
In this section we prove that a catenary Noetherian normal local domain there exists a nontrivial principal radical ideal. This result can be found in .

Lemma 15.105.1. Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension one, and let $x\in \mathfrak m$ be an element not contained in any minimal prime of $R$. Then

1. the function $P : n \mapsto \text{length}_ R(R/x^ n R)$ satisfies $P(n) \leq n P(1)$ for $n \geq 0$,

2. if $x$ is a nonzerodivisor, then $P(n) = nP(1)$ for $n \geq 0$.

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 10.61.3). 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

$R/x^{n-1}R \xrightarrow {x} R/x^ nR \to R/xR \to 0$

where $x$ denotes the multiplication by $x$ map. Since length is additive (Algebra, Lemma 10.51.3), 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 $n \geq 1$. $\square$

Lemma 15.105.2. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $1$. Let $x \in \mathfrak m$ be an element not contained in any minimal prime of $R$. Let $t$ be the number of minimal prime ideals of $R$. Then $t \leq \text{length}_ R(R/xR)$.

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

$0 \to R \to \prod \nolimits _{i = 1}^ t R/\mathfrak p_ i \to Q \to 0$

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

$R_{\mathfrak p_ j} \to M_{\mathfrak p_ j} = \left(\prod \nolimits _{i=1}^ t R/\mathfrak p_ i\right)_{\mathfrak p_ j} = (R/\mathfrak p_ j)_{\mathfrak p_ j}$

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 10.61.3). Since $\text{Supp}(Q) = \{ \mathfrak m\}$ we can pick an $n \gg 0$ such that $x^ n$ acts as $0$ on $Q$ (Algebra, Lemma 10.61.4). Now consider the diagram

$\xymatrix{ 0 \ar[r] & R \ar[r] \ar[d]^-{x^ n} & M \ar[r] \ar[d]^-{x^ n} & Q \ar[r] \ar[d]^-{x^ n} & 0 \\ 0 \ar[r] & R \ar[r] & M \ar[r] & Q \ar[r] & 0 }$

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:

$0 \to Q \to R/x^ nR \to M/x^ nM \to Q \to 0$

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 15.105.1 and the fact that $x$ is a nonzerodivisor on $R$ and $R_ i$, we conclude that

$n \text{length}_ R(R/xR) = \sum \nolimits _{i = 1}^ t n \text{length}_{R_ i}(R_ i/x R_ i)$

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

Lemma 15.105.3. Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension $d > 1$, let $f \in \mathfrak m$ be an element not contained in any minimal prime ideal of $R$, and let $k\in \mathbf{N}$. Then there exist elements $g_1, \ldots , g_{d - 1} \in \mathfrak m^ k$ such that $f, g_1, \ldots , g_{d - 1}$ is a system of parameters.

Proof. We have $\dim (R/fR) = d - 1$ by Algebra, Lemma 10.59.12. Choose a system of parameters $\overline{g}_1, \ldots , \overline{g}_{d - 1}$ in $R/fR$ (Algebra, Proposition 10.59.8) 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. $\square$

Lemma 15.105.4. Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension two, and let $f \in \mathfrak m$ be an element not contained in any minimal prime ideal of $R$. Then there exist $g \in \mathfrak m$ and $N \in \mathbf{N}$ such that

1. $f,g$ form a system of parameters for $R$.

2. If $h \in \mathfrak m^ N$, then $f + h, g$ is a system of parameters and $\text{length}_ R (R/(f, g)) = \text{length}_ R(R/(f + h, g))$.

Proof. By Lemma 15.105.3 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 10.31.5. 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

$(f + h, g) = (f + af + bg, g) = ((1 + a)f, g) = (f, g)$

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. $\square$

Lemma 15.105.5. Let $R$ be a Noetherian local normal domain of dimension $2$. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be pairwise distinct primes of height $1$. There exists a nonzero element $f \in \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ such that $R/fR$ is reduced.

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 10.118.7 or Algebra, Lemma 10.151.4. 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 10.62.19 and 10.151.6.

Choose and fix $g$ and $N$ as in Lemma 15.105.4. 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 10.59.10 and 10.151.6). 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 15.105.2 we have

\begin{align*} t(f + h) & \leq \text{length}_{R/(f + h)}(R/(f + h, g)) \\ & = \text{length}_ R(R/(f + h, g)) \\ & = \text{length}_ R(R/(f, g)) \end{align*}

see Algebra, Lemma 10.51.5 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

$\text{div}(f + h) = \sum \nolimits _{j = 1}^ s \mathfrak q_ j + \sum \nolimits _{k = 1}^ v e_ k \mathfrak r_ k$

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 10.14.2) 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

$h = \prod \nolimits _{m_ j = 1} a_ j^2 \times \prod \nolimits _{m_ j > 1} a_ j$

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 $j$. $\square$

Lemma 15.105.6. Let $(A, \mathfrak m, \kappa )$ be a Noetherian normal local domain of dimension $2$. If $a \in \mathfrak m$ is nonzero, then there exists an element $c \in A$ such that $A/cA$ is reduced and such that $a$ divides $c^ n$ for some $n$.

Proof. Let $\text{div}(a) = \sum _{i = 1, \ldots , r} n_ i \mathfrak p_ i$ with notation as in the proof of Lemma 15.105.5. Choose $c \in \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ with $A/cA$ reduced, see Lemma 15.105.5. 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 10.151.6. $\square$

In the rest of this section we prove the result in dimension $> 2$.

