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\begin{document}

\section{Primary ideals and isolated
components}\label{primary-ideals-and-isolated-components}

\emph{Written by GPT-6.1 Sol (OpenAI) in Codex at Ultra reasoning
effort, October 2026. Self-checked by the writing AI, GPT-6.1 Sol, at
Ultra. No independent review is claimed. Public domain (CC0).}

An irreducible intersection decomposition need not be unique. We now ask
what can still be recovered without choosing a decomposition.
Localization singles out groups of primary components. A second
construction groups associated primes by containment; a third groups the
actual connected pieces of the spectrum. These are different kinds of
separation.

We assume a Noetherian commutative ring \(R\) with identity. We use the
finite irreducible decompositions proved in
\href{../NOE-IDL-01.html}{Chain conditions and irreducible
decompositions}. The earlier lesson
\href{../prerequisites/AG-CA/AG-CA-04.html}{Associated primes and
primary decomposition} contains actual proofs of associated-prime
existence and zero divisors (Theorem 1.2), submodule and direct-sum
inclusions (Proposition 2.1), finiteness (Theorem 2.2), localization and
minimal support (Theorems 3.1--3.2), primary modules and ideals
(Proposition 4.1 and Theorem 4.2), primary-decomposition existence
(Theorem 5.3), and uniqueness (Theorems 6.1--6.2). We use those results
with their Noetherian and finite-module hypotheses. Finite prime
avoidance and the Chinese remainder construction are proved below.
\href{../prerequisites/AG-CA/AG-CA-01.html}{Spectra of rings}, Theorem
1.2, Proposition 2.1 and Theorems 4.1 and 4.3, proves the radical
correspondence, closed-set operations and finite irreducible components;
\href{../prerequisites/AG-CA/AG-CA-03.html}{Noetherian and Artinian
rings}, Proposition 2.2, proves nilradical nilpotence. Freely accessible
references are Noether's English working edition and the Stacks project.

\subsection{1. The chain argument that makes a component
primary}\label{the-chain-argument-that-makes-a-component-primary}

An ideal \(Q\ne R\) is \textbf{primary} if \(ab\in Q\) and \(b\notin Q\)
imply \(a^n\in Q\) for some \(n\ge1\). Its radical is a prime. For a
submodule \(N\ne M\), the corresponding condition is that \(av\in N\),
\(v\notin N\) imply \(a^nM\subset N\). Equivalently, multiplication by
each scalar on \(M/N\) is either injective or nilpotent. These
conditions agree for ideals, using finite generation to pass from
individual powers to a common power.

\textbf{Theorem 1.1 (Noether's primary lemma).} An irreducible submodule
of a Noetherian module over a commutative ring is primary.

\textbf{Proof.} Pass to \(U=M/N\), so zero is irreducible and \(U\) is
uniform. Fix \(a\in R\) whose multiplication map has a nonzero kernel.
The ascending chain \(0:_U a^j\) stabilizes, say at \(j=h\ge1\). The
submodules \(a^hU\) and \(0:_U a^h\) intersect trivially: if \(a^hu\)
lies in the latter, then \(a^{2h}u=0\), so stabilization implies
\(a^hu=0\). The kernel submodule is nonzero. Uniformity therefore forces
\(a^hU=0\). Thus every noninjective scalar is nilpotent on \(U\), as
required. \(\square\)

In the ideal version, if \(ab\in Q\) with \(b\notin Q\), choose \(h\)
with \((Q:a^h)=(Q:a^{2h})\). Then

\[
Q=(Q+a^hR)\cap(Q+bR).
\]

For if \(z=q+a^hr=q'+bs\), multiplication by \(a\) gives
\(a^{h+1}r\in Q\), since \(ab\in Q\). Stabilization gives \(a^hr\in Q\),
so \(z\in Q\). Irreducibility and \(b\notin Q\) force \(a^h\in Q\). This
is the scalar form of the same proof.

Together with finite irreducible decompositions this gives Noether's
existence route to primary decomposition. The associated-prime route and
the classical uniqueness statements are the cited prerequisite theorems.
We now prove the further uniqueness statements specific to this lesson.

