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It is well known that if we have two groups/rings/vector spaces (over same field), then we can embed them in a common corresponding object. It is natural to consider for fields.

However, if we take two fields of different characteristics, then we can not embed both in a common field. With this remark, I will shorten my problem in a simple case.

Question: If $F$ is a field, and $E_1,E_2$ are any two extensions of $F$, then does there always exists a field $E$ which is extension of both $E_1$ and $E_2$?

Groups
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    The ring $R = E_1 \otimes_F E_2$ contains a maximal ideal $m$. The quotient works - agree? – peter a g Oct 15 '15 at 02:05
  • with any maximal ideal it works? – Groups Oct 15 '15 at 02:23
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    If $m$ is maximal, the quotient is a field $E$, and the canonical maps from $E_i \to E$ are ring homomorphisms - as the $E_i$ are fields, the maps are injective - so everything seems OK to me. I was wondering whether you wanted something more constructive - if so I would have chickened out. – peter a g Oct 15 '15 at 02:31
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    Actually, there is an objection to the argument. Why does my $E_1\otimes_FE_2$, the tensor product, necessarily contain $1$? Your example: it certainly wouldn't had $F$ been the integers (I know - not a field), and the $E_i$ been $\mathbb F_p$ for differing $p$. So my argument - if correct ! - is not complete. – peter a g Oct 15 '15 at 02:50
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    @peter: the only way your argument fails is if the tensor product is zero, which it can't be in this case. (I don't know what yo mean by "contain $1$," but the zero ring is the only ring that can receive a morphism from a field that isn't injective.) The point is that $E_1$ and $E_2$ are both nonzero vector spaces over the same field $F$, so their tensor product over $F$ has the same property. – Qiaochu Yuan Oct 15 '15 at 03:12
  • @QiaochuYuan I actually just logged back on to clarify - "contained a non-zero 1" - I meant, as you wrote, that the tensor product was non-trivial. And yes, your vector space argument is great. Thanks for the rescue... – peter a g Oct 15 '15 at 03:22
  • @QiaochuYuan - I added an answer below, and acknowledged you. – peter a g Oct 15 '15 at 03:40

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To summarize the comments above:

The commutative ring $ R = E_1 \otimes_F E_2$ is non-trivial, as it is the tensor product of two vector spaces over the field $F$ (thank you, Qiaochu Yuan, for saving me from stupidity, if not from embarrassment - see above). Hence, $R$ contains a maximal ideal $m$, and we can form the quotient field $E = R /m$. Any ring homomorphism $\pi$ from a field to a non-trivial ring is injective (because if $x \ne 0$, then $\pi (x ) \pi (x^{-1}) =1$, so $\pi(x)\ne 0$).

Therefore the canonical homomorphisms $E_i \to E$ are the desired embeddings.

peter a g
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