$\mathbb{R} ^ \mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.

Question

let $R=\mathbb{R}^ \mathbb{R}$ (all the functions like $f:\mathbb{R} \rightarrow \mathbb{R}$). For each $f, g \in R$ and $a \in R$:

$$(f+g)(a):=f(a)+g(a)$$ $$(fg)(a):=f(a)g(a)$$

I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.

Answer 1

Let's prove it's neither Noetherian nor Artinian.

To each $A \subseteq \mathbb{R}$, we can assign a set $\varphi(A) \subseteq \mathbb{R}^\mathbb{R}$ as follows:

$$\varphi(A) = \{f \in \mathbb{R}^\mathbb{R} : \forall x \in A, f(x) = 0\}.$$

Exercise 0. Show that $\varphi(A)$ is an ideal for all sets $A \subseteq \mathbb{R}$.

Conclude that $$\varphi : \mathcal{P}(\mathbb{R}) \rightarrow \mathcal{P}(\mathbb{R}^\mathbb{R})$$ can be viewed as a function $$\mathcal{P}(\mathbb{R}) \rightarrow \mathrm{Ideal}(\mathbb{R}^\mathbb{R}).$$

Exercise 1. Show that $\varphi : \mathcal{P}(\mathbb{R}) \rightarrow \mathrm{Ideal}(\mathbb{R}^\mathbb{R})$ is injective and order-reversing.

Exercise 2. Find an order-reversing injection $\psi : \mathbb{Z} \rightarrow \mathcal{P}(\mathbb{R})$.

Conclude that $\varphi \circ \psi : \mathbb{Z} \rightarrow \mathrm{Ideal}(\mathbb{R}^\mathbb{R})$ is an order-preserving injection.

Conclude that $\mathbb{R}^\mathbb{R}$ is neither Artinian nor Noetherian.