\subsection{2. Selecting an entire isolated
set}\label{selecting-an-entire-isolated-set}

\textbf{Lemma 2.0 (finite prime avoidance).} Let \(J,I_1,\ldots,I_r\) be
ideals of a commutative ring, with all but at most two of the \(I_i\)
prime. If \(J\not\subset I_i\) for every \(i\), there is
\(a\in J\setminus\bigcup_i I_i\). In particular, if an ideal is
contained in a finite union of prime ideals, it is contained in one of
them.

\textbf{Proof.} For \(r=0\), choose \(a=0\); for \(r=1\), the assertion
is the hypothesis. For \(r=2\), choose \(x\in J\setminus I_1\) and
\(y\in J\setminus I_2\). If neither already avoids both, then
\(x\in I_2\) and \(y\in I_1\), and \(x+y\) avoids both by subtraction.

Induct on \(r\). Delete an ideal contained in another: this changes
neither the union nor the desired conclusion. If deletion reduces the
list, use induction. Otherwise there are no inclusions, and for
\(r\ge3\) we may put a prime ideal last. Induction gives \(x\in J\)
outside \(I_1,\ldots,I_{r-1}\). If \(x\notin I_r\), it works. Otherwise
choose \(b\in J\setminus I_r\) and, for each \(i<r\), choose
\(b_i\in I_i\setminus I_r\), using noncontainment. Set
\(y=b\prod_{i<r}b_i\). Then \(y\in J\cap\bigcap_{i<r}I_i\), while
primality gives \(y\notin I_r\). For \(i<r\), membership of \(x+y\) in
\(I_i\) would force \(x\in I_i\); membership in \(I_r\) would force
\(y\in I_r\). Thus \(x+y\) avoids the entire list. The last assertion is
the contrapositive for a list of prime ideals. \(\square\)

Fix a primary decomposition

\[
I=\bigcap_{\mathfrak p\in P}Q_{\mathfrak p},
\qquad P=\operatorname{Ass}(R/I),
\]

with distinct radicals and no redundant component. A subset
\(S\subset P\) is \textbf{isolated} if
\(\mathfrak q\subset\mathfrak p\), \(\mathfrak q\in P\),
\(\mathfrak p\in S\) imply \(\mathfrak q\in S\). This is downward
closure within the finite associated set; it does not ask that every
prime of \(R\) below \(\mathfrak p\) belong to \(P\).

\textbf{Theorem 2.1.} Put
\(T=R\setminus\bigcup_{\mathfrak p\in S}\mathfrak p\). Then

\[
I_S:=\bigcap_{\mathfrak p\in S}Q_{\mathfrak p}
=\{r\in R:r/1\in I R_T\}.
\]

In particular \(I_S\) is independent of the chosen primary
decomposition.

\textbf{Proof.} The set \(T\) is multiplicatively closed. For a
\(\mathfrak q\)-primary ideal \(Q\), if
\(T\cap\mathfrak q\ne\varnothing\), a power of an element of this
intersection lies in \(Q\), so \(QR_T=R_T\). If
\(T\cap\mathfrak q=\varnothing\), the contraction of \(QR_T\) is exactly
\(Q\): from \(tr\in Q\) and \(t\notin\mathfrak q\), primaryness gives
\(r\in Q\).

The primes of \(P\) disjoint from \(T\) are exactly those in \(S\). One
inclusion is immediate. For the other, disjointness says
\(\mathfrak q\subset\bigcup_{\mathfrak p\in S}\mathfrak p\). Lemma 2.0
puts \(\mathfrak q\) inside one of those primes, and isolation puts it
in \(S\). Localization preserves finite intersections, so only the
selected components survive, with their contractions unchanged. This
proves the formula. If \(S=\varnothing\), take \(T=R\), which contains
zero; the localization is zero and the contraction is \(R\), agreeing
with the empty intersection. \(\square\)

For \((x^2,xy)\), the associated primes are \((x)\subset(x,y)\). The set
consisting of \((x)\) is isolated, and localization gives
\(I_{\{(x)\}}=(x)\). The singleton consisting of \((x,y)\) is not
isolated; the component \((x^2,y+ax)\) can vary.

