Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 4.
Primary Decomposition
Exercise
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Exercise
Proof. If $a=⋂_{i=1}q_{i}$ is a minimal primary decomposition of $a$, and $q_{i}$ is $p_{i}$-primary, then $a=r(a)=r\mleft(i=1⋂m q_{i}\mright)=i=1⋂m p_{i}.$
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Exercise
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Exercise
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Exercise
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Exercise
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Exercise
1. | $a[x]$ is the extension of $a$ to $A[x]$. |
2. | If $p$ is a prime ideal in $A$, then $p[x]$ is a prime ideal in $A[x]$. |
3. | If $q$ is a $p$-primary ideal in $A$, then $q[x]$ is a $p[x]$-primary ideal in $A[x]$. |
4. | If $a=⋂_{i=1}q_{i}$ is a minimal primary decomposition in $A$, then $a[x]=⋂_{i=1}q_{i}[x]$ is a minimal primary decomposition in $A[x]$. |
5. | If $p$ is a minimal prime ideal of $a$, then $p[x]$ is a minimal prime ideal of $a[x]$. |
Proof.
1. | Trivial. |
2. | If $p$ is a prime ideal in $A$, then $A/p$ is integral. Hence $A[x]/p[x]≃(A/p)[x]$ is integral, and $p[x]$ is prime in $A[x]$. |
3. | If $q$ is primary in $A$, then all zero divisors in $A/q$ are nilpotents; so if $f∈A[x]/q[x]≃(A/q)[x]$ is a zero divisor, there exists $a∈A$ such that $af=0$; thus every coefficients of $f$ are zero divisors, hence nilpotents, and therefore $f$ is nilpotent (Exercise 1.2). This means $q[x]$ is primary in $A[x]$. Also, in $A/q$ the nilradical is $p/q$, so by Exercise 1.2 again, the nilradical of $A[x]/q[x]$ is $p[x]/q[x]$. So $q[x]$ is $p[x]$-primary. |
4. | Obviously $a[x]=⋃_{i=1}q[x]$, and $q[x]$ are primary by iii). Hence it’s a primary decomposition of $a[x]$. Obviously it’s a minimal primary decomposition of $a[x]$. |
5. | Obvious by iii), iv). $□$ |
Exercise
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Exercise
Proof. If $p∈D(A)$ and $x∈p$, then there exists $a∈A$, such that $p$ is minimal in prime ideals containing $(0:a)$; so $p/(0:a)$ is a minimal prime ideal in $A/(0:a)$. By Exercise 3.9 (minimal prime ideals only contains zero divisors), there exists $t∈A$ such that $xt∈(0:a)$; that is, $xta=0$. So $x$ is a zero divisor.
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Let $S$ be a multiplicatively closed subsetof $A$, and identify $Spec(S_{−1}A)$ with its image in $Spec(A)$. Show that $D(S_{−1}A)=D(A)∩Spec(S_{−1}A).$
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If the zero ideal has a primary decomposition, show that $D(A)$ is the set of associated prime ideals of $0$.
Proof. If $0=i=1⋂n q_{i}$where $q_{i}$ is $p_{i}$-primary, then every $p_{i}$ is $r(0:a)$ for some $a$; hence $p_{i}∈D(A)$.
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Exercise
1. | $S_{p}(0)⊆p$. |
2. | $r(S_{p}(0))=p⟺p$ is a minimal prime ideal of $A$. |
3. | If $p⊇p_{′}$, then $S_{p}(0)⊆S_{p_{′}}(0)$. |
4. | $⋂_{p∈D(A)}S_{p}(0)=0$, where $D(A)$ is defined in Exercise 9. |
Proof.
