Infinite field of characteristic p elementary equivalent to field with transcendental element over prime subfield I'm trying to show that if F is an infinite field of characteristic p then it's elementary equivalent to a field F' of char p which contains an element transcendental over its prime subfield (the prime subfield is the one generated by 1).
I seem to recall that every infinite field of char p has a subfield isomorphic to $F_p$, but not sure how to go from there. Any hint of how to show the elementary equivalence?
 A: There is a straightforward proof using the Upward Löwenheim–Skolem Theorem.
Let $T$ be the theory, over the language of fields, whose axioms are all sentences true in our field $F$. Then $T$ has models of arbitrarily high cardinality, since it has an infinite model.
Any uncountable model $E$ of $T$ must contain objects transcendental over the prime field, because the algebraic closure of the prime field is countably infinite.  This completes the proof.  
We can prove a stronger result. Add to the language of field theory a constant symbol for every element of $F$, and let $T_F$ be the theory over this language whose axioms are all sentences true in $F$, with the natural interpretation of the new constants. The theory with axioms $T_F$ has models of arbitrarily large cardinality $\kappa$. Choose $\kappa$ greater than the cardinality of $F$, and let $E$ be a model of $T_F$ of cardinality $\kappa$. Without loss of generality, we can make $E$ an extension of $F$. Then $E$ contains an element transcendental over $F$, again by cardinality considerations.
Thus any infinite field $F$ has an elementary extension that has objects transcendental over $F$.
Comment: If desired, we can then use the Downward Löwenheim–Skolem Theorem, or other techniques, to produce, in the first situation, a countably infinite field elementarily equivalent to $F$, but with an element transcendental over the prime field. And in the generalization, we can make an elementary extension $E$ of $F$ with an object transcendental over $F$, such that $E$ has the same cardinality as $F$.
For a different-looking proof of the same facts, using basically the same idea, we can use a suitable ultrapower of $F$.  That may feel more concrete to some.
