Let $K$ be a field having two field extensions $L\supseteq K$ and $M\supseteq K$. Does there exist a field $N$ along with embeddings $L\to N$ and $M\to N$, such that the diagram $$ \require{AMScd} \begin{CD} K @>>> L\\ @V V V @VV V\\ M @>>> N \end{CD} $$ is commutative? To put it less formally, do $L$ and $M$ have a common field extension $N$ (with $K$ lying in the intersection)?

If yes, consider this side question (but do leave an answer even if you can only answer the main question!): Does the above property of fields generalise to the following, stronger property? Let $p$ be either $0$ or a prime number. Does there exist a sequence of fields $$ \mathbb F_p = L_0\subseteq L_1\subseteq L_2\subseteq L_3\subseteq L_4\subseteq\cdots$$ (where I use the convention $\mathbb F_0 = \mathbb Q$) such that any field $K$ of characteristic $p$ has an extension field among the $L_\alpha$? This sequence should be understood as enumerated by ordinal numbers. In other words, what I need is a function that assigns to each ordinal number $\alpha$ a field $L_\alpha$ containing all $L_\beta$ with $\beta < \alpha$. (Intuitively, I suspect that this might depend on the Axiom of Choice.)

If the second property holds, it shows that you can essentially only extend a field in one "direction." It also shows that the class (it is obviously not a set) $\mathbb M_p = \bigcup_\alpha L_\alpha$ is a field (class). We can define all the usual field operations (addition, multiplication, division) here since all pairs of elements lie in $L_\alpha$ for a sufficiently large $\alpha$. This "monster field" of characteristic $p$ then contains all other set fields of that characteristic.

  • $\begingroup$ Are you looking for an example such as $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $M=\mathbb{Q}(\sqrt{3})$ and $N=\mathbb{Q}(\sqrt{2},\sqrt{3})$ $\endgroup$ – Anurag A Jan 10 '15 at 8:28
  • $\begingroup$ No, I'm looking for the answer as to whether it is in general possible. If yes, I would like a proof or a reference to a proof. If no, I would like a counterexample. $\endgroup$ – Gaussler Jan 10 '15 at 8:33
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    $\begingroup$ There are fields of arbitrarily large cardinal, so no set of fields (in particular, no sequence of fields) contains an extension of every field. $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '15 at 8:44
  • $\begingroup$ But yes, your example catches the idea of what I am talking about. $\endgroup$ – Gaussler Jan 10 '15 at 8:44
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    $\begingroup$ Most humans will understand that by sequence in your question you meant what is most commonly named a sequence. If you intended otherwise, you should be explicit; you want ordinals and not cardinals. I was about to write an answer, but you have managed to demotivate me completely. $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '15 at 8:53

I believe the answer to both questions is yes. I will assume we fix one characteristic $p$ (possibly $0$) and stick to it, because there are no morphisms/embeddings between fields of different characteristic.

First note that we may embed every field in an algebraic closure (in fact even in an algebraic closure of the same cardinality if the field is infinite) and restrict ourselves to algebraically closed fields.

One should show the following lemma:

Lemma. If $X,Y$ are algebraically closed fields with transcendence degree $\kappa_X$ and $\kappa_Y$, where $|X|\leq|Y|$ and $\kappa_X \leq \kappa_Y$, then there exists an embedding $X\to Y$.

To see this, adapt the proof of Theorem 1 here. What is different is that the map $f$ from the proof of the theorem is now injective and one needs a theorem that an embedding of fields can be extended to an embedding of their algebraic closures. This is for instance the content of theorem V.2.8 in Lang's Algebra.

With this result it follows easily that both $L$ and $M$ (by assumption algebraically closed) embed into a common algebraically closed field of cardinality $\mathrm{max}(|K|,|L|)$ and transcendence degree $\mathrm{max}(\kappa_K,\kappa_L)$.

For your second question, let $\alpha \geq\omega $ be an ordinal number and define $L_\alpha$ to be an algebraically closed field of cardinality and transcendence degree both equal to $|\alpha|$. (This will certainly exist, for instance the algebraic closure of $\Bbb F_p(\{t_i\mid i\in\alpha\})$. Define $L_\alpha = \mathbb F_p$ if $\alpha < \omega$. Then by the lemma and the observation that the transcendence degree cannot exceed the cardinality of the field, every field of characteristic $p$ must be embedded in one of these fields.