1. | $x∈S_{p}(0)⟺∃t∈/p,tx=0$. Since $p$ is prime, $tx=0∈p$ and $t∈/p$, there must be $x∈p$, so $S_{p}(0)⊆p$. |
2. | $r(S_{p}(0))=p⟺$ the image of $p$ in $A_{p}$ is the nilradical of $A_{p}⟺$ there is no prime ideal contained in $p_{e}$ in $A_{p}⟺p$ is minimal in $A$. |
3. | If $p⊇p_{′}$, then $A_{p_{′}}$ is localization of $A_{p}$ at $p_{p}$, so the map $A→A_{p_{′}}$ can be factor into $A→A_{p}$ and $A_{p}→A_{p_{′}}$, and $ker(A→A_{p_{′}})⊇ker(A→A_{p})$, i.e. $S_{p}(0)⊆S_{p_{′}}(0)$. |
4. | If $x∈S_{p}(0)$, then there exists $t∈/p$ such that $xt=0$; that is, $(0:x)⊆p$. If $x∈⋂_{p∈D(A)}S_{p}(0)$, then there is no minimal prime ideal containing $(0:x)$, so $(0:x)=A$, and $x=0$. $□$ |
Exercise
Proof. Since $p$ is a minimal prime ideal, $A_{p}$ has unique prime ideal $p$; so in $A_{p}$ every zero divisor is nilpotent, and so is $A/S_{p}(0)≃\im(A→A_{p})⊆A_{p}$. Hence $S_{p}(0)$ is primary. By Exercise 10 ii), $S_{p}(0)$ is $p$-primary.
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Let $a$ be the intersection of the ideals $S_{p}(0)$ as $p$ runs through the minimal prime ideals of $A$. Show that $a$ is contained in the nilradical of $A$.
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Support that the zero ideal is decomposable. Prove that $a=0$ if and only if every prime ideal of $0$ is isolated.
Proof. By Exercise 9, $D(A)$ is all minimal prime containing $(0:a)$ for some $a∈A$. but $(0:a)⊆p⟺a∈/S_{p}(0)$; and $D(A)$ is exactly the set of associated prime ideals of $0$.
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Exercise
1. | $S(a)∩S(b)=S(a∩b)$. |
2. | $S(r(a))=r(S(a))$. |
3. | $S(a)=(1)⟺a$ meets $S$. |
4. | $S_{1}(S_{2}(a))=(S_{1}S_{2})(a)$. |
If $a$ has a primary decomposition, prove that the set of ideals $S(a)$ (where $S$ runs through all multiplicatively closed subsets of $A$) is finite.
Proof.
1. | $S_{−1}(a∩b)=S_{−1}a∩S_{−1}b$, so $S(a∩b)=(S_{−1}a∩S_{−1}b)_{c}=S(a)∩S(b)$. |
2. | $S(r(a))=(S_{−1}(r(a)))_{c}=(r(S_{−1}a))_{c}=r(S(a))$. |
3. | $S(a)=(1)$ if and only if $S_{−1}a=1$, so it’s equivalent to $a$ meets $S$. |
4. | By definition, $y∈S(a)⟺∃s∈S$ such that $sy∈a$. So clearly $y∈S_{1}(S_{2}(a))⟺∃s_{1}∈S_{1},s_{2}∈S_{2},s_{1}s_{2}y∈a⟺y∈(S_{1}S_{2})(a)$. |
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Exercise
1. | $p_{(n)}$ is a $p$-primary ideal; |
2. | if $p_{n}$ has a primary decomposition, then $p_{(n)}$ is its $p$-primary component; |
3. | if $p_{(m)}p_{(n)}$ has a primary decomposition, then $p_{(m+n)}$ is its $p$-primary component; |
4. | $p_{(n)}=p_{n}⟺p_{n}$ is $p$-primary. Misprint in the book: $⋯⟺p_{(n)}$ is $p$-primary; but by i) it’s always $p$-primary. |
Proof.
