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I have seen a Theorem:

every field contains exactly one prime subfield $K_0$.

and next:

$K_0$ has to contain 1 and all its multiples: $n \cdot 1 = 1 + \ldots + 1$

Is this because it has to contain 0 and 1 by definition, and then it must also holds that $1 + 1$ belongs to K, and $(1 + 1) + 1$ too because of two-arguments $+$ operation of its abelian group?

so we can prove it by induction?

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Minor remark: simple subfield is customary called prime field. – Orat Feb 24 '13 at 15:42
thank you, changed – where_is_tftp Feb 24 '13 at 15:54
up vote 5 down vote accepted

Yes, that is correct.

The base case ($n=1$) is trivial.

For the inductive step let $n\in \Bbb N$ such that $n\cdot 1_K \in K$.

By definition we have $(n+1)\cdot 1_K=n\cdot 1_K + 1_K$. Since $n\cdot 1_K\in K$ (by our hypothesis) and since $1_K\in K$, from the fact that $K$ is a field it follows that $n\cdot 1_K + 1_K\in K$, that is $(n+1)\cdot 1_K\in K$, which completes the induction.

Edit: The fact that $K$ with $+$ is abelian is irrelevant. You just use the fact that it is a group.

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or even just that it is subgroup with its + ? because 1+1 must belong to K0 – where_is_tftp Feb 24 '13 at 16:04
@cf16 Every subgroup is a group. But you got the right idea. – Git Gud Feb 24 '13 at 16:05

If $A$ is a group and $a\in A$, then there is a unique group homomorphism $f\colon \mathbb Z\to A$ with $f(1)=a$. We write $a^n$ or - if $A$ is written additively - $n\cdot a$ for the element $f(n)\in A$ with $f$ defined like this,

Apply this to the additive group of $K$ and $a=1$.

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yes, but my proof is more low level I think xD – where_is_tftp Feb 24 '13 at 16:06
@cf16 I think Hagen just meant to give you an alternative view of this problem since I had already answered the question in the most elementary way. – Git Gud Feb 24 '13 at 16:10
@Hagen von Eitzen and Git Gud, thank you both very much – where_is_tftp Aug 26 '13 at 11:43

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