It is a 20th century result that there exists only one algebraically closed field for a given characteristic $p$ and cardinality $\kappa>\aleph_0$, up to isomorphism. Is there a better way to imagine such fields, other than adjoining $\kappa$ transcendental elements to $\mathbb Q$ or $\mathbb F_p$ and taking the algebraic closure?
In characteristic 0, do these fields enjoy some nice topological properties? Can you maybe do analysis on them?
I have not found much about fields of cardinality greater than $\mathfrak c$, not even about the "next simplest case" $2^\mathfrak c$. Do you know of an references?

$\begingroup$ (This is just an uninformed comment) I expect there to be problems in "doing analysis" on such fields, in the same way as it's preferable to do analysis on $\mathbb C$ (which is not only algebraically closed, but also complete with respect to a norm) rather than on $\bar{\mathbb Q}$. $\endgroup$– QfwfqMar 24 '18 at 16:11

$\begingroup$ @Qfwfg That's exactly my question. I agree with you that if such a field were useful for analysis, someone would probably be doing it, but I know almost nothing about these fields, so I cannot say for sure. $\endgroup$– FusRoDahMar 24 '18 at 16:15

$\begingroup$ @Qfwfq not sure what you mean by "preferable", but one can also do analysis on such fields as $\mathbf{C}_p$ (completed algebraic closure of $p$adics) or Puiseux series, and there are good reasons to do so. (Of course the complex numbers are more important for plenty of reasons, but fortunately mathematicians don't have an exclusive contract.) $\endgroup$– YCorMar 24 '18 at 16:57

2$\begingroup$ As YCor notes, one can do analysis on $\mathbb C_p$, and people certainly do. The relevant term to look up is rigid analysis. However, one reason analysis on $\mathbb C$ is "nicer" than on $\mathbb C_p$ is because the former is locally compact, while the latter is not. Thus many proof in complex analysis that depend on compactness arguments don't work over $\mathbb C_p$, and in many cases the statements are no longer valid. For this reason people often work with Berkovich space, which is a locally compact, uniquely pathconnected topological space that contains a copy of $\mathbb C_p$. $\endgroup$– Joe SilvermanMar 24 '18 at 22:32

1$\begingroup$ @Noah Schweber: yep, sorry for nitpicking! :) $\endgroup$– QfwfqMar 25 '18 at 13:20
Here is a different way to think about how these fields arise.
In characteristic zero, all such fields in any uncountable cardinality $\kappa$ can be viewed as arising as a nonstandard version of the algebraic closure the rationals.
That is, take any nonstandard model of arithmetic $M$ of uncountable size $\kappa$, or an $\omega$nonstandard model of set theory with $\kappa$ many natural numbers, if you like. Let $\mathbb{Q}^*$ be the nonstandard version of the rational field as $M$ sees it, and let $\overline{\mathbb{Q}^*}$ be the algebraic closure of $\mathbb{Q}^*$ as it is constructed in $M$. This is an algebraically closed field of characteristic zero and of the desired cardinality $\kappa$.
In particular, even the complex field $\mathbb{C}$ itself can be seen as the algebraic closure of the rationals $\mathbb{Q}^*$ taken inside a nonstandard model of arithmetic or set theory, and $\mathbb{Q}^*$ is the quotient field there of its integer part $\mathbb{Z}^*$.
In this sense, every uncountable algebraically closed field of characteristic zero is exactly a nonstandard version of the algebraic closure of $\mathbb{Q}$.
Note that the algebraic closure of a field as computed inside $M$ is more generous than the actual algebraic closure, since a nonstandard model will be wanting to include all the roots of the various nonstandard polynomials, as well as the standard ones.
I made a blog post, Nonstandard models of arithmetic arise in the complex numbers, when this phenomenon arose a few weeks ago at our seminar.