1. | $r(S_{p}(p_{n}))=S_{p}p$ is the maximal ideal in $A_{p}$, hence $S_{p}(p_{n})$ is $S_{p}p$-primary. So its contraction $p_{(n)}$ is $p$-primary. |
2. | If $p_{n}$ has a primary decomposition $p_{n}=⋂_{i=1}q_{i}$, where $q_{i}$ is $p_{i}$-primary; then $p=r(p_{n})=⋂_{i=1}p_{i}$; so there exists some $i$ such that $p_{i}=p$; and all other $p_{i}⊋p$. Take $S_{p}(p_{n})$, we know $p_{(n)}$ is its $p$-primary decomposition. (Theorem 4.11) |
3. | Similar to ii), since $r(p_{(m)}p_{(n)})=p$, it’s sufficient to prove $S_{p}(p_{(m)}p_{(n)})=p_{(m+n)}$. For any $x∈p_{(n)}p_{(m)}$, by definition there exists $s∈S_{p}$ such that $sx∈p_{n+m}$. Hence $p_{(n)}p_{(m)}⊂p_{(n+m)}$, so $p_{n+m}⊆p_{(n)}p_{(m)}⊆p_{(n+m)}.$Take $S_{p}(−)$ we have $S_{p}(p_{(n)}p_{(m)})=p_{(n+m)}$. |
4. | If $p_{(n)}=p_{n}$, then by i) it’s $p$-primary. Conversely if $p_{n}$ is $p$-primary, then $p_{n}=⋂{p_{n}}$ is a primary decomposition of $p_{n}$; so by ii) $p_{n}=p_{(n)}$. $□$ |
Exercise
Proof. First we show $p=(a:x_{0})$ must be prime. If $y∈/(a:x_{0})$, then $(a:x_{0})⊆(a:x_{0}y)=(1)$; so there must be $(a:x_{0})=(a:x_{0}y)$; i.e. if $yz∈(a:x_{0})$, then $z∈(a:x_{0}y)=(a:x_{0})$. So $p=(a:x_{0})$ is prime.
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Exercise
Proof. If $a=⋂_{i=1}q_{i}$, where $q_{i}$ is $p_{i}$-primary, and $Σ={p_{1},…,p_{p}}$. Since $f∈p_{i}⟺p∈/Σ$, so $S_{f}∩p_{i}=∅⟺p∈/Σ$. Thus $S_{f}(a)=⋂_{i=1}S_{f}(q_{i})=⋂_{i=1}q_{i}=q_{Σ}$.
If $f∈p_{i}$, then by $r(q_{i})=p_{i}$, we know there exists $N_{i}$ such that $f_{N_{i}}∈q_{i}$. Let $N=max(N_{p+1},…,N_{m})$, then for all $n≥N$, $(a:f_{n})=i=1⋂m (q_{i}:f_{n})=i=1⋂p q_{i}=q_{Σ}$$□$
Exercise
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Exercise
(L1) For every ideal $a=(1)$ in $A$ and every prime ideal $p$, there exists $x∈/p$ such that $S_{p}(a)=(a:x)$, where $S_{p}=A−p$.
Then every ideal in $A$ is an intersection of (possibly infinitely many) primary ideals.
Proof. Let $a$ be an ideal $=(1)$ in $A$, let $p_{1}$ be a minimal prime ideal containing $a$. By Exercise 11 (apply to $A/a$), $q_{1}=S_{p_{1}}(a)$ is a $p_{1}$-primary ideal. By (L1) property, there exists $x∈/p_{1}$ such that $q_{1}=(a:x)$. Now $q_{1}∩(a+(x))⊇a$, and if $bx∈q_{1}$, then by $x∈/p_{1}$, $b∈q_{1}$; so $bx∈a$. This means $a=q_{1}∩(a+(x))$.
Let $a_{1}$ be a maximal element of the set of ideals $b$ such that: $b⊇a+(x)$ and $q_{1}∩b=a$. Then $a_{1}⊆p_{1}$. Repeat the construction starting with $a_{1}$, and so on. After $n$ steps we have $a=q_{1}∩⋯∩q_{n}∩a_{n}$.
