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I read that any field $F$ has a unique smallest subfield $F_0$ (Dummit and Foote : Exercise 7.5.3).

Consider the field $F = F_p \times F_p$. $(F_p,0)$ is a sub-field of $F$ since it has a unit $(1,0)$ and every element has an inverse. Obviously this set is not unique. Other subfields are $\{(0,0), (1,1), ... ,(p,p)\}$ and $(0, F_p)$.

What am I missing here ? Do all subfields of $F$ contain $(1,1)$ ?

UnadulteratedImagination
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    $F$ is not a field: $(1,0)\cdot (0,1) = (0,0)$, so that $F$ is not even an integral domain. – rogerl Dec 12 '14 at 02:11

1 Answers1

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  1. The product of two fields is not a field, as it is not even an integral domain (as pointed out in a comment).

  2. But also, yes, a subfield has to contain the multiplicative identity of the original field (this is even the case for subrings, at least this is a common though not universal convention). Thus, what you consider are not even subrings (except for the one containing $(1,1)$). Furthermore this smallest subfield is the field generated by the multiplicative identity. It is either, isomorphic to $Z/pZ$ for a prime $p$ or to the rationals.

quid
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