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In set-theory, one of the standard result (Bernstein's theorem) is that if there is an injection from $A$ to $B$ and an injection from $B$ to $A$, then there is a bijection from $A$ to $B$.

Consider similar situation on sets with some algebraic structures, say $A$ and $B$ are fields. If there is an injective homomorphism from $A$ into $B$ and also from $B$ into $A$, can we conclude that $A$ and $B$ are isomorphic fields?

p Groups
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  • I don't know about fields, but this is false for groups. Let $A$ be the free group on two letters $x,y$, and $B$ be the free group on three letters $a,b,c$. There is an obvious injective homomorphism from $A$ to $B$. Further, mapping $a\to x^2, b\to y^2, c \to xy$ is an injective homomorphism from $B$ to $A$. But these two free groups are not isomorphic. – florence Jul 06 '16 at 06:00
  • yes, free group of rank $2$ contains free groups of more rank, in short. – p Groups Jul 06 '16 at 06:00
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    See the discussion here. There are counterexamples in the comments. – Michael Albanese Jul 06 '16 at 06:01
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    Also here. As long as the kernel of the isogeny is not the full torsion subgroup we get non-isomorphic fields both embeddable into the other. – Jyrki Lahtonen Jul 06 '16 at 06:58
  • Actually Martin Brandenburg's answer in MathOverflow makes the possibility more obvious in that you won't need any background in elliptic curves. Thanks to user26857 for digging it up. – Jyrki Lahtonen Jul 06 '16 at 07:09
  • See also http://math.stackexchange.com/questions/803332/embeddings-a-%E2%86%92-b-%E2%86%92-a-but-a-not-cong-b?noredirect=1&lq=1, http://math.stackexchange.com/questions/257650 – Watson Aug 19 '16 at 12:17

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