# On the existence of a “universal” field without algebraic elements

Let $\mathfrak{M}$ be an infinite cardinal. Consider all fields $F$ which have the following properties:

(1) $F$ contains $\mathbb{Q}$.

(2) $F$ has cardinality $\leqslant \mathfrak{M}$.

(3) All elements of $F \setminus \mathbb{Q}$ are transcendental over $\mathbb{Q}$.

(Such a field need not be a purely transcendental extension of $\mathbb{Q}$.)

Does there exist a field that satisfies (1)-(3) and contains an isomorphic copy of any field which has properties (1)-(3)?

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## 1 Answer

It's a guess only:

what about taking $\mathfrak M$ number of transcendentals $(\xi_i)_{i<\mathfrak M}$ over $\mathbb Q$, and consider the algebraic closure of $\mathbb Q(\xi_i)_i$?

Edit: Instead of the algebraic closure, consider only (all the roots of) all irreducible polynomials that has at least one transcendental over $\mathbb Q$ among its coefficients..

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This contains the entire algebraic closure of $\mathbb{Q}$. –  Chris Eagle Sep 24 '12 at 11:33
I assume you mean irreducible monics, since otherwise $\xi x^2-2\xi$ is such an irreducible with $\sqrt{2}$ as a root. In that case, what you describe isn't even a field: it contains $\xi$ (root of $x-\xi$) and $\xi+\sqrt{2}$ (root of $x^2-2\xi x+\xi^2-2$) but not $\sqrt{2}$. –  Chris Eagle Sep 24 '12 at 11:49
You're right. As I indicated, it was a guess only. –  Berci Sep 24 '12 at 20:22