\subsection{3. Relative primality is
directional}\label{relative-primality-is-directional}

Define the ideal quotient \((B:A)=\{r:rA\subset B\}\). Say that \(A\) is
\textbf{relatively prime to \(B\)} if \((B:A)=B\). This means that the
entire ideal \(A\) kills no nonzero element of \(R/B\). Say
\textbf{mutually relatively prime} when both directions hold. These
words do not mean comaximal.

\textbf{Theorem 3.1.} For \(B\ne R\), the following are equivalent:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \((B:A)=B\).
\item
  \(A\) is contained in no prime of \(\operatorname{Ass}(R/B)\).
\item
  No prime of \(\operatorname{Ass}(R/A)\) is contained in a prime of
  \(\operatorname{Ass}(R/B)\).
\end{enumerate}

\textbf{Proof.} If \(A\subset\mathfrak p=\operatorname{Ann}(v)\) for a
nonzero \(v\in R/B\), then \(Av=0\), so the ideal quotient is larger
than \(B\). Conversely, if \(K=0:_{R/B}A\ne0\), it has an associated
prime \(\mathfrak p\), and an element of \(K\) with that annihilator is
also in \(R/B\). Necessarily \(A\subset\mathfrak p\). This proves
equivalence of the first two conditions. For any prime \(\mathfrak p\)
containing \(A\), there is a minimal prime \(\mathfrak q\) over \(A\)
contained in \(\mathfrak p\) (Spectra of rings, Lemma 4.2), and this
\(\mathfrak q\) is associated to \(R/A\) (Associated primes and primary
decomposition, Theorem 3.2). Conversely every associated prime of
\(R/A\) contains \(A\). These facts prove equivalence with the third
condition, including \(A=R\), whose associated set is empty. \(\square\)

Lemma 2.0 also provides a single element of \(A\) outside all associated
primes of \(R/B\) when the conditions hold. Such an element acts
injectively on \(R/B\).

For \(A=(x^2,y)\) and \(B=(x)\), the element \(y\in A\setminus(x)\)
proves relative primality to \(B\). The reverse fails: \(x\notin A\),
but \(xB\subset A\). Equivalently the prime \((x)\) associated to \(B\)
lies inside \((x,y)\), which is associated to \(A\).

\subsection{4. A graph on the associated
primes}\label{a-graph-on-the-associated-primes}

Form the finite \textbf{comparability graph} on
\(P=\operatorname{Ass}(R/I)\): connect distinct primes when one contains
the other. Let its connected components be \(S_1,\ldots,S_t\). Every
\(S_i\) is isolated, because any containment is an edge. Thus each
\(I_{S_i}\) in Theorem 2.1 is canonical.

\textbf{Theorem 4.1 (Noether's relative decomposition).} Every proper
ideal has a unique unordered decomposition

\[
I=I_{S_1}\cap\cdots\cap I_{S_t}
\]

into mutually relatively prime proper ideals which admit no further such
decomposition. Its components are indexed by the connected components of
the comparability graph.

\textbf{Proof.} Group a primary decomposition by the \(S_i\). Each
grouped decomposition stays irredundant within its group, since deleting
a primary component there would also delete it from the original
decomposition. The first uniqueness theorem for primary decomposition
therefore gives \(\operatorname{Ass}(R/I_{S_i})=S_i\). There are no
containments between different \(S_i\), so Theorem 3.1 proves mutual
relative primality.

We verify that every mutually relatively prime decomposition comes from
grouping in this fashion. Suppose \(I=\bigcap_i A_i\), with all
\(A_i\ne R\) and mutually relatively prime. Put
\(B_i=\bigcap_{j\ne i}A_j\). For any
\(\mathfrak p\in\operatorname{Ass}(R/A_i)\), every \(A_j\), \(j\ne i\),
is not contained in \(\mathfrak p\). Choosing one element outside
\(\mathfrak p\) from each and multiplying shows
\(B_i\not\subset\mathfrak p\). If \(v\in R/A_i\) has annihilator
\(\mathfrak p\), choose \(b\in B_i\setminus\mathfrak p\). Then
\(bv\ne0\), still with annihilator \(\mathfrak p\). Its representative
in \(B_i/I\subset R/I\) proves \(\mathfrak p\in P\). The diagonal
injection \(R/I\hookrightarrow\bigoplus R/A_i\) proves the reverse
inclusion for their union. Theorem 3.1 also says their associated sets
are disjoint and have no containment between them. Hence those sets
partition \(P\) into unions of entire graph components; each is
isolated.

