ring. For any set $S$ we let $F(S)$ denote the free $R$-module on $S$. Then any left $R$-module has the following two step resolution

The first map is given by the rule

Lemma 115.4.5. Let $S$ be a multiplicative set of $A$. Then the map

induced by the canonical ring map $A \to S^{-1}A$ is a homeomorphism onto its image and $\mathop{\mathrm{Im}}(f) = \{ \mathfrak p \in \mathop{\mathrm{Spec}}(A) : \mathfrak p\cap S = \emptyset \} $.

**Proof.**
This is a duplicate of Algebra, Lemma 10.17.5.
$\square$

Lemma 115.4.6. Let $A \to B$ be a finite type, flat ring map with $A$ an integral domain. Then $B$ is a finitely presented $A$-algebra.

**Proof.**
Special case of More on Flatness, Proposition 38.13.10.
$\square$

Lemma 115.4.7. Let $R$ be a domain with fraction field $K$. Let $S = R[x_1, \ldots , x_ n]$ be a polynomial ring over $R$. Let $M$ be a finite $S$-module. Assume that $M$ is flat over $R$. If for every subring $R \subset R' \subset K$, $R \not= R'$ the module $M \otimes _ R R'$ is finitely presented over $S \otimes _ R R'$, then $M$ is finitely presented over $S$.

**Proof.**
This lemma is true because $M$ is finitely presented even without the assumption that $M \otimes _ R R'$ is finitely presented for every $R'$ as in the statement of the lemma. This follows from More on Flatness, Proposition 38.13.10. Originally this lemma had an erroneous proof (thanks to Ofer Gabber for finding the gap) and was used in an alternative proof of the proposition cited. To reinstate this lemma, we need a correct argument in case $R$ is a local normal domain using only results from the chapters on commutative algebra; please email stacks.project@gmail.com if you have an argument.
$\square$

Lemma 115.4.8. Let $A \to B$ be a ring map. Let $f \in B$. Assume that

$A \to B$ is flat,

$f$ is a nonzerodivisor, and

$A \to B/fB$ is flat.

Then for every ideal $I \subset A$ the map $f : B/IB \to B/IB$ is injective.

**Proof.**
Note that $IB = I \otimes _ A B$ and $I(B/fB) = I \otimes _ A B/fB$ by the flatness of $B$ and $B/fB$ over $A$. In particular $IB/fIB \cong I \otimes _ A B/fB$ maps injectively into $B/fB$. Hence the result follows from the snake lemma applied to the diagram

with exact rows. $\square$

Lemma 115.4.9. If $R \to S$ is a faithfully flat ring map then for every $R$-module $M$ the map $M \to S \otimes _ R M$, $x \mapsto 1 \otimes x$ is injective.

**Proof.**
This lemma is a duplicate of Algebra, Lemma 10.82.11.
$\square$

Remark 115.4.10. This reference/tag used to refer to a Section in the chapter Smoothing Ring Maps, but the material has since been subsumed in Algebra, Section 10.127.

Lemma 115.4.11. Let $(R, \mathfrak m)$ be a reduced Noetherian local ring of dimension $1$ and let $x \in \mathfrak m$ be a nonzerodivisor. Let $\mathfrak q_1, \ldots , \mathfrak q_ r$ be the minimal primes of $R$. Then

**Proof.**
Special (very easy) case of Chow Homology, Lemma 42.3.2.
$\square$

Lemma 115.4.12. Let $A$ be a Noetherian local normal domain of dimension $2$. For $f \in \mathfrak m$ nonzero denote $\text{div}(f) = \sum n_ i (\mathfrak p_ i)$ the divisor associated to $f$ on the punctured spectrum of $A$. We set $|f| = \sum n_ i$. There exist integers $N$ and $M$ such that $|f + g| \leq M$ for all $g \in \mathfrak m^ N$.

**Proof.**
Pick $h \in \mathfrak m$ such that $f, h$ is a regular sequence in $A$ (this follows from Algebra, Lemmas 10.157.4 and 10.72.7). We will prove the lemma with $M = \text{length}_ A(A/(f, h))$ and with $N$ any integer such that $\mathfrak m^ N \subset (f, h)$. Such an integer $N$ exists because $\sqrt{(f, h)} = \mathfrak m$. Note that $M = \text{length}_ A(A/(f + g, h))$ for all $g \in \mathfrak m^ N$ because $(f, h) = (f + g, h)$. This moreover implies that $f + g, h$ is a regular sequence in $A$ too, see Algebra, Lemma 10.104.2. Now suppose that $\text{div}(f + g ) = \sum m_ j (\mathfrak q_ j)$. Then consider the map

where $\mathfrak q_ j^{(m_ j)}$ is the symbolic power, see Algebra, Section 10.64. Since $A$ is normal, we see that $A_{\mathfrak q_ i}$ is a discrete valuation ring and hence

