A classmate posed a question in class as to if there existed an infinite field $F$ for which every subring $R \subseteq F$ was a subfield. We'd already determined that if $F$ was a finite field, then every subring was a subfield. Moreover, we confirmed that $\mathbb{Z}$ can be embedded into any field of characteristic $0$, so any candidate must have positive characteristic. I believe I've determined a solution under the assumption of some weak choice principles. Let $\mathcal{M} \subseteq \mathbb{N}$ be a nonprincipal ultrafilter, and set $K_{k} = \mathbb{F}_{p^{k}}$. Let $F = \prod_{k \in \mathbb{N}} K_{k} / \mathcal{M}$, i.e. the ultrapower of $(K_{k})_{k \in \mathbb{N}}$.
First, to show that the set is infinite, assume for contradiction $F$ is finite with size $N$. Choose $\ell \in \mathbb{N}$ such that $p^{\ell} > N$. Then consider the set of elements $\left( a_{k}^{(c_{0}, \ldots, c_{\ell - 1})} \right)_{k \in \mathbb{N}} \in \prod_{k \in \mathbb{N}} K_{k}$ such that \begin{align*} a_{k} = \begin{cases} 0 & k < \ell \\ \sum_{j = 0}^{\ell - 1} c_{j}x^{j} & k \geq \ell . \end{cases} \end{align*} We have that $\left( a_{k}^{(c_{0}, \ldots, c_{\ell - 1})} \right)_{k \in \mathbb{N}} \equiv \left( a_{k}^{(d_{0}, \ldots, d_{\ell - 1})} \right)_{k \in \mathbb{N}}$ in the sense of $\mathcal{M}$ if and only if $c_{0} = d_{0}, \ldots, c_{\ell - 1} = d_{\ell - 1}$, as otherwise they disagree on a cofinite subset of $\mathbb{N}$, so $F$ has at least $p^{\ell}$ elements.
Can I use some sort of transfer principle to say that since every $K_{k}$ has the property that all subrings are subfields, it follows that the same holds for $F$? That is, if I can say for each $k$ that $$ (\forall R \subseteq K_{k})([[(\exists r, s \in R)(r \neq s)] \land [(\forall r, s \in R)(rs \in R)] \land [(\forall r, s \in R)(r + s \in R)] \land [(\forall r \in R)(\exists s \in R)(r + s = 0)]] \implies [(1 \in R) \land (\forall r \in R)(\exists s \in R)(rs = 1)]) , $$ then $$ (\forall R \subseteq F)([[(\exists r, s \in R)(r \neq s)] \land [(\forall r, s \in R)(rs \in R)] \land [(\forall r, s \in R)(r + s \in R)] \land [(\forall r \in R)(\exists s \in R)(r + s = \mathbf{0})]] \implies [(\mathbf{1} \in R) \land (\forall r \in R)(\exists s \in R)(rs = \mathbf{1})]) , $$ where \begin{align*} \mathbf{0} & = \left\{ (a_{k})_{k \in \mathbb{N}} \in \prod_{k \in \mathbb{N}} K_{k} : (a_{k})_{k \in \mathbb{N}} \equiv (0, 0, 0, \ldots) \mod{\mathcal{M}} \right\} \\ \mathbf{1} & = \left\{ (a_{k})_{k \in \mathbb{N}} \in \prod_{k \in \mathbb{N}} K_{k} : (a_{k})_{k \in \mathbb{N}} \equiv (1, 1, 1, \ldots) \mod{\mathcal{M}} \right\} . \end{align*} Is this correct?
You are greatly overcomplicating things...
Let $p$ be a prime and let $F$ be the algebraic closure of the Galois field $\mathbb F_p$. Let $R\subseteq F$ be a subring. Let $x\in R\setminus0$. Since $x$ is algebrac over $\mathbb F_p$, the subring $\mathbb F_p[x]$ of $R$ is a field, so $x$ is invertible in $R$. $R$ is thus a field.