Commutative Algebra/Noetherian rings

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Rings as modules[edit]

Theorem 14.1:

We had already observed that a ring is a module over itself, where the module operation is given by multiplication and the addition by ring addition. In this context, we further have that the submodules of are exactly the ideals.

Proof: Being a submodule means being an additive subgroup closed under the module operation. In the above context, this is exactly the definition of ideals.

Transfer of the properties[edit]

Definition 14.2:

Let be a (commutative) ring. is called Noetherian if and only if every ascending chain of ideals of

eventually becomes stationary.

From theorems 6.7 and 14.1, we obtain the following characterisation of Noetherian rings:

Theorem 14.3:

The following are equivalent:

  1. is Noetherian.
  2. Every ideal of is finitely generated.
  3. Every set of ideals of has a maximal element with respect to inclusion.

In analogy to theorem 6.11, we further obtain

Theorem 14.4:

If is Noetherian, is another ring and is a surjective ring homomorphism, then is Noetherian.

Proof 1: Proceed in analogy to theorem 6.11, using the isomorphism theorem of rings.

Proof 2: Use theorem 6.11 directly.

New properties in the ring setting[edit]

When rings are considered, several new properties show themselves in the noetherian case.

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Noetherian rings and constructions[edit]

In this section we will prove theorems involving Noetherian rings and module or localisation-like structures over them.

Theorem 14.4:

Let be Noetherian and let be a finitely generated -module. Then is Noetherian.

Theorem 14.5 (Hilbert's basis theorem):

Let be a Noetherian ring. Then the polynomial ring over , , is also Noetherian.

Proof 1:

Consider any ideal . We form the ideal , that shall contain all the leading coefficients of any polynomials in ; that is

.

Since is Noetherian, as a finite set of generators; call those generators . All belong to a certain as a leading coefficient; let thus be the degree of that polynomial for all . Set

.

We further form the ideals and of and claim that

.

Indeed, certainly and thus (see the section on modules). The other direction is seen as thus: If , , then we can set to be the leading coefficient of , write for suitable and then subtract , to obtain

so long as . By repetition of this procedure, we subtract a polynomial of to obtain a polynomial in , that is, .

However, both and are finitely generated ideals ( is finitely generated as an -module and hence Noetherian by the previous theorem, which is why so is as a submodule of a Noetherian module). Since the sum of finitely generated ideals is clearly finitely generated, is finitely generated.

Exercises[edit]

  • Let be a Noetherian ring, and let be an -module. Prove that is Noetherian if and only if it is finitely generated. (Hint: Is there any surjective ring homomorphism , where is the number of generators of ? If so, what does the first isomorphism theorem say to that?)

Noetherian spaces[edit]