If $a_{n}=(1)$ for some $n$, then $a=q_{1}∩⋯∩q_{n}$is a finite intersection of primary ideals. Otherwise define $a_{ω}=n=1⋃∞ a_{n},$then $a=a_{ω}∩⋂_{n=1}q_{i}$. Similarly, for any ordinal $λ$, construct $a_{λ+1}$ from $a_{λ}$ by the construction above (such that $a_{λ}=q_{λ}∩a_{λ+1}$, and $a_{λ}⊊a_{λ+1}$), and for any limit ordinal $λ$ define $a_{λ}=⋃_{β<λ}a_{β}$. Then for any ordinal $λ$, we have $a=a_{λ}∩β<λ⋂ q_{β}.$And $a_{λ}⊊a_{λ+1}$. By transfinite induction, it’s easy to show $∣a_{λ}∣≥∣λ∣$ (if $a_{λ}=(1)$). So for some ordinal larger than $∣A∣$, there must be $a_{λ}=(1)$, and $a=β<λ⋂ q_{β}.$$□$
Exercise
(L2) Given an ideal $a$ and a descending chain $S_{1}⊇⋯⊇S_{n}⊇…$ of multiplicatively closed subsets of $A$, there exists an integer $n$ such that $S_{n}(a)=S_{n+1}(a)=…$.
Prove that the following are equivalent:
1. | Every ideal in $A$ has a primary decomposition. |
2. | $A$ satisfies (L1) and (L2). |
Proof. i)$⟹$ii): For every ideal $a=(1)$ in $A$ and every prime ideal $p$, if $a$ has a primary decomposition $a=⋂_{i=1}q_{i}$, where $q_{i}$ is $p_{i}$-primary, then $S_{p}(a)=p_{i}⊆p⋂ q_{i}$Let $b=⋂_{p_{i}⊆p}p_{i}$, then $b⊆p$; so there exists an element $f∈b∖p$, i.e. for any $i=1,…,n,f∈p_{i}⟺p_{i}⊆p$. By Exercise 15), $S_{p}(a)=(a:f_{n})$ for all large $n$. So (L1) holds.
For every ideal $a$, since $a$ has a primary decomposition, so by Exercise 12), the set of all $S(a)$ is finite (where $S$ runs through all multiplicatively closed subsets of $A$). So any descending chain $S_{1}(a)⊇⋯⊇S_{n}(a)⊇…$ must be stationary. Thus (L2) holds.
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Exercise
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Proof. Induction on $n$. The case $n=1$ is trivial (let $a=p_{1}$).
Suppose $n>1$, and $p_{n}$ is maximal in ${p_{1},…,p_{n}}$. By inductive hypothesis, there exists an ideal $b$ whose associated ideals are $p_{1},…,p_{n−1}$. If $b⊆S_{p_{n}}(0)$, choose a minimal ideal $p$ contained in $p_{n}$, then $b⊆S_{p_{n}}(0)⊆S_{p}(0)$. Take radicals we have $⋂_{i=1}p_{i}⊆p$, hence $p_{i_{0}}⊆p$ for some $i_{0}$. Since $p$ is minimal, $p_{i_{0}}=p$, contradiction (cause none of $p_{1},…,p_{n−1}$ is minimal). So $b⊆S_{p_{n}}(0)$, hence $b⊆q_{n}$ for some $p_{n}$-primary ideal $q_{n}$.
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Primary decomposition of modules
Practically the whole of this chapter can be transposed to the context of modules over a ring $A$. The following exercises indicate how this is done.
Exercise
State and prove the formulas for $r_{M}$ analogous to (1.13).
