The Stacks project

Proof. Since $\dim (R) = 1$, we have $\dim (R/x^ n R) = 0$ and so $\text{length}_ R(R/x^ n R)$ is finite for each $n$ (Algebra, Lemma 10.62.3). To show the lemma we will induct on $n$. Since $x^0 R = R$, we have that $P(0) = \text{length}_ R(R/x^0R) = \text{length}_ R 0 = 0$. The statement also holds for $n = 1$. Now let $n \geq 2$ and suppose the statement holds for $n - 1$. The following sequence is exact

where $x$ denotes the multiplication by $x$ map. Since length is additive (Algebra, Lemma 10.52.3), we have that $P(n) \leq P(n - 1) + P(1)$. By induction $P(n - 1) \leq (n - 1)P(1)$, whence $P(n) \leq nP(1)$. This proves the induction step.

If $x$ is a nonzerodivisor, then the displayed exact sequence above is exact on the left also. Hence we get $P(n) = P(n - 1) + P(1)$ for all $n \geq 1$. $\square$

Proof. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the minimal prime ideals of $R$. Set $R' = R/\sqrt{0} = R/(\bigcap _{i = 1}^ t \mathfrak p_ i)$. We claim it suffices to prove the lemma for $R'$. Namely, it is clear that $R'$ has $t$ minimal primes too and $\text{length}_{R'}(R'/xR') = \text{length}_ R(R'/xR')$ is less than $\text{length}_ R(R/xR)$ as there is a surjection $R/xR \to R'/xR'$. Thus we may assume $R$ is reduced.

Assume $R$ is reduced with minimal primes $\mathfrak p_1, \ldots , \mathfrak p_ t$. This means there is an exact sequence

Here $Q$ is the cokernel of the first map. Write $M = \prod _{i = 1}^ t R/\mathfrak p_ i$. Localizing at $\mathfrak p_ j$ we see that

is surjective. Thus $Q_{\mathfrak p_ j} = 0$ for all $j$. We conclude that $\text{Supp}(Q) = \{ \mathfrak m\} $ as $\mathfrak m$ is the only prime of $R$ different from the $\mathfrak p_ i$. It follows that $Q$ has finite length (Algebra, Lemma 10.62.3). Since $\text{Supp}(Q) = \{ \mathfrak m\} $ we can pick an $n \gg 0$ such that $x^ n$ acts as $0$ on $Q$ (Algebra, Lemma 10.62.4). Now consider the diagram

where the vertical maps are multiplication by $x^ n$. This is injective on $R$ and on $M$ since $x$ is not contained in any of the $\mathfrak p_ i$. By the snake lemma (Algebra, Lemma 10.4.1), the following sequence is exact:

Hence we find that $\text{length}_ R(R/x^ nR) = \text{length}_ R(M/x^ nM)$ for large enough $n$. Writing $R_ i = R/\mathfrak p_ i$ we see that $\text{length}(M/x^ nM) = \sum _{i = 1}^ t \text{length}_ R(R_ i/x^ nR_ i)$. Applying Lemma 15.125.1 and the fact that $x$ is a nonzerodivisor on $R$ and $R_ i$, we conclude that

Since $\text{length}_{R_ i}(R_ i/x R_ i) \geq 1$ the lemma is proved. $\square$

Proof. We have $\dim (R/fR) = d - 1$ by Algebra, Lemma 10.60.13. Choose a system of parameters $\overline{g}_1, \ldots , \overline{g}_{d - 1}$ in $R/fR$ (Algebra, Proposition 10.60.9) and take lifts $g_1, \ldots , g_{d - 1}$ in $R$. It is straightforward to see that $f, g_1, \ldots , g_{d - 1}$ is a system of parameters in $R$. Then $f, g_1^ k, \ldots , g_{d - 1}^ k$ is also a system of parameters and the proof is complete. $\square$

Proof. By Lemma 15.125.3 there exists a $g \in \mathfrak m$ such that $f, g$ is a system of parameters for $R$. Then $\mathfrak m = \sqrt{(f, g)}$. Thus there exists an $n$ such that $\mathfrak m^ n \subset (f, g)$, see Algebra, Lemma 10.32.5. We claim that $N = n + 1$ works. Namely, let $h \in \mathfrak m^ N$. By our choice of $N$ we can write $h = af + bg$ with $a, b \in \mathfrak m$. Thus

because $1 + a$ is a unit in $R$. This proves the equality of lengths and the fact that $f + h, g$ is a system of parameters. $\square$

