Commutative Algebra/Noetherian rings

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