Lemma 15.105.7. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $d$, let $g_1, \ldots , g_ d$ be a system of parameters, and let $I = (g_1, \ldots , g_ d)$. If $e_ I/d!$ is the leading coefficient of the numerical polynomial $n \mapsto \text{length}_ R(R/I^{n+1})$, then $e_ I \leq \text{length}_ R(R/I)$.

Proof. The function is a numerical polynomial by Algebra, Proposition 10.58.5. It has degree $d$ by Algebra, Proposition 10.59.8. If $d = 0$, then the result is trivial. If $d = 1$, then the result is Lemma 15.105.1. To prove it in general, observe that there is a surjection

$\bigoplus \nolimits _{i_1, \ldots , i_ d \geq 0,\ \sum i_ j = n} R/I \longrightarrow I^ n/I^{n + 1}$

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

$\text{length}_ R(R/I^{n + 1}) - \text{length}_ R(R/I^ n) \leq \text{length}_ R(R/I) {n + d - 1 \choose d - 1}$

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. $\square$

Lemma 15.105.8. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $d$, let $t$ be the number of minimal prime ideals of $R$ of dimension $d$, and let $(g_1,\ldots ,g_ d)$ be a system of parameters. Then $t \leq \text{length}_ R(R/(g_1,\ldots ,g_ n))$.

Proof. If $d = 0$ the lemma is trivial. If $d = 1$ the lemma is Lemma 15.105.2. 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 15.105.2 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

$0 \to R \to \prod \nolimits _{i = 1}^ t R/\mathfrak p_ i \to Q \to 0$

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

$R_{\mathfrak p_ j} \to M_{\mathfrak p_ j} = \left(\prod \nolimits _{i=1}^ t R/\mathfrak p_ i\right)_{\mathfrak p_ j} = (R/\mathfrak p_ j)_{\mathfrak p_ j}$

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 $\dim (\text{Supp}(Q))10.61.6. By Algebra, Lemma 10.58.10 (which applies as$R$does not have finite length) the polynomial $n \longmapsto \text{length}_ R(M/I^ nM) - \text{length}_ R(R/I^ n) - \text{length}_ R(Q/I^ nQ)$ 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 10.61.6) 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 15.105.7.$\square$Lemma 15.105.9. Let$(R, \mathfrak m)$be a Noetherian local ring of dimension$d$, and let$f \in \mathfrak m$be an element not contained in any minimal prime ideal of$R$. Then there exist elements$g_1, \ldots , g_{d - 1} \in \mathfrak m$and$N \in \mathbf{N}$such that 1.$f, g_1, \ldots , g_{d - 1}$form a system of parameters for$R$2. If$h \in \mathfrak m^ N$, then$f + h, g_1, \ldots , g_{d - 1}$is a system of parameters and we have$\text{length}_ R R/(f, g_1, \ldots , g_{d-1}) = \text{length}_ R R/(f + h, g_1, \ldots , g_{d-1})$. Proof. By Lemma 15.105.3 there exist$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)$, see Algebra, Lemma 10.31.5. We claim that$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 \begin{align*} (f + h, g_1, \ldots , g_{d - 1}) & = (f + af + \sum b_ ig_ i, g_1, \ldots , g_{d - 1}) \\ & = ((1 + a)f, g_1, \ldots , g_{d - 1}) \\ & = (f, g_1, \ldots , g_{d - 1}) \end{align*} because1 + 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$Proposition 15.105.10. Let$R$be a catenary Noetherian local normal domain. Let$J \subset R$be a radical ideal. Then there exists a nonzero element$f \in J$such that$R/fR$is reduced. Proof. The proof is the same as that of Lemma 15.105.5, using Lemma 15.105.8 instead of Lemma 15.105.2 and Lemma 15.105.9 instead of Lemma 15.105.4. We can use Lemma 15.105.8 because$R$is a catenary domain, so every height one prime ideal of$R$has dimension$d - 1$, and hence$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$is a discrete valuation ring, see Algebra, Lemma 10.118.7 or Algebra, Lemma 10.151.4. We denote by$\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 10.62.19 and 10.151.6. Choose and fix$g_2, \ldots , g_{d - 1}$and$N$as in Lemma 15.105.9. 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 10.59.10 and 10.151.6). 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$\dim (R/\mathfrak p) = d. Hence by Lemma 15.105.8 we have \begin{align*} t(f + h) & \leq \text{length}_{R/(f + h)}(R/(f + h, g_1, \ldots , g_{d - 1})) \\ & = \text{length}_ R(R/(f + h, g_1, \ldots , g_{d - 1})) \\ & = \text{length}_ R(R/(f, g_1, \ldots , g_{d - 1})) \end{align*} see Algebra, Lemma 10.51.5 for the first equality. Therefore we see thatt(f + h)$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 $\text{div}(f + h) = \sum \nolimits _{j = 1}^ s \mathfrak q_ j + \sum \nolimits _{k = 1}^ v e_ k \mathfrak r_ k$ 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 10.14.2) 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 $c = \sum \nolimits _{j = 1, \ldots , s} b_ j \times \prod \nolimits _{j' \not= j} a_{j'}$ 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 $h = c \times \prod \nolimits _{m_ j = 1} a_ j \times y$ where$y \in \mathfrak m^ N$is an element which is not contained in$\mathfrak q_ j$for all$j$.$\square$## Comments (0) ## Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi\$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.