$\begingroup$ Very nice answer. So in essence, is using the compactness theorem to add constant symbols $t_{\alpha}$ for $\alpha<\kappa$, $\kappa$ some infinite cardinal to the language in order to build the nonstandard model, the same as introducing independent variables over $\mathbb{Q}$? Maybe the similarity is simply formal... $\endgroup$ Mar 24 '18 at 21:12

$\begingroup$ @16278263789 Yes, I think that is right. I find it surprising that one can add the transcendental elements in such a way so as to get an ordered field that forms a nonstandard version of the rationals. $\endgroup$ Mar 24 '18 at 21:17

$\begingroup$ It seems to trace back to creating a chain of elementary endextensions of the model $M$ of some length $\kappa$ which as it goes builds the non standard ordered field. $\endgroup$ Mar 24 '18 at 21:23

$\begingroup$ By the way just as an aside (maybe of interest to OP) by Lefschetz Principle in algebraic geometry all truth can be reduced to looking at the model $\mathbb{C}$ since the Principle states that a statement is true over any algebraically closed field of characteristic $0$ if and only if it is true over $\mathbb{C}$ $\endgroup$ Mar 24 '18 at 21:31

$\begingroup$ I agree, but the point of my answer is that we can take on the additional expressive power of M and view the field as being an algebraic extension of the rationals, as M sees it. $\endgroup$ Mar 24 '18 at 21:35
As a purely algebraic alternative to model theoretic approaches to nonstandard algebraic closure in characteristic $0$, we can also constructively build them in the regular old universe out of the ordinals.
Let $\alpha>0$ be an ordinal, and consider the field of fractions $Frac(\mathfrak{G}(\omega^{\omega^\alpha}))=\mathbb{Q}^{\omega^{\omega^\alpha}}$ of the Grothendieck ring $\mathfrak{G}(\omega^{\omega^\alpha})=\mathbb{Z}^{\omega^{\omega^\alpha}}$ of the $\delta$number $\omega^{\omega^\alpha}$ under natural (Hessenberg) operations. This is a non Archimedean ordered field (thusly it properly contains the rationals) with no nontrivial roots where all the ordinals up to $\omega^{\omega^\alpha}$ are now treated as the 'natural numbers' and $\mathbb{Z}^{\omega^{\omega^\alpha}}$ is treated as the 'integer subring' of our field.
We can then move into an algebraic closure $\overline{\mathbb{Q}}^{\omega^{\omega^\alpha}}$, where we allow 'polynomials' to really be elements of the monoid algebra $\mathbb{Q}^{\omega^{\omega^\omega}}(X^{\omega^{\omega^\alpha}})$  these are expressions of the form $$\sum_{i<n}q_iX^{\alpha_i}$$ with $q_i\in\mathbb{Q}^{\omega^{\omega^\omega}}$ and $\alpha_i<\omega^{\omega^\alpha}$ for all $i<n$. This gives us '$\beta^{th}$ roots' for members of $\mathbb{Q}^{\omega^{\omega^\alpha}}$ for all $\beta<\omega^{\omega^\alpha}$, as opposed to only finite roots which are provided by algebraic closure using standard polynomials.
There is some subtlety here when trying to approach this without model theory guaranteeing the existence of a way to evaluate $q_iX^{\alpha_i}$ as a function on $\mathbb{Q}^{\omega^{\omega^\alpha}}$ when $\omega\leq\alpha_i$  we can get around this by embedding $\mathbb{Q}^{\omega^{\omega^\alpha}}$ in the Surreals $N_0$ and looking at what the Surreal exponential function does to its image.**
You can also do modular arithmetic in $\omega^{\omega^\alpha}$ for ordinals $\beta\geq\omega$ and proceed to produce a nonstandard $\mathbb{Q}^{\omega^{\omega^\alpha}}_p$ for $p\geq\omega$ prime, and then algebraically close that too. We could also just use all of the ordinals instead of stopping at $\omega^{\omega^\alpha}$, in which case we produce a proper class sized version of the above.
I'm currently researching these notions to better understand the Surreals, but they seem to provide a constructive model of many nonstandard constructions usually accessed through model theoretic techniques.
*This used to claim that $\mathbb{Q}^{\omega^{\omega^\alpha}}$ was essentially a nonstandard model of the rationals, which is incorrect (thanks to nombre on this).
** I am currently checking if $\mathbb{Q}^{\omega^{\omega^\alpha}}$ embedded in $N_0$ 'preserves exponentiation' to welldefine it on the preimage; I believe it should, but this is just intuition for now.