Proof. Let $f \in \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ be a nonzero element. We will modify $f$ slightly to obtain an element that generates a radical ideal. The localization $R_\mathfrak p$ of $R$ at each height one prime ideal $\mathfrak p$ is a discrete valuation ring, see Algebra, Lemma 10.119.7 or Algebra, Lemma 10.157.4. We denote by $\text{ord}_\mathfrak p(f)$ the corresponding valuation of $f$ in $R_{\mathfrak p}$. Let $\mathfrak q_1, \ldots , \mathfrak q_ s$ be the distinct height one prime ideals containing $f$. Write $\text{ord}_{\mathfrak q_ j}(f) = m_ j \geq 1$ for each $j$. Then we define $\text{div}(f) = \sum _{j = 1}^ s m_ j\mathfrak q_ j$ as a formal linear combination of height one primes with integer coefficients. Note for later use that each of the primes $\mathfrak p_ i$ occurs among the primes $\mathfrak q_ j$. The ring $R/fR$ is reduced if and only if $m_ j = 1$ for $j = 1, \ldots , s$. Namely, if $m_ j$ is $1$ then $(R/fR)\mathfrak q_ j$ is reduced and $R/fR \subset \prod (R/fR)_{\mathfrak q_ j}$ as $\mathfrak q_1, \ldots , \mathfrak q_ j$ are the associated primes of $R/fR$, see Algebra, Lemmas 10.63.19 and 10.157.6.

Choose and fix $g$ and $N$ as in Lemma 15.125.4. For a nonzero $y \in R$ denote $t(y)$ the number of primes minimal over $y$. Since $R$ is a normal domain, these primes are height one and correspond $1$-to-$1$ to the minimal primes of $R/yR$ (Algebra, Lemmas 10.60.11 and 10.157.6). For example $t(f) = s$ is the number of primes $\mathfrak q_ j$ occurring in $\text{div}(f)$. Let $h \in \mathfrak m^ N$. By Lemma 15.125.2 we have

see Algebra, Lemma 10.52.5 for the first equality. Therefore we see that $t(f + h)$ is bounded independent of $h \in \mathfrak m^ N$.

By the boundedness proved above we may pick $h \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ such that $t(f + h)$ is maximal among such $h$. Set $f' = f + h$. Given $h' \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ we see that the number $t(f' + h') \leq t(f + h)$. Thus after replacing $f$ by $f'$ we may assume that for every $h \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ we have $t(f + h) \leq s$.

Next, assume that we can find an element $h \in \mathfrak m^ N$ such that for each $j$ we have $\text{ord}_{\mathfrak q_ j}(h) \geq 1$ and $\text{ord}_{\mathfrak q_ j}(h) = 1 \Leftrightarrow m_ j > 1$. Observe that $h \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$. Then $\text{ord}_{\mathfrak q_ j}(f + h) = 1$ for every $j$ by elementary properties of valuations. Thus

for some pairwise distinct height one prime ideals $\mathfrak r_1, \ldots , \mathfrak r_ v$ and $e_ k \geq 1$. However, since $s = t(f) \geq t(f + h)$ we see that $v = 0$ and we have found the desired element.

Now we will pick $h$ that satisfies the above criteria. By prime avoidance (Algebra, Lemma 10.15.2) for each $1 \leq j \leq s$ we can find an element $a_ j \in \mathfrak q_ j$ such that $a_ j \not\in \mathfrak q_{j'}$ for $j' \not= j$ and $a_ j \not\in \mathfrak q_ j^{(2)}$. Here $\mathfrak q_ j^{(2)} = \{ x \in R \mid \text{ord}_{\mathfrak q_ j}(x) \geq 2\} $ is the second symbolic power of $\mathfrak q_ j$. Then we take

Then $h$ clearly satisfies the conditions on valuations imposed above. If $h \not\in \mathfrak m^ N$, then we multiply by an element of $\mathfrak m^ N$ which is not contained in $\mathfrak q_ j$ for all $j$. $\square$

