The converse and composition series

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

Ideal factorization appears to be a conclusion about multiplication. In fact, it also forces finite generation, dimension one and integral closedness. We first recover these properties from unique factorization, then read prime multiplicities from the simple constituents of quotient modules. Finally, we prove that the existence of factorizations already forces uniqueness. This connects ideal theory to a general invariant, module length.

Read Noether's axioms for Dedekind domains first. We use basic localization, the Chinese remainder theorem, and Jordan–Hölder from Noetherian and Artinian rings, Theorem 3.2: two composition series of the same module have isomorphic simple factors with the same multiplicities. For the factorization theorem by local DVRs, compare Discrete valuation rings and Dedekind domains, Theorem 3.2. Basic references are [Noether], [Stacks] and [Milne].

1. Invertibility cannot be assumed in the converse

Let \(R\) be a domain with identity in which every nonzero integral ideal has a unique finite factorization into nonzero prime ideals. We have not assumed any chain condition. The equality \(\mathfrak p=\mathfrak p^2\) is impossible: its two sides would be distinct factorizations of the same nonzero ideal. Thus choose \(x\in\mathfrak p\setminus\mathfrak p^2\).

Lemma 1.1. Every nonzero prime \(\mathfrak p\) is a factor of a nonzero principal ideal, and hence is invertible.

Proof. For any \(y\in\mathfrak p\), factor \((x,y^2)\). At least one factor is contained in \(\mathfrak p\), by primality. At most one is: two such factors would put \(x\) in \(\mathfrak p^2\). After localization at \(\mathfrak p\), all the other factors become unit ideals. Therefore \((x,y^2)R_{\mathfrak p}\) is a prime ideal. It contains \(y^2\), hence \(y\), so

Since \(by\) belongs to the maximal ideal, \(1-by\) is a unit. It follows that \(y\in(x)R_{\mathfrak p}\). As this works for every \(y\in\mathfrak p\),

Now factor \((x)\). Exactly one factor \(\mathfrak q\) is contained in \(\mathfrak p\), by the same argument; therefore \(\mathfrak qR_{\mathfrak p}=(x)R_{\mathfrak p}=\mathfrak pR_{\mathfrak p}\). Contraction of a localized prime contained in \(\mathfrak p\) recovers that prime, giving \(\mathfrak q=\mathfrak p\). Write \((x)=\mathfrak pJ\). Then \(\mathfrak p(x^{-1}J)=R\), as required. \(\square\)

This argument uses localization only to find a principal multiple. The forward factorization theorem in the previous lesson remains a global argument. The next steps of the converse also take place globally.

Lemma 1.2. An invertible fractional ideal is finitely generated.

Proof. If \(IJ=R\), choose a finite expression \(1=\sum_{i=1}^n a_i b_i\), where \(a_i\in I\), \(b_i\in J\). For \(a\in I\),

and every coefficient \(ab_i\) belongs to \(R\). Hence the \(a_i\) generate \(I\). \(\square\)

All nonzero ideals factor into invertible primes, so all are invertible and finite. The zero ideal is finite too. Thus \(R\) is Noetherian.

Lemma 1.3. Every nonzero prime is maximal.

Proof. Suppose \(0\ne\mathfrak p\subseteq\mathfrak m\), with \(\mathfrak m\) maximal. Since \(\mathfrak m\) is invertible, \(J=\mathfrak p\mathfrak m^{-1}\) is a nonzero integral ideal. Then \(\mathfrak p=J\mathfrak m\). Unique factorization forces \(J=R\) and \(\mathfrak p=\mathfrak m\). \(\square\)

2. Integral closedness from exponents

Factor fractional ideals by clearing a denominator, exactly as in the preceding lesson. For a nonzero prime \(\mathfrak p\), define \(v_{\mathfrak p}(I)\) to be its exponent in the fractional ideal \(I\), and write \(v_{\mathfrak p}(x)=v_{\mathfrak p}((x))\) for \(x\in K^\times\). Product exponents add. Furthermore,

For the forward direction \(IJ^{-1}\) is integral and so has nonnegative exponents; the reverse direction follows by factoring it. In particular, \(x\in R\) if and only if all its exponents are nonnegative.