(The situation is in fact simpler if one is willing to apply a result that two algebraically close fields of the same characteristic and the same uncountable cardinality are isomorphic, so you can stop worrying about the transcendence degree in larger fields.)

Finally, all of this (not just the second question) depends heavily on the axiom of choice. Maybe not crucially so (see this MO question) but in the absense of choice and with weaker axioms in place this clearly becomes much harder.

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  • $\begingroup$ Keeping $p$ fixed was my intention as well, as formulated in the question. Nice answer! $\endgroup$ – Gaussler Jan 10 '15 at 13:21
  • $\begingroup$ Your lemma does not seem true to me. What about $X=\mathbb{Q}(t,\sqrt{2})$ and $Y=\mathbb{Q}(r,s)$? Then $X$ has transcendence degree $\kappa_X=1$ over its prime field, and $Y$ has transcendence degree $\kappa_Y=2$, yet there cannot be any field homomorphism $X\to Y$, because where would $\sqrt{2}$ go? $\endgroup$ – Sal Jan 10 '15 at 16:06
  • $\begingroup$ @Sal but these fields are not algebraically closed right? $\endgroup$ – Myself Jan 10 '15 at 16:07
  • $\begingroup$ @Myself: Sorry, I hadn't read the preceding comment carefully enough - thanks for the clarification. $\endgroup$ – Sal Jan 10 '15 at 16:08
  • $\begingroup$ I noticed that it wasn't in the statement of the lemma so I have added it. You are certainly correct that it doesn't make much sense for general fields, I'm not even sure if the transcendence degree is well defined in that case. $\endgroup$ – Myself Jan 10 '15 at 16:12

The $K$-algebra $A=L\otimes_K M$ has, like all non-zero rings, a maximal ideal $\mathfrak m\subset A$ .
The quotient algebra $N=A/\mathfrak m $ is then an extension field of both $L$ and $M$, via the composed morphisms $L\to A\to N$ and $M\to A\to N$ .
[Recall that any morphism of rings with source a field is injective.]

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  • $\begingroup$ Surprisingly elegant solution! $\endgroup$ – Gaussler Jan 10 '15 at 14:28

It is not clear what is meant by the question, because it is not clear where $M,L$ live and what it means to have a common extension field. As stated, I would be inclined to say the answer to the question is no, by the following example. Let $K=\Bbb R$, and let inside the quaternions $L=\Bbb C=\Bbb R[\mathbf i]$ and $M=\Bbb R[\mathbf j]$, isomorphic to but distinct from $L$. Then no extension field of $\Bbb R$ can contain both $L$ and $M$, because the polynomial $X^2+1$ would have at least $4$ roots. But of course you can find an extension of $\Bbb R$ that contains subfields isomorphic to $L,M$.

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  • $\begingroup$ WHen I speak about abstract algebra, "isomorphic" is more or less the same as "equal". That $N$ is a common extension field of $L$ and $M$ simply means that $M$ and $L$ can be embedded into $N$, and that something isomorphic to $K$ lies in $M\cap L\subseteq N$. $\endgroup$ – Gaussler Jan 10 '15 at 11:41
  • $\begingroup$ Then please edit the question to say that clearly (after which I will delete this answer). $\endgroup$ – Marc van Leeuwen Jan 10 '15 at 11:42
  • $\begingroup$ Has been done. Sorry for the confusion. I'm used to being a bit abstract when dealing with algebra. $\endgroup$ – Gaussler Jan 10 '15 at 11:46
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    $\begingroup$ @Gaussler I think that what Marc van Leeuwen really means is a warning that even if the answer is yes, you have no control over how both field will 'interact' inside a larger field that contains them both. This is for instance why S. Lang says that the compositum cannot be defined if there is no common field in which both fields are embedded. $\endgroup$ – Myself Jan 10 '15 at 11:54
  • $\begingroup$ But for that matter, you can keep the answer. It's a good and instructive example of why one has to take care about becoming too abstract. $\endgroup$ – Gaussler Jan 10 '15 at 11:54

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