Proof. Let $\text{div}(a) = \sum _{i = 1, \ldots , r} n_ i \mathfrak p_ i$ with notation as in the proof of Lemma 15.125.5. Choose $c \in \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ with $A/cA$ reduced, see Lemma 15.125.5. For $n \geq \max (n_ i)$ we see that $-\text{div}(a) + \text{div}(c^ n)$ is an effective divisor (all coefficients nonnegative). Thus $c^ n/a \in A$ by Algebra, Lemma 10.157.6. $\square$

Proof. The function is a numerical polynomial by Algebra, Proposition 10.59.5. It has degree $d$ by Algebra, Proposition 10.60.9. If $d = 0$, then the result is trivial. If $d = 1$, then the result is Lemma 15.125.1. To prove it in general, observe that there is a surjection

sending the basis element corresponding to $i_1, \ldots , i_ d$ to the class of $g_1^{i_1} \ldots g_ d^{i_ d}$ in $I^ n/I^{n + 1}$. Thus we see that

Since $d \geq 2$ the numerical polynomial on the left has degree $d - 1$ with leading coefficient $e_ I / (d - 1)!$. The polynomial on the right has degree $d - 1$ and its leading coefficient is $\text{length}_ R(R/I)/ (d - 1)!$. This proves the lemma. $\square$

Proof. If $d = 0$ the lemma is trivial. If $d = 1$ the lemma is Lemma 15.125.2. Thus we may assume $d > 1$. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ be the minimal prime ideals of $R$ where the first $t$ have dimension $d$, and denote $I = (g_1, \ldots , g_ n)$. Arguing in exactly the same way as in the proof of Lemma 15.125.2 we can assume $R$ is reduced.

Assume $R$ is reduced with minimal primes $\mathfrak p_1, \ldots , \mathfrak p_ t$. This means there is an exact sequence

Here $Q$ is the cokernel of the first map. Write $M = \prod _{i = 1}^ t R/\mathfrak p_ i$. Localizing at $\mathfrak p_ j$ we see that

is surjective. Thus $Q_{\mathfrak p_ j} = 0$ for all $j$. Therefore no height $0$ prime of $R$ is in the support of $Q$. It follows that the degree of the numerical polynomial $n \mapsto \text{length}_ R(Q/I^ nQ)$ equals $\dim (\text{Supp}(Q)) < d$, see Algebra, Lemma 10.62.6. By Algebra, Lemma 10.59.10 (which applies as $R$ does not have finite length) the polynomial

has degree $< d$. Since $M = \prod R/\mathfrak p_ i$ and since $n \to \text{length}_ R(R/\mathfrak p_ i + I^ n)$ is a numerical polynomial of degree exactly(!) $d$ for $i = 1, \ldots , t$ (by Algebra, Lemma 10.62.6) we see that the leading coefficient of $n \mapsto \text{length}_ R(M/I^ nM)$ is at least $t/d!$. Thus we conclude by Lemma 15.125.7. $\square$

Proof. By Lemma 15.125.3 there exist $g_1, \ldots , g_{d - 1} \in \mathfrak m$ such that $f, g_1, \ldots , g_{d - 1}$ is a system of parameters for $R$. Then $\mathfrak m = \sqrt{(f, g_1, \ldots , g_{d - 1})}$. Thus there exists an $n$ such that $\mathfrak m^ n \subset (f, g)$, see Algebra, Lemma 10.32.5. We claim that $N = n + 1$ works. Namely, let $h \in \mathfrak m^ N$. By our choice of $N$ we can write $h = af + \sum b_ ig_ i$ with $a, b_ i \in \mathfrak m$. Thus

because $1 + a$ is a unit in $R$. This proves the equality of lengths and the fact that $f + h, g_1, \ldots , g_{d - 1}$ is a system of parameters. $\square$

Proof. The proof is the same as that of Lemma 15.125.5, using Lemma 15.125.8 instead of Lemma 15.125.2 and Lemma 15.125.9 instead of Lemma 15.125.4. We can use Lemma 15.125.8 because $R$ is a catenary domain, so every height one prime ideal of $R$ has dimension $d - 1$, and hence the spectrum of $R/(f + h)$ is equidimensional. For the convenience of the reader we write out the details.