Since $V(f + g, h) = \{ \mathfrak m\} $ this implies that $c$ becomes an isomorphism on inverting $h$ (small detail omitted). Since $h$ is a nonzerodivisor on $A/(f + g)$ we see that the length of $A/(f + g, h)$ equals the Herbrand quotient $e_ A(A/(f + g), 0, h)$ as defined in Chow Homology, Section 42.2. Similarly the length of $A/(h, \mathfrak q_ j^{(m_ j)})$ equals $e_ A(A/\mathfrak q_ j^{(m_ j)}, 0, h)$. Then we have

The equalities follow from Chow Homology, Lemmas 42.2.3 and 42.2.4 using in particular that the cokernel of $c$ has finite length as discussed above. It is straightforward to prove that $e_ A(\mathfrak q^{(m)}/\mathfrak q^{(m + 1)}, 0, h)$ is at least $1$ by Nakayama's lemma. This finishes the proof of the lemma. $\square$

Lemma 115.4.13. Let $A \to B$ be a flat local homomorphism of Noetherian local rings. If $A$ and $B/\mathfrak m_ A B$ are Gorenstein, then $B$ is Gorenstein.

**Proof.**
Follows immediately from Dualizing Complexes, Lemma 47.21.8.
$\square$

Lemma 115.4.14. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $s$ be an integer. Assume

$A$ has a dualizing complex,

if $\mathfrak p \not\in V(I)$ and $V(\mathfrak p) \cap V(I) \not= \{ \mathfrak m\} $, then $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim (A/\mathfrak p) > s$.

Then there exists an $n > 0$ and an ideal $J \subset A$ with $V(J) \cap V(I) = \{ \mathfrak m\} $ such that $JI^ n$ annihilates $H^ i_\mathfrak m(M)$ for $i \leq s$.

**Proof.**
According to Local Cohomology, Lemma 51.9.4 we have to show this for the finite $A$-module $E^ i = \text{Ext}^{-i}_ A(M, \omega _ A^\bullet )$ for $i \leq s$. The support $Z$ of $E^0 \oplus \ldots \oplus E^ s$ is closed in $\mathop{\mathrm{Spec}}(A)$ and does not contain any prime as in (2). Hence it is contained in $V(JI^ n)$ for some $J$ as in the statement of the lemma.
$\square$

Lemma 115.4.15. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. Assume

$A$ has a dualizing complex,

$\text{cd}(A, I) \leq d$,

if $\mathfrak p \not\in V(I)$ then $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) > s$ or $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim (A/\mathfrak p) > d + s$.

Then the assumptions of Algebraic and Formal Geometry, Lemma 52.10.4 hold for $A, I, \mathfrak m, M$ and $H^ i_\mathfrak m(M) \to \mathop{\mathrm{lim}}\nolimits H^ i_\mathfrak m(M/I^ nM)$ is an isomorphism for $i \leq s$ and these modules are annihilated by a power of $I$.

**Proof.**
The assumptions of Algebraic and Formal Geometry, Lemma 52.10.4 by the more general Algebraic and Formal Geometry, Lemma 52.10.5. Then the conclusion of Algebraic and Formal Geometry, Lemma 52.10.4 gives the second statement.
$\square$

Lemma 115.4.16. In Algebraic and Formal Geometry, Situation 52.10.1 we have $H^ s_\mathfrak a(M) = \mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM)$.

**Proof.**
This is immediate from Algebraic and Formal Geometry, Theorem 52.10.8. The original version of this lemma, which had additional assumptions, was superseded by the this theorem.
$\square$

Lemma 115.4.17. Let $A$ be a Noetherian ring. Let $f \in \mathfrak a$ be an element of an ideal of $A$. Let $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

$A$ has a dualizing complex and is complete with respect to $f$,

$A_ f$ is $(S_2)$ and for every minimal prime $\mathfrak p \subset A$, $f \not\in \mathfrak p$ and $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$ we have $\dim ((A/\mathfrak p)_\mathfrak q) \geq 3$.

Then the completion functor

is fully faithful on the full subcategory of finite locally free objects.

**Proof.**
This lemma is a special case of Algebraic and Formal Geometry, Lemma 52.15.6.
$\square$

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