Let F and E be finite fields of the same characteristic. In Milne's notes exercise. It says it is not true that There is a ring homomorphism of F into E iff $|E|$ is a power of $|F|$. But I cannot find such an example. So to see it clearly. I am searching for such an example. Could someone give one? Thanks.
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1$\Bbb F_4$ is a subfield of $\Bbb F_8$, isn't it? – Hagen von Eitzen May 31 '17 at 16:39
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@HagenvonEitzen Sorry I haven't heard about that. May I ask why is that true？ – non-abelian group of order 9 May 31 '17 at 16:48
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2@HagenvonEitzen $\Bbb F_4$ is not a subfield of $\Bbb F_8$. – Angina Seng May 31 '17 at 16:58
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1Hagen von Eitzen just posed that as a question for OP to ponder. I would add $\Bbb{F}2$ and $\Bbb{F}{16}$ to the list as well. What inclusions are there and how do the cardinalities relate? Another hint: this is sometimes stated as a field of of $p^d$ elements embeds in a field with $p^{d'}$ elements iff $d|d'$ – sharding4 May 31 '17 at 17:00
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Related question https://math.stackexchange.com/questions/91087/subfields-of-finite-fields – Birch Bryant May 31 '17 at 17:01
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@sharding4 I have just read it. Note that my question is why we also have a homomorphism when the order in NOT a power of another. – non-abelian group of order 9 May 31 '17 at 17:09
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There's always the trivial map I guess. – sharding4 May 31 '17 at 17:15
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Please check the claim. A ring homomorphism from one field to another is always injective. So the domain is isomorphic to a subfield of the range. I haven't checked but I would be somewhat surprised if ring homomorphisms don't map the unit to the unit for Milne. – Jyrki Lahtonen May 31 '17 at 21:01