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Let $A,B$ be two fields. Let $\phi:A\rightarrow B$ and $\psi:B\rightarrow A$ be two morphisms of fields. Can i conclude that $A$ and $B$ are isomorphic fields?

My guess is yes, because every morphism of fields is injective, hence in this case $B$ contains an isomorphic copy of $A$, which in turns contains one copy of $B$. If this is right, how can i formalize it?

Federica Maggioni
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  • Unless you require a morphism of fields to be a morphism of unitary rings carrying the multiplicative unit to the mult. unit, the morphisms can be the zero ones and not injective... – DonAntonio Apr 05 '13 at 14:09
  • @DonAntonio I require that – Federica Maggioni Apr 05 '13 at 14:15
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    The question is a bit subtle. It is possible for a field that is infinite dimensional over its prime field to be isomorphic to its proper subfield. – Jyrki Lahtonen Apr 05 '13 at 14:20
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    See Martin Brandenburg's answer here. –  Apr 05 '13 at 14:26
  • The tag "morphism" is a great achivement of this site! –  Apr 05 '13 at 14:39
  • @DonAntonio: Excuse me for the impolite tone. I do still think that although one can define lots of things, some lead to bizarre situations. Defining unital rings as a full subcategory of non-unital ones can be done (not requiring subrings to share their $1$), but many results require a more restricted notion; one could use $\Bbb Z$-algebras for that (but calling $\Bbb Z$-algebras unital rings, like $\Bbb Z$-modules are Abelian groups, has its charm). However if morphisms of fields can be zero, then images of field morphsims would not be subfields; this is very uncommon in algebraic categories – Marc van Leeuwen Apr 05 '13 at 17:03
  • @DonAntonio: OK I've read, and I'll remove my comment. I also just realised that what I said about $\Bbb Z$-algebras is nonsense; the definition of $R$-algebra does not even require the existence of a $1$. What I meant is unital $\Bbb Z$-algebras, taken to also imply that homomorphisms preserve units... – Marc van Leeuwen Apr 05 '13 at 17:17
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    Dear @DonAntonio, I'm sorry that you don't like the ring with $0=1$ but I'm afraid you'll have to coexist with it: I don't think that in 2013 there is one active algebraic geometer or commutative algebraist in the world who would deny that this ring exists . Excluding it would exclude the empty scheme from scheme theory, which is exactly as (un)reasonable as excluding the empty set from set theory. On the other hand I don't think there has ever existed one book on field theory which claimed that this ring with $0=1$ is a field. – Georges Elencwajg Apr 05 '13 at 19:01
  • Oh, I don't care about it almost at all, dear @GeorgesElencwajg, though I'm sure a huge lot of mathematicians plainly and sheerly erase it from their schedule, and even algebraic geometers won't cry a lot, me thinks, if the victim of denying axiomatically $,0=1,$ in ring theory is a such a non-transcendental being as the empty scheme...OTOH I can't tell about all the books on field theory, but I think you're right in that, and good if it is so. – DonAntonio Apr 05 '13 at 19:35
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    Dear @YACP, do you mean yourself? I quite agree and we are in very good company: Bourbaki calls these structures without unit element "pseudorings" and Jacobson calls them "rngs": probably a joke based on the fact that the capital letter "I" typographically looks like the unity $1$ of a ring, so that a "ring" without unity should be denoted by the word "RING" without the letter "I" . Again, I don't think that there are any algebraic geometers or commutative algebraists (of whom you are a distinguished representative) who would call "rings" these structures without $1$. – Georges Elencwajg Apr 05 '13 at 21:27
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    Possible duplicate of Example of two non-isomorphic fields which embed inside each other – Watson Aug 19 '16 at 12:16

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This is an occasion, when instincts developed over finite extensions of (prime) fields lead one astray.

The first counterexamples that come to mind need a bit of background from the theory of elliptic curves. It is quite possible for there to be isogenies going back and forth between two non-isomorphic elliptic curves, $E_1$ and $E_2$. The isogenies give rise to embeddings between the corresponding function fields $K(E_1)$ and $K(E_2)$ (take for example $K=\mathbb{C}$ to avoid several algebraic pitfalls). Yet, if the two elliptic curves are not isomorphic, the functions fields won't be isomorphic either.

Jyrki Lahtonen
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