1$\begingroup$ For *, this does not qualify at least in the usual sense that this is not an elementary extension. If it were, then the set of squares in $\mathbb{Q}^{\omega^{\omega^{\alpha}}}$ would be dense in the subset of positive elements, but here no square is close to $\omega$. Note that it is unclear what a $\omega$th square root of $1$ would be in No since no "surcomplex" exponential is known and $\omega$ is both $2.\omega$ and $4.\omega$, so adding zeroes of functions $X^{\alpha}+q$ for $q>0$ and taking the algebraic closure successively does not provide an anwser to this. $\endgroup$– nombreMar 25 '18 at 11:07

$\begingroup$ @nombre Thank you as always, I'll edit accordingly. I hadn't considered transfinite roots of $1$ at all actually before this, that is a fascinating question. $\endgroup$ Mar 25 '18 at 12:08

$\begingroup$ You're welcome. To be more accurate about roots of $1$, there is a way to define a "natural" exponential for surreals using the formula $exp(a+bi)=exp(a)(cos(b)+isin(t))$ and defining $cos$ and $sin$ as $2 \pi$periodic functions (with respect to Conway integers) coinciding with the standard analytic functions on $[0;2\pi[$, with this convention $(1)^{\frac{1}{\omega}}$ could be any $exp(\frac{2k+1}{\omega}\pi),k\in \mathbb{Z}$. But I am not sure this is interesting or useful. What precisely do you mean by **? The exponential of a non zero rationnal will not lie in the image for instance. $\endgroup$– nombreMar 25 '18 at 12:34

$\begingroup$ @nombre In essence, if $f:\mathbb{Q}^{\omega^{\omega^\alpha}}\rightarrow N_0$ is the embedding we discussed, I am wondering whether $f^{1}(f(q)^{f(n)})$ is well defined for $q\in\mathbb{Q}^{\omega^{\omega^\alpha}}$ and $n\in\mathbb{Z}^{\omega^{\omega^\alpha}}$ — at the least I would suspect that we need to move to an $\varepsilon$number instead of just a $\delta$number, but is there a problem all the way down in the rationals? $\endgroup$ Mar 25 '18 at 12:45

$\begingroup$ Oh yes I got confused with the exponential here. Is this condition really necessary? Because in any event the "roots" will lie outside of the image. If you apply the formula $exp(\omega^a)=\omega^{\omega^{g(a)}}$ for $a>0$ with $g(a)=\{v a;g(a') \  \ g(a'')\}$ (where $v$ is the natural valuation ranging in No) to $\omega^{\omega}$, this already goes outside of the image. More previsions: $log(\omega)=\omega^{\frac{1}{\omega}}$ and $g(\frac{1}{2^n})= \frac{1}{2^n}$ for $n\in \mathbb{N}$. $\endgroup$– nombreMar 25 '18 at 12:59
In the caracteristic zero case, I suggest one way to look at this is to choose a prescribed real closed subfield $R_{\kappa}$ of said algebraically closed field $F_{\kappa}$ of cardinal $\kappa$ that one finds particularly interesting, so that geometric intuition (for instance basically that of a plane) or "analytic" intuition may apply to $F_{\kappa}$.
Of course each $R_{\kappa}$ appears "as an isomorphic copy" in each algebraically closed field of cardinal $\kappa$, but not all of them are isomorphic, so this is a means to distinguish ordered pairs $(F_{\kappa},R_{\kappa})$ between them.
Each choice naturally transfers some structure from $R_{\kappa}$ to $F_{\kappa}=R_{\kappa}\oplus R_{\kappa}.\sqrt{1}$, be it an absolute value $F_{\kappa}\rightarrow R_{\kappa}^{\geq 0}$ (and thus a nice sequential uniform structure), a valuation extending that of $R_{\kappa}$, a Hahn series structure, a transfer principle, order related saturation properties, an exponential map...
That said I don't think there is a distinguished real closed subfield of $F_{\kappa}$ in a fieldtheoretic standpoint in general.