Definition 12.1. Minimal Prime.
Let \(I\) be an ideal in a ring \(R\text{.}\) A prime \(P\) is a minimal prime of \(I\) if \(P\) is minimal in \(V(I)\text{.}\) A minimal prime of \(R\) is a minimal element in \(\Spec(R)\text{.}\)
Section 12.1 Minimal Primes and Support
“There is no support so strong as the strength that enables one to stand alone.”―Ellen Glasgow
Definition 12.2. Minimal Prime.
Let \(I\) be an ideal in a ring \(R\text{.}\) A minimal prime of \(I\) is a minimal element (with respect to containment) in \(V(I)\text{.}\) More precisely, \(P\) is a minimal prime of \(I\) if the following hold: - \(P\) is a prime ideal, - \(P \supseteq I\text{,}\) and - if \(Q\) is also a prime ideal and \(I \subseteq Q \subseteq P\text{,}\) then \(Q = P\text{.}\)
The set of minimal primes of \(I\) is denoted \(\Min(I)\)
FIX
Definition 12.3. Nilradical.
The nilpotent elements of a ring \(R\) are exactly the elements in every minimal prime of \(R\text{.}\) The radical of \((0)\) is often called the nilradical of \(R\text{,}\) denoted \(\mathcal{N}(R)\).
Example 12.4. Minimal Primes and Radical Ideals.
If \(P\) is prime, then \(\Min(P)=\{P\}\text{.}\) Also, since \(V(I)=V(\sqrt{I})\text{,}\) we have \(\Min(I)=\Min(\sqrt{I})\text{.}\)
Example 12.5. Minimal Primes of \((x^2,xy)\).
Let \(k\) be a field and \(R = k[x,y]\text{.}\) Every prime containing \(I = (x^2, xy)\) must contain \(x^2\text{,}\) and thus \(x\text{.}\) On the other hand, \((x)\) is a prime ideal containing \(I\text{.}\) Therefore, \(I\) has a unique minimal prime, and \(\Min(I) = \{ (x) \}\text{.}\)
Example 12.6. Nilradical of \((x^2,xy)\).
Let \(k\) be a field. The radical of the ideal \(I = (x^2, xy)\) in the ring \(R = k[x,y]\) is \(\sqrt{I} = (x).\) We saw before that \(\Min(I) = \{ (x) \}\text{.}\) The nilradical of \(R = k[x,y]/(x^2,xy)\) corresponds to the radical of \((x^2,xy)\) is \(k[x,y]\text{,}\) so it is the ideal \((x)/(x^2,xy)\text{.}\)
Theorem 12.7. Finite Minimal Primes.
Over a noetherian ring \(R\text{,}\) every ideal \(I\) has finitely many minimal primes, and thus \(\sqrt{I}\) is a finite intersection of primes.
Proof.
Let \(S=\{ \text{ideals} \ I \subseteq R \ | \ \Min(I) \ \text{is infinite} \}\text{,}\) and suppose, to obtain a contradiction, that \(S\neq \es\text{.}\) Since \(R\) is noetherian, \(S\) has a maximal element \(J\text{.}\) If \(J\) was a prime ideal, then \(\Min(J) = \{ J \}\) would be finite, by , so \(J\) is not prime. However, \(\Min(J)=\Min(\sqrt{J})\text{,}\) and thus \(\sqrt{J} \supseteq J\) is also in \(S\text{,}\) so we conclude that \(J\) is radical. Since \(J\) is not prime, we can find some \(a, b \notin J\) with \(ab \in J\text{.}\) Then \(J \subsetneq J + (a) \subseteq \sqrt{J + (a)}\) and \(J \subsetneq \sqrt{J + (b)}\text{.}\) Since \(J\) is maximal in \(S\text{,}\) we conclude that \(\sqrt{J + (a)}\) and \(\sqrt{J + (b)}\) have finitely many minimal primes, so we can write
\begin{equation*}
\sqrt{J + (a)} = P_1 \cap \cdots \cap P_t \textrm{ and } \sqrt{J + (b)} = P_{t+1} \cap \cdots \cap P_s
\end{equation*}
for some prime ideals \(P_i\text{.}\)
Let \(f \in \sqrt{J + (a)} \cap \sqrt{J + (b)}\text{.}\) Some sufficiently high power of \(f\) is in both \(J + (a)\) and \(J + (b)\text{,}\) so there exist \(n, m \geq 1\) such that
\begin{equation*}
f^n \in J + (a) \quad \textrm{ and } \quad f^m \in J + (b).