Lemma 2.1. For fractional ideals \(I,J\),

Proof. Put \(H=I+J\). The integral ideals \(IH^{-1}\) and \(JH^{-1}\) sum to \(R\). They cannot both lie in \(\mathfrak p\). Their \(\mathfrak p\)-exponents are therefore nonnegative with at least one zero. Subtracting \(v_{\mathfrak p}(H)\) proves the formula. \(\square\)

Consequently \(v_{\mathfrak p}(x+y)\ge\min(v_{\mathfrak p}(x),v_{\mathfrak p}(y))\) when \(x+y\ne0\). If the two exponents differ, equality holds: a strict inequality, applied to \(x=(x+y)-y\), would contradict the smaller exponent. Thus a sum with one term of strictly smallest exponent cannot vanish.

Theorem 2.2, the converse. A domain with identity and unique prime factorization of nonzero ideals satisfies all five axioms of the preceding lesson.

Proof. Noetherianity and dimension at most one were proved in Section 1. For normality, suppose

If \(x\notin R\), choose \(\mathfrak p\) with \(v_{\mathfrak p}(x)<0\). Then the first term has exponent \(nv_{\mathfrak p}(x)\), strictly smaller than every nonzero later term, because \(v_{\mathfrak p}(c_i)\ge0\). The preceding observation makes the sum nonzero, a contradiction. Hence \(x\in R\). The identity and domain axioms were hypotheses, and the Artinian criterion supplies the descending chain axiom. \(\square\)

Reference: [Noether, Section 9] gives the converse; [Stacks, Tag 034X] gives the equivalent characterizations. The exponent argument above proves integral closedness without using the local normality criterion.

3. Quotients record multiplicities

We recall three isomorphisms for modules. A homomorphism \(f:M\to N\) induces \(M/\ker f\simeq\operatorname{im}f\), by sending a coset to its image. If \(U,V\subseteq M\), the same map applied to \(U\to(U+V)/V\) gives \(U/(U\cap V)\simeq(U+V)/V\). Finally, for \(U\subseteq V\subseteq M\), mapping the coset of \(m+U\) to \(m+V\) gives \((M/U)/(V/U)\simeq M/V\). These are the isomorphism theorems used below.

Lemma 3.1. If \(J\) is a nonzero invertible integral ideal and \(\mathfrak p\) a nonzero prime, then \(J/J\mathfrak p\simeq R/\mathfrak p\).

Proof. It is a vector space over \(R/\mathfrak p\). Write \(1=\sum a_i b_i\) with \(a_i\in J\), \(b_i\in J^{-1}\). At least one \(a_i b_i\notin\mathfrak p\). In \(R_{\mathfrak p}\) that product is a unit, and \(J_{\mathfrak p}=a_iR_{\mathfrak p}\): for \(a\in J\), \(a/a_i=(ab_i)/(a_i b_i)\) lies in \(R_{\mathfrak p}\). Therefore its quotient by \(\mathfrak p\) is one-dimensional. Localization does not change a module annihilated by the maximal ideal \(\mathfrak p\), since all inverted scalars act as nonzero field elements. The original quotient is consequently one-dimensional too. \(\square\)

Theorem 3.2. If \(I=\prod_{i=1}^r\mathfrak p_i^{e_i}\) with distinct primes, then \(R/I\) has finite length \(\sum_i e_i\). Its simple factors are \(R/\mathfrak p_i\), with multiplicity \(e_i\).

Proof. List the prime factors with repetitions as \(\mathfrak q_1,\ldots,\mathfrak q_N\), and form

Each quotient \(J_{j-1}/J_j\) is \(R/\mathfrak q_j\) by Lemma 3.1, hence simple and nonzero. Taking quotients by \(I\) gives a composition series. It has \(N=\sum e_i\) factors. Jordan–Hölder makes the multiset intrinsic. \(\square\)

The annihilator of the simple module \(R/\mathfrak p\) is exactly \(\mathfrak p\). Two such modules can be isomorphic only when their primes agree. Thus Jordan–Hölder yields a second proof of uniqueness of prime exponents once existence and invertibility are established. Length counts ideal factors, not the logarithm of the size of a quotient unless all residue fields have the same size.

