Commutative Algebra/Noetherian rings

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

In analogy to theorem 6.11, we further obtain

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

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1=Theorem 14.4:

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

Proof 1:

Consider any ideal $$I \le R[x]$$. We form the ideal $$J \le R$$, that shall contain all the leading coefficients of any polynomials in $$I$$; that is
 * $$a \in J :\Leftrightarrow \exists f \in I: f(x) = a x^m + \text{(lower terms)}$$.

Since $$R$$ is Noetherian, $$J$$ as a finite set of generators; call those generators $$j_1, \ldots, j_n$$. All $$j_k$$ belong to a certain $$f_j \in R[x]$$ as a leading coefficient; let thus $$d_k$$ be the degree of that polynomial for all $$1 \le k \le n$$. Set
 * $$d := \max_{1 \le k \le n} d_k$$.

We further form the ideals $$K := \langle f_1, \ldots, f_n \rangle$$ and $$L := \langle 1, x, x^2, \ldots, x^{d-1} \rangle \cap I$$ of $$R[x]$$ and claim that
 * $$I = K + L$$.

Indeed, certainly $$K, L \subseteq I$$ and thus $$K + L \subseteq I$$ (see the section on modules). The other direction is seen as thus: If $$g(x) \in I$$, $$\deg g = m$$, then we can set $$a \in R$$ to be the leading coefficient of $$f$$, write $$a = r_1 j_1 + \cdots + r_n j_n$$ for suitable $$r_1, \ldots, r_n \in R$$ and then subtract $$h(x) := (r_1 j_1 f_1 x^{m - d_1} + \cdots + r_n j_n f_n x^{m - d_n})$$, to obtain
 * $$\deg (g - h) < m$$

so long as $$m \ge d$$. By repetition of this procedure, we subtract a polynomial $$h'$$ of $$g$$ to obtain a polynomial in $$L$$, that is, $$g \in K + L$$.

However, both $$K$$ and $$L$$ are finitely generated ideals ($$\langle 1, x, x^2, \ldots, x^{d-1} \rangle$$ is finitely generated as an $$R$$-module and hence Noetherian by the previous theorem, which is why so is $$L$$ as a submodule of a Noetherian module). Since the sum of finitely generated ideals is clearly finitely generated, $$I$$ is finitely generated.

Exercises

 * Let $$R$$ be a Noetherian ring, and let $$M$$ be an $$R$$-module. Prove that $$M$$ is Noetherian if and only if it is finitely generated. (Hint: Is there any surjective ring homomorphism $$R[x_1, \ldots, x_n] \to M$$, where $$n$$ is the number of generators of $$M$$? If so, what does the first isomorphism theorem say to that?)