Crossposted on MathOverflow: https://mathoverflow.net/q/368381/461
Let $A$ be the $\mathbb R$-algebra of all $\mathbb R$-valued functions on $\mathbb N$, that is $$ A=\mathbb R^{\mathbb N}=\prod_{n\in \mathbb N}\mathbb R. $$ Let $\mathfrak a$ be the ideal of $A$ formed by the finitely supported $\mathbb R$-valued functions on $\mathbb N$, that is $$ \mathfrak a=\mathbb R^{(\mathbb N)}=\bigoplus_{n\in\mathbb N}\mathbb R. $$ Let $M$ be the set of those maximal ideals of $A$ which contain $\mathfrak a$.
Let us assume the Continuum Hypothesis.
I will make four statements, and my question will be: are these statements true? (I think they are true, but I'm far from being sure.)
(1) The fields $A/\mathfrak m$ and $A/\mathfrak n$ are isomorphic for all $\mathfrak m,\mathfrak n\in M$.
(2) Statement (1) follows from Corollary 6.1.2 in the book Model Theory (1990) by Keisler, H. Jerome and Chen-Chung Chang.
Here is the statement of Corollary 6.1.2:
And here is a corollary to the corollary:
Assume the continuum hypothesis. Suppose $\mathcal A$ is a model for a countable language $\mathcal L$ and $|A|\le\omega_1$. Then we have $\prod_D\mathcal A\cong\prod_E\mathcal A$ for all nonprincipal ultrafilters $D, E$ over $\omega$.
(3) However, it does not follow from Corollary 6.1.2 that $A/\mathfrak m$ and $A/\mathfrak n$ are isomorphic as extensions of $\mathbb R$ for all $\mathfrak m,\mathfrak n\in M$. This because in the case of extensions of $\mathbb R$ the language $\mathcal L$ is uncountable.
(4) More precisely the question of knowing if $A/\mathfrak m$ and $A/\mathfrak n$ are isomorphic as extensions of $\mathbb R$ for all $\mathfrak m,\mathfrak n\in M$ is an open problem.
Are Statements (1), (2), (3) and (4) above true?
Note that, since the fields $A/\mathfrak m$ and $A/\mathfrak n$ have $2^{2^{\aleph_0}}$ distinct structures of extensions of $\mathbb R$, as shown by Eric Wofsey, a field isomorphism between these two fields will not be $\mathbb R$-linear in general.
For the sake of completeness let me spell out the condition "$A/\mathfrak m$ and $A/\mathfrak n$ are isomorphic as extensions of $\mathbb R$". Let $\Delta:\mathbb R\to A$ be the diagonal morphism defined by $$ (\Delta(c))(x)=c\quad(\forall\ x\in\mathbb R), $$ let $\pi_{\mathfrak m}:A\to A/\mathfrak m$ and $\pi_{\mathfrak n}:A\to A/\mathfrak n$ be the canonical projections, and let $\phi:A/\mathfrak m\to A/\mathfrak n$ be an isomorphism of fields. Then $\phi$ is an isomorphism of extensions of $\mathbb R$ if and only if we have $$ \phi\circ\pi_{\mathfrak m}\circ\Delta=\pi_{\mathfrak n}\circ\Delta. $$ EDIT. One can easily prove in ZFC that the transcendence degree of $A/\mathfrak m$ over $\mathbb R$ is $2^{\aleph_0}$; in particular it does not depend on $\mathfrak m$. Indeed, for $x\in\mathbb R$ define $f_x\in\mathbb R^{\mathbb N}=A$ by $f_x(n)=e^{e^{nx}}$ [or $f_x(n)=\exp(\exp(nx))$ if you don't like small letters] and let $g_x$ be the image of $f_x$ in $A/\mathfrak m$. Then a straightforward argument shows that the subset $\{g_x\ |\ x>0\}\subset A/\mathfrak m$ is algebraically independent over $\mathbb R$.