Let \(S=\operatorname{Ass}(R/A_i)\) and localize at
\(T=R\setminus\bigcup_{\mathfrak p\in S}\mathfrak p\). The other factors
become the unit ideal: a minimal prime over any such factor cannot be
contained in any member of \(S\); Lemma 2.0 gives an element of that
factor in \(T\). The factor \(A_i\) contracts unchanged, by its primary
decomposition and Theorem 2.1 applied to the full set \(S\). Thus
\(A_i=I_S\). A factor with two graph components splits further by the
already constructed decomposition. A factor with one component cannot
split, since the same argument would separate that connected graph. This
proves both irreducibility and uniqueness. \(\square\)

For an explicit pair of comparable pairs, use

\[
I=(x^2,xy)\cap((x-1)^2,(x-1)(y-1))\subset k[x,y].
\]

The first associated pair is \((x)\subset(x,y)\); the second is
\((x-1)\subset(x-1,y-1)\). The two factors contain \(x^2\) and
\((x-1)^2\), respectively. These polynomials are coprime in \(k[x]\), so
Bézout makes the factors comaximal. The quotient is their product, and
its associated set is exactly the two indicated pairs. The relative
decomposition has the two displayed factors.

\subsection{5. Separating connected
spaces}\label{separating-connected-spaces}

\textbf{Lemma 5.0 (Chinese remainders).} Pairwise comaximal ideals
\(J_1,\ldots,J_s\) satisfy

\[
R/\bigcap_iJ_i\cong\prod_iR/J_i,
\qquad \bigcap_iJ_i=\prod_iJ_i.
\]

\textbf{Proof.} For two ideals \(A+B=R\), choose \(a\in A,b\in B\) with
\(a+b=1\). If \(z\in A\cap B\), then \(z=za+zb\in AB\), so
\(A\cap B=AB\). Given residues represented by \(u,v\), the element
\(ub+va\) realizes them modulo \(A,B\); the map's kernel is their
intersection. For each \(i\) in a finite pairwise comaximal list, choose
\(b_{ij}\in J_j\) congruent to \(1\) modulo \(J_i\), for \(j\ne i\).
Their product \(e_i\) is \(1\) modulo \(J_i\) and \(0\) modulo every
other ideal. Thus \(\sum_i e_i u_i\) realizes arbitrary residues. The
same products show \(J_i+\prod_{j\ne i}J_j=R\), so the two-ideal
intersection identity proves the product formula by induction.
\(\square\)

An idempotent \(e\) is \textbf{primitive} if it is nonzero and cannot be
written as \(f+g\) with nonzero idempotents \(f,g\) satisfying \(fg=0\).

\textbf{Theorem 5.1 (Noether's coprime decomposition).} A proper ideal
\(I\) has a unique unordered decomposition as a product of pairwise
comaximal proper ideals having no further such factorization. These
factors correspond to the connected components of
\(\operatorname{Spec}(R/I)\) and to its primitive nonzero idempotents.

\textbf{Proof.} Write \(A=R/I\). A Noetherian spectrum has finitely many
irreducible components, the sets \(V(\mathfrak p)\) for its minimal
primes. Form a graph on these components, with an edge when they
intersect. The union in each graph component is connected: each
irreducible space is connected, and adjoining a connected space meeting
the preceding union preserves connectedness. These finitely many unions
are mutually disjoint closed subsets, hence also open. They are exactly
the connected components.

