MathOverflow is a question and answer site for professional mathematicians. 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

Suppose $F$ is a field, and $F_1, F_2$ are two extension fields of $F$. Is it always the case that there is a field $L$, containing three subfields $F, K_1, K_2$ and two ring isomorphisms $\varphi_{i}:F_i\rightarrow K_1$ fixing $F$?

Note 1: We lose no generality assuming $F$, rather than an isomorphic copy of $F$, is a subfield of $L$.

I ask this because I was wondering if there is a way to combine the reals and the $p$-adic numbers into a single extension of $\mathbb{Q}$.

Note 2: I seem to recall someone telling me this couldn't be done (perhaps with additional topological data preserved). But I cannot seem to remember the reason why. In any case, I want to know if there is something other than topology which prevents it.

share|cite|improve this question
Concerning the p-adics, reals: We have $\mathbb{R} \subseteq \mathbb{C}$, $\mathbb{Q}_p \subseteq \mathbb{C}_p$ and $\mathbb{C} \cong \mathbb{C}_p$. Using this isomorphism you can embedd $\mathbb{Q}_p$ into $\mathbb{C}$. Of course this iso. isn't defined in a constructible way. – Ralph Feb 6 '12 at 20:12
Regarding Note 2: There is no topological field containing topological copies of $\mathbb{R}$ and $\mathbb{Q}_p$, since each of these induce distinct topologies on $\mathbb{Q}$. The isomorphism Ralph describes is not continuous. – Kevin Ventullo Feb 7 '12 at 3:05
up vote 28 down vote accepted

The tensor product $F_1 \otimes_F F_2$ is not 0, hence it has a quotient which is a field. This contains the images of both $F_i$.

share|cite|improve this answer
Years ago, I first learnt the solution in David's answer and could not really find anything enlightening or memorable about it. Later the argument using tensor products was used in a text on algebraic geometry (namely the lemma that $|X \times_S Y| \to |X| \times_{|S|} |Y|$ is surjective) and of course now everything was clear as crystal. As a side remark, both proofs use variants of the axiom of choice (even twice). – Martin Brandenburg Feb 7 '12 at 11:00
(This also reminds me of the concise tensor product construction of the algebraic closure of a field: In $k' := \bigotimes_{0 \neq f \in k[x]} k[x]/(f)$ every polynomial as a root, thus the colimit of $k \subseteq k' \subseteq k'' \subseteq k''' \subseteq ...$ is an algebraic closure of $k$. One can show $k''=k'$, but this is not trivial.) – Martin Brandenburg Feb 7 '12 at 11:03

Sure. Find $T_i$ between $F$ and $F_i$ such that $T_i/F$ is pure transcendental and $F_i/T_i$ is purely algebraic. Let $T_i = F(S_i)$, with the $S_i$ algebraically independent. Without loss of generality, suppose that the cardinality of $S_1$ is less than or equal to that of $S_2$. Then the algebraic closure of $F(S_2)$ is a suitable $L$.

share|cite|improve this answer

In the language of Model Theory, your question can be rewritten as: "does the theory of fields have the amalgamation property? and the answer is yes.
Well known examples of theories with the amalgamation property include: fields, ordered fields, groups, abelian groups and boolean algebras.

share|cite|improve this answer
+1 since you provide the general background, but it would be even a better answer if you add a specific reference where the amalgamation property for fields is proved. I believe that model theorists argue as in David's answer or equivalently with "$\kappa$-categorical" arguments. – Martin Brandenburg Feb 7 '12 at 11:07
@Martin: To be honest, I´ve never seen a published proof of the fact that fields have the amalgamation property. Model Theory textbooks (e.g. Hodges, Chang & Keisler) just mention the fact and use it to prove other things, like quantifier elimination for algebraically closed fields. – Ramiro de la Vega Feb 7 '12 at 11:53

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.