For \(\mathbb Z/12\), the chain \(\mathbb Z\supset2\mathbb Z\supset4\mathbb Z\supset12\mathbb Z\) gives factors \(\mathbb Z/2,\mathbb Z/2,\mathbb Z/3\). In \(\mathbb Z[i]\), \((2)=(1+i)^2\) as ideals because \(2\) and \((1+i)^2\) differ by a unit. Hence \(\mathbb Z[i]/(2)\) has two factors \(\mathbb F_2\); its cardinality is four and its length is two.

4. Non-normal orders and the stronger theorem

Let \(R=\mathbb Z[s]\), \(s^2=-3\). The reduction modulo \(2\) is \(\mathbb F_2[s]/((s+1)^2)\), with its unique prime \(\mathfrak p=(2,1+s)\). Direct multiplication gives

since \((1+s)^2=2(s-1)=2(s+1)-4\). If \((2)\) were a product of primes, every factor would contain \((2)\) and therefore equal \(\mathfrak p\). Exponent one fails because \(1+s\notin(2)\). For \(n\ge2\), \(\mathfrak p^n=2^{n-1}\mathfrak p\subseteq2\mathfrak p\), and \(2\notin2\mathfrak p\). Thus no exponent works.

Similarly let \(R=k[t^2,t^3]\) and \(\mathfrak m=(t^2,t^3)\). The quotient \(R/(t^2)\) is \(k[t^3]/((t^3)^2)\), so \(\mathfrak m\) is the only prime containing \((t^2)\). The ideal \((t^2)\) is not \(\mathfrak m\), because \(t^3\notin(t^2)\). It is not \(\mathfrak m^n\) for \(n\ge2\), because those powers have no term of degree two. Again factorization fails.

Existence alone forces the full converse

Sections 1 and 2 began with unique factorization. To remove that hypothesis, we must establish invertibility by another route: the argument \(\mathfrak p\ne\mathfrak p^2\) is not yet available. The following observation gives a starting point even in a domain without any chain condition.

Lemma 4.1, principal factors. Let \(A\) be any domain with identity. If a nonzero principal ideal has a finite factorization into prime ideals, all those factors are invertible. Moreover, any two such factorizations of that principal ideal have the same prime factors with the same multiplicities.

Proof. If \((c)=\mathfrak q_1\cdots\mathfrak q_r\), the fractional ideal

is an inverse of \(\mathfrak q_i\). The product with no factors means \(A\). This proves invertibility without assuming that the primes are finitely generated.

For uniqueness, compare two factorizations. If both are nonempty, choose a prime minimal under inclusion among the finitely many primes in both lists, and put it first in the list in which it occurs. Since the equal product from the other list is contained in this prime, primality forces one of its factors to be contained in the chosen prime. Minimality makes the two primes equal. Multiply the equality by their fractional inverse to cancel them, and repeat. If one list becomes empty, its product is \(A\); a nonempty product of proper ideals is contained in each of its factors and cannot equal \(A\). Hence both lists finish together. This cancellation argument also proves uniqueness for any ideal whenever all factors in the two proposed factorizations are invertible. \(\square\)

Lemma 4.2, the local obstruction. Suppose \((A,\mathfrak m)\) is a local domain with identity and every nonzero proper ideal is a finite product of prime ideals. Every nonzero invertible prime ideal of \(A\) equals \(\mathfrak m\).

Proof. Let \(\mathfrak p\) be such a prime. Write \(1=\sum_i u_i v_i\), with \(u_i\in\mathfrak p\) and \(v_i\in\mathfrak p^{-1}\). Some \(u_i v_i\) is a unit, since otherwise their sum would lie in \(\mathfrak m\). For every \(u\in\mathfrak p\),

Thus \(\mathfrak p=(\pi)\), where \(\pi=u_i\ne0\). Suppose \(a\in\mathfrak m\setminus\mathfrak p\). Factor the two nonzero proper ideals

Each factor contains its product, hence contains \(\mathfrak p\). Every \(\mathfrak q_i\) also contains \(a\); every \(\mathfrak r_j\) contains \(a^2\), and therefore \(a\) by primality. Consequently all these primes strictly contain \(\mathfrak p\).