We give the algebraic passage from a clopen partition
\(\operatorname{Spec}(A)=V(J)\sqcup V(K)\). Disjointness yields
\(J+K=A\), and covering yields \(JK\subset\sqrt0\). The nilradical of a
Noetherian ring is nilpotent: it is finitely generated by nilpotents,
and a sufficiently long product of its generators vanishes. Choose \(h\)
with \((JK)^h=0\). The ideals \(J^h,K^h\) remain comaximal, since expand
\((a+b)^{2h-1}=1\) for \(a+b=1\), \(a\in J,b\in K\). Every term belongs
to one of those powers. Their intersection is their product, which is
zero. Chinese remainders therefore give \(A\cong A/J^h\times A/K^h\).
The element corresponding to \((1,0)\) is idempotent.

Conversely an idempotent \(e\) gives \(A\cong eA\times(1-e)A\) and the
clopen partition \(D(e)\sqcup D(1-e)\). The idempotent for a specified
clopen subset is unique. If idempotents \(e,f\) have the same values at
every prime, then \(e(1-f)\) and \(f(1-e)\) are nilpotent idempotents
and therefore zero, so \(e=f\). Thus connected components give unique
orthogonal primitive idempotents \(e_i\), with sum one. Put
\(J_i=\ker(A\to e_iA)=(1-e_i)A\), and take their inverse images in
\(R\). They are pairwise comaximal, their intersection and product are
\(I\), and their quotients are connected. Any other factorization gives
another clopen partition and hence must group these components;
irreducibility of its factors forbids grouping more than one.
\(\square\)

Relative irreducibility implies coprime irreducibility, because
comaximal factors are mutually relatively prime. The reverse fails for
\((xy)\). Its associated primes \((x),(y)\) are incomparable, so
\((xy)=(x)\cap(y)\) is a relative decomposition. Yet its spectrum is
connected: the two axes meet at the origin. Thus \((xy)\) has no
nontrivial comaximal factorization.

For

\[
I=(x)\cap(y)\cap(x-1,y-1),
\]

the axes and the point \((1,1)\) are separate connected components. The
coprime decomposition is \(I=(xy)(x-1,y-1)\): modulo the point ideal,
\(xy=1\), so the factors are comaximal. In \(R/I\), \(xy\) is the
idempotent equal to zero on the axes and one at the point; \(1-xy\) is
the other primitive idempotent. The relative decomposition has three
factors, since its three associated primes are incomparable. The
comaximal decomposition has two.

For the Artinian ring
\(k[\epsilon]/(\epsilon^2)\times k[\delta]/(\delta^3)\), the two
primitive idempotents are \((1,0),(0,1)\). Its zero ideal is the
intersection and product of the two projection kernels. Each factor
quotient is local and therefore connected.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={A containment graph and two crossing axes with a separate point}]{assets/two-kinds-of-separation.png}}
\caption{A containment graph and two crossing axes with a separate
point}
\end{figure}

\emph{Figure 1.} Left: the comparability graph for
\((x^2y,xy^2)=(x)\cap(y)\cap(x^2,y^2)\); the embedded origin joins the
two minimal primes into one relative group. Right: the geometric example
of the axes and \((1,1)\) from Section 5 has two connected pieces. These
depict different ideals and different tests. The graph edges mean prime
containment; the crossing axes mean actual intersection of closed sets.
The proofs are Theorems 4.1 and 5.1. Original diagram, CC0; rendered
DejaVu glyphs retain their font licence.

\subsection{6. Exercises}\label{exercises}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \textbf{Basic.} For \((x^2,xy)\), verify the family
  \((x)\cap(x^2,y+ax)\), identify its associated primes, and state which
  singleton set is isolated.
\item
  \textbf{Intermediate.} Derive Theorem 3.1 using the zero-divisor
  theorem and prime avoidance, rather than associated primes of
  \(0:_{R/B}A\).
\item
  \textbf{Intermediate.} Find the comaximal decomposition and primitive
  idempotents for the axes together with \((1,1)\).
\item
  \textbf{Intermediate.} Construct an ideal with exactly three
  associated primes \(\mathfrak p_1\subsetneq\mathfrak p_2\) and
  \(\mathfrak p_3\) incomparable with both; compute its relative
  decomposition.
\item
  \textbf{Advanced.} Prove that \(I\) has no nontrivial comaximal
  factorization exactly when \(\operatorname{Spec}(R/I)\) is connected.
  Explain why the same conclusion does not use the comparability graph
  alone.
\end{enumerate}

\subsection{7. Solutions}\label{solutions}

\textbf{1.} The calculation in the preceding lesson gives equality and
irredundancy. The first factor is prime, and the second quotient is
\(k[x]/(x^2)\), so its unique associated prime is \((x,y)\). The first
uniqueness theorem gives associated set \(\{(x),(x,y)\}\). Only
\(\{(x)\}\) is an isolated singleton. Its component is fixed; the other
varies.

