Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question

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..

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.