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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.

1 Answers1

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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.

goblin GONE
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  • I get the distinct sense this website is slowly dying... – goblin GONE Jan 08 '19 at 09:44
  • Thank you so much. I'm still a little bit confused but I try to understand it... –  Jan 08 '19 at 12:58
  • @t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now) – goblin GONE Jan 08 '19 at 14:15
  • This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $\phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $n\in \mathbb N^+$ ordered by inclusion." – rschwieb Jan 08 '19 at 15:05
  • @rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $\varphi$ function I've just defined and then reconceptualizing things in your own terms? – goblin GONE Jan 09 '19 at 08:49
  • Seems to not be worth the repetition. I guess my comment will have to suffice. – rschwieb Jan 09 '19 at 12:53