\textbf{2.} If \(A\) is contained in none of the finitely many
associated primes of \(R/B\), avoidance gives \(a\in A\) outside their
union. The zero-divisor theorem makes multiplication by \(a\) injective.
Thus any element killed by the whole ideal \(A\) is zero. If
\(A\subset\mathfrak p\) for an associated prime, its witnessing element
is killed by all of \(A\), giving the converse. Minimal primes over
\(A\) supply the third equivalent condition as in Theorem 3.1.

\textbf{3.} The factors are \((xy)\) and \((x-1,y-1)\). Since \(1-xy\)
belongs to the point ideal, their sum is \(R\). Their quotients are
connected (crossing axes and a point). Modulo \(I\),
\((xy)^2-xy=xy(xy-1)\in I\), so \(xy\) is idempotent; \(1-xy\) is its
complement. Both are nonzero and primitive because their factors have
connected spectra.

\textbf{4.} Take \(I=(x^2,xy)\cap(x-1,y-1)\). The factors are comaximal
because \(x^2\) evaluates to one at \((1,1)\). The first quotient has
associated primes \((x)\subset(x,y)\), and the second has \((x-1,y-1)\).
No containment links the point prime to the first pair: \(x\) does not
vanish at the point, while \(x-1\) is in neither prime of that pair.
Thus Theorem 4.1 gives exactly the displayed two relative factors.

\textbf{5.} A comaximal factorization gives a nontrivial product of
quotient rings and hence a clopen partition. A clopen partition gives a
nontrivial idempotent by the explicit nilpotence and Chinese-remainder
construction in Theorem 5.1. Its projection kernels lift to a nontrivial
comaximal factorization. The comparability graph misses intersections of
incomparable components: \((x)\) and \((y)\) are incomparable but their
closed sets meet at \((x,y)\). Connectedness of the spectrum uses
intersections of irreducible components, not only containments among
associated primes.

\subsection{In Noether's words}\label{in-noethers-words}

Sections 4--8 of
\href{https://github.com/KokunoYumeto/emmy-noether-en/blob/main/source/Noether_English_ED0014.tex}{Ideal
Theory in Ring Domains, work 19} distinguish primary components,
directional relative primality and coprimality. Definition V and
Theorems X--XII govern the relative decomposition; Section 7 treats
isolated ideals; Section 8 treats coprime factors. The modern
localization formulation above fixes which direction of prime
containment is meant. \href{https://doi.org/10.5281/zenodo.21908301}{The
German authority} accompanies the English edition.

\subsection{References}\label{references}

\begin{itemize}
\tightlist
\item
  Emmy Noether, \emph{Idealtheorie in Ringbereichen}, 1921, Sections
  4--8.
  \href{https://github.com/KokunoYumeto/emmy-noether-en/blob/main/source/Noether_English_ED0014.tex}{Free
  English working edition, work 19}, with the original bibliographic
  information and machine-assisted translation provenance.
\item
  The Stacks project authors,
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html\#section-ass}{associated
  primes, Tag 00L9},
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html\#lemma-silly}{prime
  avoidance, Tag 00DS},
  \href{https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html\#lemma-product-local}{Artinian
  decomposition, Tag 00JA}. These AI Integrated Stacks Project links use
  the edition described in the course introduction. No
  primary-decomposition theorem is attributed to a Stacks tag.
\end{itemize}

\subsection{Editable sources}\label{editable-sources}

\href{ideal-theory-in-rings-source.zip}{Complete source archive} ·
\href{ideal-theory-in-rings.tex}{Complete course in LaTeX} ·
\href{NOE-IDL-02.tex}{This lesson in LaTeX}. The archive includes all
five lesson texts, the figures and their reproducible source.

\end{document}