SYSTEMATIC MATHEMATICS

Commutative Algebra

What does it mean to study a ring near one prime, why do finite generators prevent infinite ideal drift, how do prime chains measure dimension, and how can normalization repair a singular coordinate ring without erasing its arithmetic? This course answers those questions in fifty-three visible bilingual chapters across ten content-sized units of lengths 6, 5, 6, 5, 6, 5, 5, 6, 5, and 4. It constructs localization of rings and modules, prime spectra and the Zariski topology, exact sequences, Hom and tensor products, flatness, Noetherian and Artinian conditions, integral dependence, lying-over and going-up, normalization, primary decomposition and associated primes, Krull dimension, local rings, Nakayama’s lemma, completions, regular local rings, depth and Cohen–Macaulay structure, Gröbner evidence, Nullstellensatz, and one complete cusp-ring dossier. Every ring, identity convention, ideal, module scalar, denominator set, finiteness hypothesis, prime chain, localization, completion topology, and computational-versus-structural boundary is named. The course does not treat a generated ideal list as a proof of primality, infer global equality from one localization, or claim to replace homological algebra, algebraic geometry, algebraic number theory, computational algebra systems, or scheme theory.

Before this course: Completed Abstract Algebra, Proof, Logic & Set Theory, and Linear Algebra. Students should know commutative rings with identity, ideals, quotient rings, ring homomorphisms, polynomial rings, fields, vector spaces, bases, linear maps, kernels, images, determinants, equivalence relations, and proof by induction. Galois Theory & Finite Fields and Number Theory provide useful motivation but are not prerequisites. No category theory, homological algebra, algebraic geometry, algebraic number theory, Gröbner-basis software, or scheme theory is assumed.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

Units, zero divisors, nilpotents, and idempotents reveal different failures of field behavior

Objective: Why does 3̄·4̄=0̄ not make either factor nilpotent?

A commutative ring permits addition, subtraction, and multiplication but not division by every nonzero element. Four element types answer different questions: can it be inverted, can it kill something nonzero, does a power vanish, or does it split the ring into two persistent pieces? Begin by naming the commutative ring, its identity, every ideal or multiplicative set, the module scalars, and the map whose kernel, image, localization, or universal property is being used. The exact result is: Every idempotent e gives a ring isomorphism R≅eR×(1−e)R by r↦(er,(1−e)r); nontrivial idempotents therefore encode product decompositions.

The object being calculated is: A commutative ring R has 1≠0 unless stated otherwise. A unit u has uv=1 for some v. A nonzero zero divisor a has ab=0 for some nonzero b. A nilpotent n satisfies n^k=0 for some k≥1. An idempotent e satisfies e²=e. An integral domain has no nonzero zero divisors. Commutative algebra replaces illegal division by exact maps and universal properties, then checks finite generation, prime support, and local hypotheses before drawing geometric or arithmetic conclusions.

Rebuild the exact problem “In R=Z/12Z, list the units, exhibit a nonzero nilpotent, find every idempotent, and use e=4 to identify the product decomposition.” without copying a diagram or software factorization. Verify each ideal containment, denominator condition, module relation, and induced map, then test this limitation: A zero divisor need not be nilpotent, and a nilpotent need not be idempotent. The zero ring and rings without identity require adjusted conventions, which this course will name when they appear.

Every idempotent e gives a ring isomorphism R≅eR×(1−e)R by r↦(er,(1−e)r); nontrivial idempotents therefore encode product decompositions.

Because e²=e and e(1−e)=0, the sets eR and (1−e)R are rings with identities e and 1−e, and the displayed map respects sums and products.

If er=0 and (1−e)r=0, then r=er+(1−e)r=0, so the map is injective.

For (ea,(1−e)b), the element r=ea+(1−e)b maps back to that pair because the cross products vanish; hence the map is surjective.

In R=Z/12Z, list the units, exhibit a nonzero nilpotent, find every idempotent, and use e=4 to identify the product decomposition.

  1. A residue class is a unit exactly when it is coprime to 12, giving {1̄,5̄,7̄,11̄}; also 6̄²=36̄=0̄, so 6̄ is nonzero nilpotent.
  2. Checking x(x−1)≡0 mod 12, or combining the idempotents modulo 3 and modulo 4, gives e∈{0̄,1̄,4̄,9̄}.
  3. For e=4̄, eR={0̄,4̄,8̄}≅Z/3Z and (1−e)R=9̄R={0̄,3̄,6̄,9̄}≅Z/4Z, so R≅Z/3Z×Z/4Z.

Result: R^×={1̄,5̄,7̄,11̄}; 6̄ is nilpotent; the idempotents are 0̄,1̄,4̄,9̄; and 4̄ yields Z/12Z≅Z/3Z×Z/4Z.