Proof. $r_{M}(N) ={x∈A:x_{q}M⊆Nfor someq>0}={x∈A:x_{q}(M/N)=0for someq>0}=r(Ann(M/N)) $And clearly:
1. | $r(r_{M}(N))=r_{M}(N)$. |
2. | $r_{M}(N_{1}∩N_{2})=r_{M}(N_{1})∩r_{M}(N_{2})$. |
3. | $r_{M}(N)=(1)⟺N=M$. |
4. | $r_{M}(N+P)⊇r(r_{M}(N)+r_{M}(P))$. |
5. | if $P⊆N⊆M$ then $r_{M}(N)=r_{M/P}(N/P)$. $□$ |
Exercise
Show that if $Q$ is primary in $M$, then $(Q:M)$ is a primary ideal and hence $r_{M}(Q)$ is a prime ideal $p$. We say that $Q$ is $p$-primary (in $M$).
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Prove the analogues of (4.3) and (4.4).
Proof. Analogues of (4.3): If $Q_{i}(1≤i≤n)$ are $p$-primary, then $Q=⋂_{i=1}Q_{i}$ is $p$-primary.
First we have $r_{M}(Q)=⋂_{i=1}r_{M}(Q_{i})=p$. If $x$ is a zero divisor in $M/Q$, then there exists $v∈/Q$ such that $xv∈Q$. Since $Q=⋂_{i=1}Q_{i}$, there exists $j$ such that $v∈/Q_{j}$, but $xv∈Q⊆Q_{j}$, thus $x∈p=r_{M}(Q)$. Hence $Q$ is $p$-primary.
Analogues of (4.4): If $Q$ is $p$-primary submodule of $M$, then
1. | if $v∈Q$ then $(Q:v)=(1)$; |
2. | if $v∈/Q$ then $(Q:v)$ is a $p$-primary ideal. |
3. | if $x∈/p$ then $(Q:x)(:={v∈M:xv∈Q})=Q$. |
i) and iii): by definition.
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Exercise
Prove the analogue of (4.5), that the prime ideals $p_{i}$ depend only on $N$ (and $M$). They are called the prime ideals belonging to $N$ in $M$. Show that they are also the prime ideals belonging to $0$ in $M/N$.
Proof. We will prove that the prime ideals $p_{i}$ are exactly all prime ideals which occur in the set of ideals $r(N:v)(v∈M)$; so that they only depend on $N$ (and $M$).
Suppose $N=Q_{1}∩⋯∩Q_{n}$ is a minimal primary decomposition of $N$, where $Q_{i}$ is $p_{i}$-primary. For any $v∈M$, we have $r(N:v)=i=1⋂n r(Q_{i}:v)$If it’s a prime ideal $p$, then $p=r(Q_{i}:v)$ for some $1≤i≤n$. But $r(Q_{i}:v)=p_{i}$ (if not $(1)$), hence $p=p_{i}$. Conversely, for each $1≤i≤n$, there exists $v_{i}∈/Q_{i}$, $v_{i}∈⋂_{j=i}Q_{j}$. In this case we have $r(N:v_{i})=r(Q_{i}:v_{i})=p_{i}$.
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Exercise
(4.6) | Let $N$ be a decomposable submodule of $M$. Then any prime ideal $p⊇r_{M}(N)$ contains a minimal prime ideal belonging to $N$ in $M$, and thus the minimal prime ideals of $N$ in $M$ are precisely the minimal elements in the set of all prime ideals containing $r_{M}(N)$. Proof. If $p⊇r_{M}(N)=⋂_{i=1}r_{M}(Q_{i})=⋂_{i=1}p_{i}$, then $p⊇p_{i}$ for some $i$. Hence $p$ contains a minimal prime ideal of $N$ in $M$. $□$ | ||||
(4.7) | Let $N$ be a decomposable submodule of $M$, let $N=⋂_{i=1}Q_{i}$ be a minimal primary decomposition, and let $r_{M}(Q_{i})=p_{i}$. Then $i=1⋃n p_{i}={x∈A:(N:x)=N}$where $(N:x)=def{v∈M:xv∈N}$. In particular, if $N=0$, then the set $D_{M}⊆A$ of zero-divisors in $M$ of $A$ is the union of the prime ideals belonging to $0$. Proof. Without loss of generality, we can assume $N=0$. Clearly $D_{M}=⋃_{v=0∈M}r(0:v)$. For each $v=0∈M$, $r(0:v)=i=1⋂n r(Q_{i}:v)=v∈/Q_{i}⋂ p_{i}.$Hence $r(0:v)⊆p_{i}$ for some $1≤i≤n$, and thus $D_{M}⊆⋃_{i=1}p_{i}$. Conversely, for each $1≤i≤n$, there exists $v_{i}∈/Q_{i},v_{i}∈⋂_{j=i}Q_{j}$, so $p_{i}=r(0:v_{i})⊆D_{M}$. $□$ | ||||
(4.8) | Let $S$ be a multiplicatively closed subset of $A$, and let $Q⊆M$ be a $p$-primary submodule.
