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I have a question:

Let $F$ and $K$ fields such that $F\subset K$. Can we conclude that $1_K=1_F$?

I found any counterexaples when $F$ and $k$ are rings, but not necessarily fields. However when these are fields I don't get conclude nothing...

Aaron Hendrickson
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3435
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    It is built into the definition that the $1$ is the same. This is also done with rings with $1$. Perhaps the example that you mention with rings is in a context (a book or author) that considers rings without a $1$. To distinguish this case, some use the name rng. – plop Jun 08 '21 at 23:45
  • @plop: one doesn't have to consider rings without $1$ for such a containment. Indeed, if $R$ is a ring with a nontrivial idempotent $e \in R$, then $Re$ is likewise a ring with identity $e$, and we clearly have $Re \subset R$. Of course, $Re$ is not a subring of $R$ by definition, as you point out. Your comment addresses the title question of the OP, but the body appears to be asking a slightly different question. – Alex Wertheim Jun 08 '21 at 23:51
  • @AlexWertheim: what definition of field do you have in mind that does not involve $1$? – Rob Arthan Jun 08 '21 at 23:55
  • I think I was expressing myself wrong. When I mentioned the case of rings, I was referring to rings that have unity. I found that certain rings with a unit may contain some sub-ring with a different unit. This made me wonder if the same could happen with the fields ... – 3435 Jun 08 '21 at 23:58
  • @RobArthan I have no idea what you mean, since I did not say anything in my comment which would suggest that I think there is a definition of field which does not involve 1. My point was that one may reasonably ask whether there are fields $F, K$ such that $F$ is a subset of $K$ (and inherits multiplication operation from $K$), but does not share the same identity as $K$ - this question is different from whether $F$ can be a subfield of $K$ but no share the same identity, which is prohibited by definition. – Alex Wertheim Jun 08 '21 at 23:59
  • I think you should clarify what you mean by $F \subset K$. I think you mean that the inclusion of $F$ in $K$ is a homomorphism of rngs (see first comment of @plop, for the meaning rng). If that is the case, then the homomorphism will also be a homomorphism of rings and hence fields. – Rob Arthan Jun 09 '21 at 00:01
  • @AlexWertheim: see my second comment: you were reading too much into the statement "$F \subset K$". The OP needs to clarify that. – Rob Arthan Jun 09 '21 at 00:04

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$1_F \cdot 1_K = 1_F$ by definition of $1_K.$ Also $1_F =1_F\cdot 1_F$ by definition of $1_F.$ Thus $1_F\cdot 1_K =1_F \cdot 1_F.$ Since $1_F\neq 0_F$ by definition of a field, there is an inverse element $1_F^{-1}$ and multiplication with it from the left yields $1_K=1_F.$

No definition necessary. It already follows from the field axioms.

Marius S.L.
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    Indeed. One can also conclude this directly from $1_{F} = 1_{F}^{2}$, since $K$ does not have any nontrivial idempotents, and so we must have $1_{F} = 1_{K}$. – Alex Wertheim Jun 08 '21 at 23:56
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    I think you are making assumptions about what the OP means by $F \subseteq K$. – Rob Arthan Jun 08 '21 at 23:58
  • You are very right! ... I was thinking to use the fact that the fields dosen't have divisors of 0 and use the inverses, but I didn't find the suited relation – 3435 Jun 09 '21 at 00:18