\end{equation*}
Thus
\begin{equation*}
f^{n+m} \in (J + (a))(J + (b)) \subseteq J^2 + J(a) + J(b) + (\underset{\in J}{ab}) \subseteq J.
\end{equation*}
Therefore, \(f \in \sqrt{J} = J\text{.}\) This shows that
\begin{equation*}
J = \sqrt{J + (a)} \cap \sqrt{J + (b)} = P_1\cap \cdots \cap P_t \cap P_{t+1} \cap \cdots \cap P_s.
\end{equation*}
By
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Intersection of Primes are Min Primes|Lemma]], we see that \(\Min(J)\) must be a subset of \(\{P_1,\dots,P_s\}\text{,}\) so it is finite.Lemma 12.8. Min Primes Intersection All Important.
If \(\Min(I) = \{ P_1, \ldots, P_n \}\text{,}\) no \(P_i\) can be deleted in the intersection \(P_1 \cap \cdots \cap P_n\text{.}\)
Proof.
Suppose that we can delete \(P_i\text{,}\) meaning that
\begin{equation*}
\bigcap_{j=1}^n P_j = \bigcap_{j \neq i} P_j.
\end{equation*}
Then
\begin{equation*}
P_i \supseteq \bigcap_{j=1}^n P_j = \bigcap_{j \neq i} P_j \supseteq \prod_{j \neq i} P_j.
\end{equation*}
Since \(P_i\) is prime, this implies that \(P_i \supseteq P_j\) for some \(j \neq i\text{,}\) but this contradicts the assumption that the primes are incomparable.
Lemma 12.9. Intersection of Primes are Min Primes.
Let \(I\) be an ideal in \(R\text{.}\) If \(I = P_1 \cap \cdots \cap P_n\) where each \(P_i\) is prime and \(P_i \not\subseteq P_j\) for each \(i \neq j\text{,}\) then \(\Min(I)=\{ P_1, \ldots, P_n\}\text{.}\) Moreover, \(I\) must be radical.
Proof.
If \(Q\) is a prime containing \(I\text{,}\) then \(Q \supseteq (P_1 \cap \cdots \cap P_n)\text{.}\) We claim that \(Q\) must contain one of the \(P_i\text{.}\) Indeed, if \(Q \not\supseteq P_i\) for all \(i\text{,}\) then there are elements \(f_i \in P_i\) such that \(f_i \notin Q\text{,}\) so their product satisfies \(f_1 \cdots f_n \in (P_1 \cap \cdots \cap P_n)\) but \(f_1 \cdots f_n \notin Q\text{.}\) This is a contradiction, so indeed any prime containing \(I\) must contain some \(P_i\text{.}\) Therefore, any minimal prime of \(I\) must be one of the \(P_i\text{.}\) Since we assumed that the \(P_i\) are incomparable, these are exactly all the minimal primes of \(I.\) By assumption, \(I\) coincides with the intersection of its minimal primes, which is \(\sqrt{I}\) by
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Spectrum Analogue of Strong Nullstellensatz|Theorem]]. Therefore, \(I = \sqrt{I}\text{.}\)
Example 12.10. Ideal is Finite Intersection of Primes iff I is Radical.