Let $f \in J$ be a nonzero element. We will modify $f$ slightly to obtain an element that generates a radical ideal. The localization $R_\mathfrak p$ of $R$ at each height one prime ideal $\mathfrak p$ is a discrete valuation ring, see Algebra, Lemma 10.119.7 or Algebra, Lemma 10.157.4. We denote by $\text{ord}_\mathfrak p(f)$ the corresponding valuation of $f$ in $R_{\mathfrak p}$. Let $\mathfrak q_1, \ldots , \mathfrak q_ s$ be the distinct height one prime ideals containing $f$. Write $\text{ord}_{\mathfrak q_ j}(f) = m_ j \geq 1$ for each $j$. Then we define $\text{div}(f) = \sum _{j = 1}^ s m_ j\mathfrak q_ j$ as a formal linear combination of height one primes with integer coefficients. The ring $R/fR$ is reduced if and only if $m_ j = 1$ for $j = 1, \ldots , s$. Namely, if $m_ j$ is $1$ then $(R/fR)\mathfrak q_ j$ is reduced and $R/fR \subset \prod (R/fR)_{\mathfrak q_ j}$ as $\mathfrak q_1, \ldots , \mathfrak q_ j$ are the associated primes of $R/fR$, see Algebra, Lemmas 10.63.19 and 10.157.6.

Choose and fix $g_2, \ldots , g_{d - 1}$ and $N$ as in Lemma 15.125.9. For a nonzero $y \in R$ denote $t(y)$ the number of primes minimal over $y$. Since $R$ is a normal domain, these primes are height one and correspond $1$-to-$1$ to the minimal primes of $R/yR$ (Algebra, Lemmas 10.60.11 and 10.157.6). For example $t(f) = s$ is the number of primes $\mathfrak q_ j$ occurring in $\text{div}(f)$. Let $h \in \mathfrak m^ N$. Because $R$ is catenary, for each height one prime $\mathfrak p$ of $R$ we have $\dim (R/\mathfrak p) = d$. Hence by Lemma 15.125.8 we have

see Algebra, Lemma 10.52.5 for the first equality. Therefore we see that $t(f + h)$ is bounded independent of $h \in \mathfrak m^ N$.

By the boundedness proved above we may pick $h \in \mathfrak m^ N \cap J$ such that $t(f + h)$ is maximal among such $h$. Set $f' = f + h$. Given $h' \in \mathfrak m^ N \cap J$ we see that the number $t(f' + h') \leq t(f + h)$. Thus after replacing $f$ by $f'$ we may assume that for every $h \in \mathfrak m^ N \cap J$ we have $t(f + h) \leq s$.

Next, assume that we can find an element $h \in \mathfrak m^ N \cap J$ such that for each $j$ we have $\text{ord}_{\mathfrak q_ j}(h) \geq 1$ and $\text{ord}_{\mathfrak q_ j}(h) = 1 \Leftrightarrow m_ j > 1$. Then $\text{ord}_{\mathfrak q_ j}(f + h) = 1$ for every $j$ by elementary properties of valuations. Thus

for some pairwise distinct height one prime ideals $\mathfrak r_1, \ldots , \mathfrak r_ v$ and $e_ k \geq 1$. However, since $s = t(f) \geq t(f + h)$ we see that $v = 0$ and we have found the desired element.

Now we will pick $h$ that satisfies the above criteria. By prime avoidance (Algebra, Lemma 10.15.2) for each $1 \leq j \leq s$ we can find an element $a_ j \in \mathfrak q_ j \cap J$ such that $a_ j \not\in \mathfrak q_{j'}$ for $j' \not= j$. Next, we can pick $b_ j \in J \cap \mathfrak q_1 \cap \ldots \cap q_ s$ with $b_ j \not\in \mathfrak q_ j^{(2)}$. Here $\mathfrak q_ j^{(2)} = \{ x \in R \mid \text{ord}_{\mathfrak q_ j}(x) \geq 2\} $ is the second symbolic power of $\mathfrak q_ j$. Prime avoidance applies because the ideal $J' = J \cap \mathfrak q_1 \cap \ldots \cap q_ s$ is radical, hence $R/J'$ is reduced, hence $(R/J')_{\mathfrak q_ j}$ is reduced, hence $J'$ contains an element $x$ with $\text{ord}_{\mathfrak q_ j}(x) = 1$, hence $J' \not\subset \mathfrak q_ j^{(2)}$. Then the element

is an element of $J$ with $\text{ord}_{\mathfrak q_ j}(c) = 1$ for all $j = 1, \ldots , s$ by elementary properties of valuations. Finally, we let

where $y \in \mathfrak m^ N$ is an element which is not contained in $\mathfrak q_ j$ for all $j$. $\square$