I found two definitions for a prime subfield $K$ of a field $F$.
1. Wolfram $-$ $K$ is the subfield of $F$ generated by the multiplicative identity $1$ of $F$.
2. ProofWiki $-$ $K$ is the intersection of all the subfields, say $\{ K_c \}$, of $F$.

I am aware that in the first definition, $\langle 1 \rangle = \{ n \cdot 1 : n \in \mathbb Z \}$. Since $1$ is in each $K_c$, $\langle 1 \rangle \subset K$. But, how are they equal? I tried two approaches and they ran into the same problem.

One, prove directly that $K \subset \langle 1 \rangle$. But how do I know each $x \in K$ is of the form $1 + 1 ... +1$? Another approach is to show that $\langle 1 \rangle$ is a subfield of $F$, ie an element in $\{ K_c \}$. But how do I show that the inverse, $x^{-1}$, of each $x \in \langle 1 \rangle$ is also of the form $1 + 1 ... +1$?

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    $\begingroup$ Well, "generate" in field means the composition of addition, mutiplication, inverse $\endgroup$ – Yilong Zhang Jun 25 '15 at 2:30
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    $\begingroup$ The field axioms only imply that the thing all subfields have in common is $1$, hence... $\endgroup$ – Adam Hughes Jun 25 '15 at 2:31
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    $\begingroup$ Also, "generated by" means you get to include inverses for free, since it's field generation, not ring generation. $\endgroup$ – Adam Hughes Jun 25 '15 at 2:33
  • $\begingroup$ Are you all suggesting this is another case of "abusing" the language? Ugh .... $\endgroup$ – Andy Tam Jun 25 '15 at 2:45

Let's write $K$ for the field "generated by $1$" and $L$ for the intersection of all subfields of $F$.

It should be clear that $L\subseteq K$. This is because $K$ is a subfield of $F$, and is therefore one of the fields in the intersection that created $L$.

What does it mean to be "generated by $1$", and why is $K\subseteq L$?

Saying that $K$ is generated by $1$ means that $K$ is the smallest subfield containing $1$. Since $K$ is closed under addition, it must contain every element of the form $1 + 1 + \cdots + 1$. Since $K$ is closed under taking negatives, it must contain every element of the form $-(1 + 1 + \cdots + 1)$. There are now two possibilities.

  1. There is some number $n$ such that $$ \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0. $$ In this case, it turns out that $n$ has to be prime and $L=\mathbb F_p$, the finite field with $p$ elements (I'll supply details if you're interested). We don't have to add multiplicative inverses in because they're already present.

  2. There is no number $n$ such that $$ \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0. $$ In this case the subring generated by $1$ is $\mathbb Z$. We think of the number $n$ as being $$ \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0. $$ You should then check that this definition makes sense with multiplication. (This is a consequence of the distributive law.) But since a subfield must contain multiplicative inverses, we're forced to add these in and we get that $L=\mathbb Q$.

(Note: we say that $F$ has characteristic $p$ in the first case and $0$ in the second.)

Finally, since $L$ is a field it contains $1$ and so it has to also contain the field generated by $1$, as that field is the smallest subfield containing $1$.

  • $\begingroup$ Ok, TY .... To answer your question, if char $p \ne 0$, it must be prime. Otherwise, with associativity and cancellation, we will end up with smaller integer, $m$, such that $m \cdot 1 = 0$, contradicting the minimality in $p$. $\endgroup$ – Andy Tam Jun 25 '15 at 12:32

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