If \(I = P_1 \cap \cdots \cap P_n\) for some primes \(P_i\text{,}\) we can always delete unnecessary components until no component can be deleted. Therefore, \(\Min(I) \subseteq \{ P_1, \ldots, P_n \}\text{.}\)
As a consequence of
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Min Primes Intersection All Important|Lemma]] and [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Intersection of Primes are Min Primes|Lemma]], if \(I\) is a radical ideal, there is a unique way to write \(I\) as a finite intersection of incomparable prime ideals. Moreover, [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Intersection of Primes are Min Primes|Lemma]], [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Finite Minimal Primes|Theorem]], and [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Spectrum Analogue of Strong Nullstellensatz|Spectrum Analogue]] also imply that an ideal \(I\) is equal to a finite intersection of primes if and only if \(I\) is radical.Definition 12.11. Support.
If \(M\) is an \(R\)-module, the support of \(M\) is
\begin{equation*}
\Supp(M) := \{ P \in \Spec(R) \ | \ M_P \neq 0\}.
\end{equation*}
Theorem 12.12. \(\Supp(M) = \ann(M)\).
Given \(M\) a finitely generated \(R\)-module over a ring \(R\text{,}\)
\begin{equation*}
\Supp(M)=V(\ann_R(M)).
\end{equation*}
In particular, \(\Supp(R/I)=V(I)\text{.}\)
Proof.
Let \(M= R m_i + \cdots + R m_n\text{.}\) We have
\begin{equation*}
\ann_R(M)=\bigcap_{i=1}^n \ann_R(m_i),
\end{equation*}
so by
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Proposition - Properties of V|Proposition]],
\begin{equation*}
V(\ann_R(M))=\bigcup_{i=1}^n V( \ann_R(m_i)).
\end{equation*}
Notice that we need finiteness here. Also, we claim that
\begin{equation*}
\Supp(M) = \bigcup_{i=1}^n \Supp(R m_i).
\end{equation*}
To show \((\supseteq)\text{,}\) notice that \((R m_i)_P \subseteq M_P\text{,}\) so
\begin{equation*}
P \in \Supp(R m_i) \implies 0 \neq (R m_i)_P \subseteq M_P \implies P \in \Supp(M).
\end{equation*}
On the other hand, the images of \(m_1, \ldots, m_n\) in \(M_P\) generate \(M_P\) for each \(P\text{,}\) and thus \(P \in \Supp(M)\) if and only if \(P \in \Supp(R m_i)\) for some \(m_i\text{.}\) Thus, we can reduce the equality \(\Supp(M)=V(\ann_R(M))\) to the case of a cyclic module \(R m\text{.}\) By
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Zero in Localization|Lemma]], \(\frac{m}{1}=0\) in \(M_P\) if and only if \((R\setminus P) \cap \ann_R (m) \neq \varnothing\text{,}\) which is equivalent to \(\ann_R (m) \not\subseteq P\text{.}\)
Example 12.13. Specific Direct Sum and Support.
Let \(k\) be a field, and \(R=k[x]\text{.}\) Take
\begin{equation*}
M=R_x / R = \bigoplus_{i>0} k \cdot x^{-i}.
\end{equation*}
With this \(k\)-vector space structure, the action is given by multiplication in the obvious way, then killing any nonnegative degree terms.
Any element of \(M\) is killed by a large power of \(x\text{,}\) so \(W^{-1}M=0\) whenever \(W\) is a multiplicatively closed subset of \(R\) with \(W \ni x\text{.}\) Therefore, if \(P \in \Supp(M)\text{,}\) then \(x \in P\text{,}\) and thus \(\Supp(M) \subseteq \{ (x)\}\text{.}\) We will soon see, in , that the support of a nonzero module is nonempty, and thus \(\Supp(M) = \{(x)\}\text{.}\)
On the other hand, the annihilator of the class of \(x^{-n}\) is \(x^n\text{,}\) so
\begin{equation*}
\ann_R(M) \subseteq \bigcap_{n \geqslant 1}(x^n) = 0.
\end{equation*}
In particular, \(V(\ann_R(M)) = \Spec(R)\text{,}\) while \(\Supp(M) = \{ (x) \} \neq \Spec(R)\text{.}\)
Example 12.14. Support of Direct Sum in \(\C[x]\).