Proof. If $S∩p=∅$, then there exists $x∈S$ satisfies $x∈p=r_{M}(Q)$, so there is $n≥1$ such that $x_{n}M⊆Q$. Hence for each $y∈S_{−1}M$, $y=(x_{n}y)/x_{n}∈S_{−1}Q$. If $S∩p=∅$. First we show $r_{S_{−1}M}(S_{−1}Q)=S_{−1}p$. Suppose $x/s∈r_{S_{−1}M}(S_{−1}Q)$, then also $x=x/1∈r_{S_{−1}M}(S_{−1}Q)$, so there is $n≥1$ such that $x_{n}(S_{−1}M)⊆(S_{−1}Q)$. In particular, for each $m∈M$, $x_{n}m/1∈S_{−1}Q$, hence exists $t∈S$ that $x_{n}tm∈Q$. So $x_{n}t∈r_{M}(Q)=p$. Since $t∈/p$, we have $x∈p$. Therefore $x/s∈S_{−1}p$ and $r_{S_{−1}M}(S_{−1}Q)⊆p$. The other direction follows immediately from definition. Suppose $x/t∈S_{−1}A,m/s∈S_{−1}M−S_{−1}Q$ but $xm/st∈S_{−1}Q$. Then there exist $u∈S$ such that $uxm∈Q$, but $m∈/Q$. hence $ux∈p$ and so $x∈p$. Thus $x/t∈S_{−1}p=r_{S_{−1}M}(S_{−1}Q)$, and $S_{−1}Q$ is a $S_{−1}p$-primary submodule. Finally, if $m/1∈S_{−1}Q$, then there exists $s∈S$ such that $sm∈Q$. But $s∈/p$, hence $m∈Q$. Therefore the preimage of $S_{−1}Q$ in $M$ is $Q$. $□$ | ||||
(4.9) | Let $S$ be a multiplicatively closed subset of $A$ and let $N$ be a decomposable submodule of $M$. Let $N=⋂_{i=1}Q_{i}$ be a minimal primary decomposition of $N$. Let $p_{i}=r_{M}(Q_{i})$ and suppose the $Q_{i}$ numbered so that $S$ meets $p_{m+1},…,p_{n}$ but not $p_{1},…,p_{m}$. Then $S_{−1}N=i=1⋂m S_{−1}Q_{i},S(N)=i=1⋂M Q_{i}$ Proof. Clearly by (4.8). $□$ | ||||
(4.10) | Let $N$ be a decomposable submodule of $M$, let $N=⋂_{i=1}Q_{i}$ be a minimal primary decomposition of $N$, and let $p_{i_{1}},…,p_{i_{m}}$ be an isolated set of prime ideals of $N$ in $M$. Then $Q_{i_{1}}∩⋯∩Q_{i_{m}}$ is independent of the decomposition. Proof. Take $S=A−⋃_{j=1}p_{i_{j}}$ in (4.9), then $Q_{i_{1}}∩⋯∩Q_{i_{m}}=S(N)$, hence independent of the decomposition. $□$ In particular, | ||||
(4.11) | : The isolated primary components ($m=1$ in (4.10), the corresponding prime ideal is minimal) are uniquely determined by $N$ (and $M$). |