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?

  • $\begingroup$ I'm afraid I don't know what the term elementary equivalent means. However, an infinite field of char $p$ may not have any elements that are transcendental over the prime field. Take an algebraic closure of $F_p$ for example. It cannot be finite because it is always possible to extend a finite field algebraically. $\endgroup$ Commented Aug 3, 2011 at 19:53
  • $\begingroup$ Two models are elementary equivalent if every sentence satisfied by one is also satisfied by the other and v.v. If two models are isomorphic then they are elementary equivalent, but the converse is not true. I think this is why your statement is consistent with the problem, there may not be an F' such that F is isomorphic to, but there is an F' such that F is elementary equivalent to. $\endgroup$
    – nullgraph
    Commented Aug 3, 2011 at 20:03
  • $\begingroup$ Hint: How does a sentence in the language of fields look like? What can you say on the formulae a transcendental element satisfies? $\endgroup$
    – Asaf Karagila
    Commented Aug 3, 2011 at 20:09
  • 1
    $\begingroup$ You need to show that there's a field which contains a transcendental and which models the theory $T$ of $F$. So add a new constant $c$ to your language, and add extra axioms to $T$ saying that $c$ is transcendental over the prime subfield. Use the compactness theorem to show that this extended theory has a model, and conclude that any such model has the properties you want. $\endgroup$ Commented Aug 3, 2011 at 20:12
  • 3
    $\begingroup$ @nullgraph - I think you might mean elementarily equivalent $\endgroup$
    – talkloud
    Commented Aug 3, 2011 at 20:55

1 Answer 1


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.


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