TITLE: Fields of arbitrary cardinality
QUESTION [16 upvotes]: So I took an introductory abstract algebra course a few semesters ago, and we were shown that groups and rings can both be made into products, i.e. if I have some group $G$ (resp. ring $R$) and some indexing set $I$, then I can make a group $G^{I}$ (resp. ring $R^{I}$) with the appropriate cardinality. However, it was also demonstrated that for fields, you cannot keep the multiplicative inverse under a product structure, i.e. if I have fields $F_{1}, F_{2}$, then $F_{1} \times F_{2}$ will still be a ring, but you will not (not just "not in general", but won't) have multiplicative inverses for all non-zero (i.e. not $(0, 0)$) elements, as you could pick $(0, x)$, where $x \neq 0$. According to Wikipedia, an ultraproduct will preserve field structure, but I'm not sure what the cardinality of $\prod_{i \in I} F_{i} / \mathscr{U}$ would be in general.
I saw another post that said you could make a field of arbitrarily large cardinality by extending a given field through throwing in a whole bunch of "transcendental" elements. The example given was that if I had $\mathbb{Q}$, I could start by dumping in the complexes, but then I suppose after that I'd just start throwing in dogs and cats or something; I'm really not sure what that responder meant, and ceteris paribus, my dogs tend to stay inside, so I'd rather keep them out of my fields, particularly the large ones the thread was interested in.
Moreover, the thread mentioned earlier seemed more concerned with just making the fields big. My question is a bit more nuanced: Given any cardinality $\kappa > 1$, can I generally construct a field $F$ of a cardinality $\kappa$? Moreover, how vague do I have to be about it (i.e. do I have to use some really choicy methods, or can I make it a bit more straightforward)?
Thanks.
EDIT: Sorry. I am aware that for finite fields, you are limited to powers of primes. I meant for infinite cardinals.

REPLY [2 votes]: A lot of info was scattered about, so here's a condensed bit of the developments in Gregory Grant's answer and the comments. The final solution was that the existence of a field structure on every infinite set $X$ is equivalent to the full axiom of choice.
Firstly, to show choice implies the existence, we borrow Gregory Grant's construction. Let $X$ be an infinite set, and let $X^* = \bigcup_{n \in \mathbb{N}} X^n$ denote the set of all finite strings on $X$. Let $\sim$ be an equivalence relation on $X^*$ where $(a_{1}, \ldots, a_{n}) \sim (b_{1}, \ldots, b_{m})$ if and only if $(\forall x \in X)( \# \{k : a_{k} = x \} = \# \{ k : b_{k} = x \} )$, i.e. the strings are equal up to rearranging the terms of the strings. Set $\hat{X} = X^{*} / \sim$, and let $R$ denote the set of formal sums over $X$. Finally, let $F$ denote the field of fractions of $R$. Then $F$ is a field equitotient with $X$.
In the other direction, if every infinite set $X$ can be given a field structure, then a fortiori every infinite set can be given a group structure, so we can use this argument.