The quotient map \(A\to B=A/\mathfrak p\) is a surjective homomorphism of rings, and \(B\) is a domain. Write \(\bar a\) for the nonzero image of \(a\). Images of the displayed factors are nonzero proper prime ideals of \(B\), and ideal multiplication commutes with this quotient map. We therefore have prime factorizations

Lemma 4.1, applied in \(B\), identifies the two factor multisets of \((\bar a^2)\). Ideals containing \(\mathfrak p\) are recovered by taking inverse images under \(A\to B\), so the original prime multisets agree too. In particular,

Since \(\pi\in H\), this gives an equation

The element \(1-u\) is a unit. From \(ba^2=\pi(1-u)\) and \(a\notin\mathfrak p\), primality forces \(b\in\mathfrak p=(\pi)\). Write \(b=\pi d\) and cancel the nonzero \(\pi\). Then \(1-u=da^2\) lies in \(\mathfrak m\), contradicting its being a unit. No such \(a\) exists, and \(\mathfrak p=\mathfrak m\). \(\square\)

The quotient argument uses uniqueness only for principal ideals whose prime factors are already invertible by Lemma 4.1. It does not assume uniqueness for arbitrary ideals in either \(A\) or \(B\).

Theorem 4.3, Matusita's strengthening. Let \(R\) be a commutative domain with identity. If every nonzero proper ideal is a finite product of prime ideals, then \(R\) is Dedekind, every nonzero prime is invertible, and the prime factorization of every nonzero ideal is unique. Fields are included, with the unit ideal represented by the empty product.

Proof. First we check that the existence hypothesis survives localization. For a maximal ideal \(\mathfrak m\), the canonical ring map \(R\to A=R_{\mathfrak m}\) is injective. Every ideal \(L\subseteq A\) is the extension of its contraction \(I\subseteq R\): if \(x/s\in L\), then \(x/1\in L\), so \(x\in I\), and conversely elements of \(IA\) lie in \(L\). If \(L\) is nonzero and proper, so is \(I\). Factor \(I\) in \(R\) and extend the product to \(A\). A prime factor not contained in \(\mathfrak m\) becomes \(A\); a prime factor contained in \(\mathfrak m\) remains a nonzero proper prime. Deleting the unit factors gives the required factorization of \(L\).

Now let \(\mathfrak p\) be a nonzero prime of \(R\), choose \(0\ne c\in\mathfrak p\), and choose a maximal ideal \(\mathfrak m\supseteq\mathfrak p\). In a factorization of \((c)\), primality of \(\mathfrak p\) gives a factor \(\mathfrak q\subseteq\mathfrak p\). Lemma 4.1 makes \(\mathfrak q\) invertible in \(R\). Its extension \(\mathfrak qA\) is a nonzero prime and is invertible: extending \(\mathfrak q\mathfrak q^{-1}=R\) gives a fractional inverse over \(A\). Lemma 4.2 consequently gives

Contraction recovers either prime. Indeed, if \(sx\) belongs to a prime contained in \(\mathfrak m\), with \(s\notin\mathfrak m\), then \(s\) is outside that prime and primality puts \(x\) in it. Contracting the equality therefore yields \(\mathfrak q=\mathfrak m\), and the inclusions \(\mathfrak q\subseteq\mathfrak p\subseteq\mathfrak m\) give equality throughout. Thus every nonzero prime is maximal and invertible.

Every nonzero ideal is now a finite product of invertible primes, or is \(R\), so it is invertible. Lemma 1.2 makes every ideal finitely generated, including the zero ideal with its empty generating set; hence \(R\) is Noetherian. The cancellation argument in Lemma 4.1 gives uniqueness for all ideal factorizations. With uniqueness established, the exponent argument of Section 2 proves integral closedness, and Theorem 3.2 gives finite length, hence the descending chain condition, for every quotient by a nonzero proper ideal. The quotient by \(R\) is zero. We have obtained all five axioms of lesson one. If \(R\) is a field, these conclusions hold directly, with no nonzero proper ideals to consider. \(\square\)

Matusita proves the existence-only implication in [Matusita, Section 2, Satz 5]. The proof above organizes the argument around principal-factor cancellation and a local contradiction. Together with the forward theorem in lesson one, it shows that existence, unique factorization and the Dedekind axioms are equivalent. Neither uniqueness nor a chain condition needs to be imposed as a separate hypothesis.