Let \(R=\C[x]\text{,}\) and \(M= \displaystyle\bigoplus_{n\in \Z} R/(x-n)\text{.}\) First, note that \(M_\fp = \displaystyle\bigoplus_{n\in \Z} (R/(x-n))_{\fp}\text{,}\) so
\begin{equation*}
\Supp(M)= \bigcup_{n\in \Z} \Supp(R/(x-n))= \bigcup_{n\in \Z} V((x-n))=\{(x-n) \ | \ n \in \Z\}.
\end{equation*}
On the other hand,
\begin{equation*}
\ann_R(M)=\bigcap_{n\in \Z} \ann_R(R/(x-n)) = \bigcap_{n\in \Z} (x-n) = 0.
\end{equation*}
Note that in this example the support is not even closed.
Lemma 12.15. SES Support.
Let \(R\) be a ring, \(L,M,N\) be modules. If
\begin{equation*}
\begin{CD}
0@>>>L@>>>M@>>>N@>>>0
\end{CD}
\end{equation*}
is exact, then \(\Supp(L) \cup \Supp(N) = \Supp(M)\text{.}\)
Proof.
localization is exact, so for any \(P\text{,}\)
\begin{equation*}
\begin{CD}
0@>>>L_{p}@>>>M_{p}@>>>N_{p}@>>>0
\end{CD}
\end{equation*}
is exact. If \(P \in \Supp(L) \cup \Supp(N)\text{,}\) then \(L_P\) or \(N_P\) is nonzero, so \(M_P\) must be nonzero as well. On the other hand, if \(P \notin \Supp(L) \cup \Supp(N)\text{,}\) then \(L_P = N_P =0\text{,}\) so \(M_P=0\text{.}\)
Example 12.16. Support’s Preserve Containment.
As a corollary of this
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - SES Support|Lemma]], given modules \(L \subseteq M\text{,}\) \(\Supp(L) \subseteq \Supp (M)\text{.}\)
Lemma 12.17. Nonzero Element Localizes.
Let \(R\) be a ring, \(M\) an \(R\)-module, and \(m\in M\text{.}\) The following are equivalent: (a) \(m=0\) in \(M\text{.}\) (b) \(\frac{m}{1}=0\) in \(M_P\) for all \(P \in \Spec(R)\text{.}\) (c) \(\frac{m}{1}=0\) in \(M_P\) for all \(P \in \mSpec(R)\text{.}\)
Proof.
The implications \(1) \Rightarrow 2) \Rightarrow 3)\) are clear. To show \(3) \Rightarrow 1)\text{,}\) we prove the contrapositive. Given \(m \neq 0\text{,}\) its annihilator is a proper ideal, which must be contained in a maximal ideal. In particular, \(V(\ann_R m)=\Supp(Rm)\) contains a maximal ideal, say \(P\text{,}\) so \(\frac{m}{1} \neq 0\) in \(M_P\text{.}\)
Corollary 12.18. Support of Nonzero Module Nonempty.
Let \(R\) be a ring, \(M\) an \(R\)-module, and \(m\in M\text{.}\) The following are equivalent: (a) \(M=0\text{.}\) (b) \(M_P=0\) in \(M_P\) for all \(P \in \Spec(R)\text{.}\) (c) \(M_P=0\) in \(M_P\) for all \(P \in \mSpec(R)\text{.}\)
Proof.
The implications \(\Rightarrow\) are clear. To show \(3) \Rightarrow 1)\text{,}\) we show the contrapositive. If \(m\neq 0\text{,}\) consider \(Rm \subseteq M\text{.}\) By
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Nonzero Element Localizes|Lemma]], there is a maximal ideal in \(\Supp(Rm)\text{,}\) and by [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - SES Support|Lemma]] applied to the inclusion \(Rm \subseteq M\text{,}\) this maximal ideal is in \(\Supp(M)\) as well.
