Noetherian and Artinian Conditions with Fractional Ideals
Noetherian and Artinian Conditions with Fractional Ideals: core definitions, structural results and verification methods in abstract algebra.
How the topic fits together
Noetherian and Artinian Modules and Rings
This source section supplies the definitions, constructions and formal results used in this article. The sequence of results is preserved so later claims are not detached from their prerequisites.
Fractional Ideals
This source section supplies the definitions, constructions and formal results used in this article. The sequence of results is preserved so later claims are not detached from their prerequisites.
Core definitions and structural results
The following cards retain the mathematical content of the source while condensing long proofs into verification strategies. Numerical examples are treated as examples, not universal requirements.
Corollary
The set B of algebraic integers in any number field L is a free Z-module of rank n = [L : Q]. Therefore B has an integral basis. The discriminant is the same for every integral basis; it is known as the field discriminant.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Definitions and Comments In this section, rings are not assumed commutative.
In this section, rings are not assumed commutative. Let M be an R-module, and suppose that one has an increasing sequence of submodules M1 ≤M2 ≤M3 ≤. . ., or a decreasing sequence M1 ≥M2 ≥M3 ≥. . .. Call the sequence stabilizes if for some t, Mt = Mt+1 = Mt+2 = . . .. The question of stabilization of sequences of submodules appears in a fundamental way in many areas of abstract algebra and its applications.
Proof / verification strategy: Use this as a definition checklist: identify the ambient object, test each stated condition separately, then compare examples and non-examples without assuming later theorems.
Proposition
The following conditions on an R-module M are equivalent, and define a Noetherian module: (1) M satisfies the acc; (2) Every nonempty collection of submodules of M has a maximal element (with respect to inclusion). The following conditions on M are equivalent, and define an Artinian module: (1′) M satisfies the dcc; (2′) Every nonempty collection of submodules of M has a minimal element.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Proposition M is Noetherian iffevery submodule of M is finitely generated.
M is Noetherian iffevery submodule of M is finitely generated.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Definition
A ring R is Noetherian [resp. Artinian] if it is Noetherian [resp. Artinian] as a module over itself. If we need to distinguish between R as a left, as opposed to right, R-module, we will refer to a left Noetherian and a right Noetherian ring, and similarly for Artinian rings.
Proof / verification strategy: Use this as a definition checklist: identify the ambient object, test each stated condition separately, then compare examples and non-examples without assuming later theorems.
Examples
1. Every PID is Noetherian. This follows from (7.5.3), since every ideal is generated by a single element. 2. Z is Noetherian (a special case of Example 1) but not Artinian.
Proof / verification strategy: Treat this as an illustration of the surrounding definitions. Recompute the stated relations directly and keep the numerical or structural choices local to the example.
Result
Remark The following observations will be useful in deriving properties of Noetherian and Artinian modules. If N ≤M, then a submodule L of M that contains N can always be written in the form K + N for some submodule K. (K = L is one possibility.) By the correspondence theorem, (K1 + N)/N = (K2 + N)/N implies K1 + N = K2 + N and (K1 + N)/N ≤(K2 + N)/N implies K1 + N ≤K2 + N.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Proposition
If N is a submodule of M, then M is Noetherian [resp. Artinian] if and only if N and M/N are Noetherian [resp. Artinian].
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Corollary
If M1, . . . , Mn are Noetherian [resp. Artinian] R-modules, then so is M1 ⊕M2 ⊕· · · ⊕Mn.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Corollary
If M is a finitely generated module over the Noetherian [resp. Artinian] ring R, then M is Noetherian [resp. Artinian]
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Definitions A series of length n for a module M is a sequence of the form
Definitions A series of length n for a module M is a sequence of the form M = M0 ≥M1 ≥· · · ≥Mn = 0. The series is called a composition series if each factor module Mi/Mi+1 is simple. [A module is simple if it is nonzero and has no submodules except itself and 0. We will study simple modules in detail in Chapter 9.] Thus we are requiring the series to have no proper refinement. Two series are equivalent if they have the same length and the same factor modules, up to isomorphism and rearrangement.
Proof / verification strategy: Use this as a definition checklist: identify the ambient object, test each stated condition separately, then compare examples and non-examples without assuming later theorems.
Jordan-H¨older Theorem For Modules If M has a composition series, then any
Jordan-H¨older Theorem For Modules If M has a composition series, then any two composition series for M are equivalent. Furthermore, any strictly decreasing sequence of submodules can be refined to a composition series.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Theorem
The R-module M has a composition series if and only if M is both Noetherian and Artinian.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Proposition
In the basic AKLB setup of (7.3.9), assume that A is integrally closed. If A is a Noetherian ring, then so is B. In particular, the ring of algebraic integers in a number field is Noetherian.
Proof / verification strategy: Start from the stated definitions, prove each required implication or inclusion separately, and use the immediately preceding structural result where it shortens the argument.
Definition
If I1, . . . , In are ideals, the product I1 · · · In is the set of all finite sums i a1ia2i · · · ani, where aki ∈Ik, k = 1, . . . , n. It follows from the definition that the product is an ideal contained in each Ij.
Proof / verification strategy: Use this as a definition checklist: identify the ambient object, test each stated condition separately, then compare examples and non-examples without assuming later theorems.
Lemma
If P is a prime ideal that contains a product I1 · · · In of ideals, then P contains Ij for some j.
