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Let $A:=\mathbb R^\mathbb N$ be the direct product of infinitely many copies of $\mathbb R$. Let $I_0$ be the ideal of all sequences of $A$ which have only finitely many nonzero entries. Then $I_0$ is a proper ideal of $A$. If $\mathfrak m$ is a maximal ideal of $A$, containing $I_0$, then can we say anything about the residue field $A/\mathfrak m$? Note that, the residue field cannot be equal to $\mathbb R$. To see this, consider the unit $(2,4,8,16,\ldots)$. If the residue field is $\mathbb R$ then there exists a real number $c$ such that the $(c,c,c,\ldots)-(2,4,8,\ldots)$ is contained in $\mathfrak m$. But at most one entry of the difference is equal to zero. If the $i$'th entry of the difference is zero, then we can add to it the sequence which has $1$ at the $i$-th entry and zero everywhere else to conclude that $\mathfrak m$ contains a unit!

sagnik chakraborty
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  • My own off-hand reaction is that that field might have large transcendence degree, to begin with... – paul garrett May 02 '23 at 17:29
  • This argument only shows that the composition of the quotient map to the residue field with the diagonal inclusion of R into the infinite product cannot be surjective. If you replace R with C, then all the residue fields will be isomorphic to C by Los’s ultraproduct theorem + cardinality. I wouldn’t be surprised if something similar holds here.

    At the very least, every residue field has continuum cardinality, lacks a square root of -1, and becomes algebraically closed of characteristic 0 as soon as such a root is adjoined.

     – Rafi May 02 '23 at 17:29
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    Strictly speaking, what you said is correct. But here we are only considering $\mathbb R$-algebra isomorphisms. Besides, since $\mathbb R$ is not isomorphic to any proper subfield of it, the argument actually shows that the residue field $A/\mathfrak m$, even as an abstract field, is not isomorphic to $\mathbb R$. – sagnik chakraborty May 02 '23 at 17:45
  • Duplicates of this and this (with many more links within these two) – math54321 May 02 '23 at 18:35
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    @Rafi The ultrapower (residue field) is not isomorphic to $\mathbb{R}$ because it is non-Archimedean: the residue class of $(1,2,3,4,\dots)$ is greater than $1$ added to itself $n$ times, for all finite $n$. To the OP: I don't understand why in the previous comment you said "$\mathbb{R}$ is not isomorphic to any proper subfield of it". The elements of the form $(r,r,r,r,\dots)$ form a subfield isomorphic to $\mathbb{R}$. – Alex Kruckman May 02 '23 at 23:26
  • @AlexKruckman: I think they meant that $\mathbb{R}$ is not isomorphic to any proper subfield of itself. – Eric Wofsey May 03 '23 at 01:07
  • Thank you Eric for the clarification. Yes, i meant that $\mathbb R$ is not isomorphic to any proper subfield of itself. – sagnik chakraborty May 03 '23 at 04:07
  • Oh, I see! But is it clear that the ultrapower is isomorphic to a proper subfield of itself? – Alex Kruckman May 03 '23 at 04:13
  • I don't know. I am not familiar with the ultrapower construction, just heard its name a few times. I only know that $\mathbb R$ is not isomorphic to any proper subfield of itself and $\mathbb C$ is isomorphic to uncountably many proper subfields of itself. – sagnik chakraborty May 03 '23 at 04:41

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This is called the ultrapower construction. There is a bijection between maximal ideals $\mathfrak{m}$ extending $I_0$ and non-principal ultrafilters on $\mathbb{N}$, as explained by Eric Wofsey here.

The resulting residue field is a real closed field of cardinality $2^{\aleph_0}$ which is $\aleph_1$-saturated, also known as a hyperreal field.

Alex Kruckman
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