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Let $F$ be a field, and let $R=\prod_{i=1}^\infty F$. In wikipedia, and more specific: enter image description here There is a claim which I suspect is wrong and here is why:

Let $m\triangleleft R$ be a maximal ideal. Hence, there are only finitly many $i\in \Bbb{N}$ such that $\rho_i(m)=0$ (where $\rho_i$ is the projection on the $i^{th}$ component), otherwise $m$ is not maximal.

For convenience let's say that $\{1,2,3\}\subset\{i\in\Bbb{N}|\rho_i(m)\neq0\}$. Let $$I_1=(a_1,0,0,...)$$ $$I_2=(a_1,a_2,0,,...)$$ $$I_3=(a_1,a_2,a_3,0,,...)$$ $$.$$ $$.$$ $$.$$ $$I_i=(a_1,a_2,a_3...,t,a_i,0,,...) \text{ (t might be zero)}$$ where $0\neq a_i\in\rho_i(m)$. There are infintly many $I_i$, and $I_i\subset I_{i+1}$, so in the ring $R_m$ the corresponding ideals (exist because $I_i\in m$) perform an ascending chain of ideals which is not stabilize. Therefore $R_m$ is not noetherian.

Am I right?

Thanks in advance

Or Shahar
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    In the ring $R=\prod_{n=0}^{\infty}{F}$, $F$ being a field, consider the maximal ideal $\mathfrak{m}$ generated by $(0,1,\ldots,1,\ldots)$. $R_{\mathfrak{m}}$ is exactly $R/\mathfrak{m}$. In particular, the ideals of $R_{\mathfrak{m}}$ are not in bijection with the ideals of $R$ contained in $\mathfrak{m}$ (the latter do surject onto the former, though). – Aphelli May 16 '22 at 23:33
  • Since Wiki does not explain why the localizations are fields, I suggest you to look for Von Neumann regular rings. Your R is such a ring. (This could be helpful: https://math.stackexchange.com/questions/2410058.) – user26857 May 17 '22 at 05:19

1 Answers1

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As was pointed out in the comments, your ring is Von Neumann regular (VNR, for short), that is, for each $x \in R$ there is $a \in R$, such that $x=ax^2$. All localizations also satisfy this property.

Now, every local ring, which is also a VNR, is in fact a field. To see that, it is enough to check that the only proper principal ideal in such ring is the zero ideal. In VNR every principal ideal is generated by an idempotent, but in local rings the only idempotents are $0$ and $1$! Thus either a principal ideal is the zero ideal, or it is the whole ring.

To finally see that local VNR is a field, suppose there is a proper ideal $I$ and a nonzero element $x \in I$. Then we just proved that the ideal $(x) \subset I$ is either zero ideal or improper, which is in both cases a contradiction.

Thus, the claim in your question is indeed true.

I hope I have not made a mistake anywhere and answered the question you asked. I would be glad to clarify the argument if anything is unclear.

user26857
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Aleksei Kubanov
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