Proof / verification strategy: Assume a non-trivial factorisation or divisibility relation and use degree, content, ideal or prime-divisibility constraints to force one factor to be a unit or to obtain a contradiction.
Proposition
If I is a nonzero ideal of the Noetherian integral domain R, then I contains a product of nonzero prime ideals.
Proof / verification strategy: Assume a non-trivial factorisation or divisibility relation and use degree, content, ideal or prime-divisibility constraints to force one factor to be a unit or to obtain a contradiction.
Corollary
If I is an ideal of the Noetherian ring R (not necessarily an integral domain), then I contains a product of prime ideals.
Proof / verification strategy: Assume a non-trivial factorisation or divisibility relation and use degree, content, ideal or prime-divisibility constraints to force one factor to be a unit or to obtain a contradiction.
Definitions
Definitions Let R be an integral domain, with K its quotient field, and let I be an R-submodule of K. Call I is a fractional ideal of R if rI ⊆R for some nonzero r ∈R. We will call r a denominator of I. An ordinary ideal of R is a fractional ideal (take r = 1), and will often be referred to an as integral ideal.
Proof / verification strategy: Use this as a definition checklist: identify the ambient object, test each stated condition separately, then compare examples and non-examples without assuming later theorems.
Lemma
(i) If I is a finitely generated R-submodule of K, then I is a fractional ideal. (ii) If R is Noetherian and I is a fractional ideal of R, then I is a finitely generated Rsubmodule of K. (iii) If I and J are fractional ideals with denominators r and s respectively, then I ∩J, I +J and IJ are fractional ideals with respective denominators r (or s), rs and rs. [The product of fractional ideals is defined exactly as in (7.6.1).]
Proof / verification strategy: Write the relevant maps explicitly, compute kernels and images at each position, and use commutativity or an induced-map argument to preserve exactness.
Definition
A Dedekind domain is an integral domain R such that (1) R is Noetherian, (2) R is integrally closed, and (3) Every nonzero prime ideal of R is maximal. Every PID is a Dedekind domain, by (7.5.5), (7.1.7), (2.6.8) and (2.6.9). We will prove that the algebraic integers of a number field form a Dedekind domain. But as we know, the ring of algebraic integers need not be a PID, or even a UFD (see the discussion at the beginning of this chapter, and the exercises in Section 7.7).
Proof / verification strategy: Use this as a definition checklist: identify the ambient object, test each stated condition separately, then compare examples and non-examples without assuming later theorems.
Lemma
Let I be a nonzero prime ideal of the Dedekind domain R, and let J = {x ∈ K : xI ⊆R}. Then R ⊂J.
Proof / verification strategy: Assume a non-trivial factorisation or divisibility relation and use degree, content, ideal or prime-divisibility constraints to force one factor to be a unit or to obtain a contradiction.
Proposition
Let I be a nonzero prime ideal of the Dedekind domain R, and let J = {x ∈K : xI ⊆R}. Then J is a fractional ideal and IJ = R.
Proof / verification strategy: Assume a non-trivial factorisation or divisibility relation and use degree, content, ideal or prime-divisibility constraints to force one factor to be a unit or to obtain a contradiction.
Quick-reference relationships
Problem-solving workflow
Fix the ring hypotheses
Record commutativity, identity, zero-divisor assumptions and whether the ring is a domain, field, PID, UFD or Euclidean domain.
Translate element questions into ideal questions
Divisibility, kernels, quotients and maximality often become clearer when expressed through generated ideals.
Choose a universal construction
For quotients, fractions or polynomial evaluation, define the candidate map and prove it is well-defined.
Separate existence from uniqueness
Division, factorisation and decomposition results often require different arguments for the two directions.
Use the strongest justified structure
Do not use field division in a general ring or unique factorisation before its hypotheses have been established.
Check the result in a concrete ring
Integers, residue rings and polynomial rings provide useful sanity checks for the abstract statement.
Worked-solution emphasis from the supplied source
The supplied worked solutions for this section repeatedly test quotient, polynomial, basis, field, module, automorphism, maximal, factor. These checks are used here as verification themes rather than copied as answer text.
Common mistakes and boundary conditions
- Using cancellation in a ring that may contain zero divisors.
- Treating every irreducible element as prime without the needed domain hypothesis.
- Assuming every ideal is principal.
- Applying polynomial root counting without an integral-domain hypothesis.
Verification checklist
- State the ambient algebraic structure and operation before applying a theorem.
- Record every hypothesis that controls the result: finiteness, commutativity, normality, primality, separability, exactness or other section-specific conditions.
- Distinguish a definition from a theorem that follows from it.
- Check whether a map is well-defined before using its kernel, image, inverse or induced map.
- Use a concrete example only as a check; do not promote an illustrative value or pattern to a universal rule.
- When a quotient, localisation or extension is constructed, identify the canonical map and what becomes equal, invertible or fixed.
Source coverage map
| Source section | Subject | PDF pages analysed |
|---|---|---|
| 7.5 | Noetherian and Artinian Modules and Rings | 142–145 |
| 7.6 | Fractional Ideals | 146–147 |
Related Mathematics pages
Source note: synthesised from the supplied abstract-algebra PDF. The complete 298-page file, including diagrams and worked solutions, was reviewed. Source-identifying author and bibliographic personal details are intentionally omitted. Formal proofs are condensed; the page does not claim requirements or values not supported by the supplied mathematics.