5. Exercises

1. Easy. Give the simple factors, their multiplicities and the length of \(\mathbb Z[i]/(6)\).
2. Medium. Explain why \(\mathfrak p\ne\mathfrak p^2\) follows from unique ideal factorization even before Noetherianity is known.
3. Medium. Prove finite generation of an invertible ideal and show that the argument actually exhibits it as a direct summand of a finite free module.
4. Medium. Prove that \((t^2)\) does not factor into prime ideals in \(k[t^2,t^3]\).
5. Hard. Reprove the integral-closedness step using prime exponents only, including the assertion that a uniquely smallest exponent prevents cancellation.

6. Solutions

1. The polynomial \(T^2+1\) is irreducible over \(\mathbb F_3\), so \((3)\) is prime with residue field \(\mathbb F_9\). The prime above \(2\) is \((1+i)\), of exponent two. Thus the factors are \(\mathbb F_2,\mathbb F_2,\mathbb F_9\), and length is three. Their cardinalities multiply to \(2\cdot2\cdot9=36\), agreeing with the rank-two lattice quotient by \(6\).

2. A nonzero prime itself is a factorization with one factor. Its square has the two-factor expression \(\mathfrak p\mathfrak p\). Equality would contradict equality of factor multisets. No finiteness assumption on the generators of \(\mathfrak p\) is involved.

3. Choose \(1=\sum a_i b_i\) as in Lemma 1.2. The map \(R^n\to I\), \((r_i)\mapsto\sum r_i a_i\), has a section \(I\to R^n\), \(a\mapsto(ab_i)_i\). Their composite on \(I\) is the identity. Thus \(I\) is finite and projective.

4. Every factor contains \((t^2)\). The quotient has only the prime \((t^3)\), so all factors are \(\mathfrak m\). The degree argument in Section 4 excludes each power, including exponent one. The empty product is \(R\) and is also excluded.

5. Lemma 2.1 gives the inequality for sums. If \(v(a)<v(b)\) and \(v(a+b)>v(a)\), apply it to \(a=(a+b)-b\) to obtain \(v(a)>v(a)\), a contradiction. Repeatedly adding higher-exponent terms leaves the smallest exponent unchanged; hence a sum with just one smallest term cannot be zero. For an integral equation and \(v(x)<0\), the exponents \(nv(x)\) and \(v(c_i)+(n-i)v(x)\) are strictly ordered with the first smallest. No such equation is possible, so all exponents of \(x\) are nonnegative and \((x)\subseteq R\).

What this lesson does not prove

Jordan–Hölder is proved in Noetherian and Artinian rings, Theorem 3.2, as specified at the start. The two converses, integral closedness, the composition series and both non-factorization examples are proved here. The proof of Theorem 4.3 includes its localization and quotient comparisons.

References

• [Noether] Emmy Noether, Abstrakter Aufbau der Idealtheorie in algebraischen Zahl- und Funktionenkörpern, Mathematische Annalen 96 (1927), 26–61, Sections 4, 9 and 10, freely accessible journal scan.
• [Stacks] The Stacks Project, Tags 034X, 00IU and 00IX, also in the corresponding sections of the AI Integrated Stacks Project English algebra reader. The edition includes AI-authored supplementary material.
• [Milne] J. S. Milne, Algebraic Number Theory, Chapter 3, author's notes.
• [Matusita] Kameo Matusita, Über ein bewertungstheoretisches Axiomensystem für die Dedekind-Noethersche Idealtheorie, Japanese Journal of Mathematics 19 (1944), 97–110, Section 2, especially Satz 5 and the argument on pp. 103–106, freely accessible journal edition.