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Thread: Subfield Isomorphic To Q

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    Senior Member vincisonfire's Avatar
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    Subfield Isomorphic To Q

    I would like to know how to show that if the characteristic of a field is 0 then it has a subfield isomorphic to $\displaystyle \mathbb Q $.

    I defined an homomorphism $\displaystyle f : \mathbb F \rightarrow \mathbb Q $
    $\displaystyle f(n) = n \cdot 1 $
    I found, $\displaystyle Ker(f) = \{0\} $ The map is thus injective.
    I thought I might find a surjective map to $\displaystyle \mathbb F/\{0\} $ and use the first isomorphism theorem to conclude that $\displaystyle \mathbb F/\{0\} $ is a subfield isomorphic to $\displaystyle \mathbb Q $.
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    Quote Originally Posted by vincisonfire View Post
    I would like to know how to show that if the characteristic of a field is 0 then it has a subfield isomorphic to $\displaystyle \mathbb Q $.

    I defined an homomorphism $\displaystyle f : \mathbb F \rightarrow \mathbb Q $
    $\displaystyle f(n) = n \cdot 1 $
    I found, $\displaystyle Ker(f) = \{0\} $ The map is thus injective.
    I thought I might find a surjective map to $\displaystyle \mathbb F/\{0\} $ and use the first isomorphism theorem to conclude that $\displaystyle \mathbb F/\{0\} $ is a subfield isomorphic to $\displaystyle \mathbb Q $.
    Define an embedding $\displaystyle f: \mathbb{Z} \to F $ by $\displaystyle f(n) = n\cdot 1$. This notation means $\displaystyle n\cdot 1 = 1 + ... + 1$, $\displaystyle n$ times if $\displaystyle n>0$ and $\displaystyle n\cdot 1 = - (1+...+1)$ if $\displaystyle n<0$ and $\displaystyle 0\cdot 1 = 0$. Therefore, $\displaystyle \mathbb{Z}$ is embedded in $\displaystyle F$, however if $\displaystyle F$ contains by an embedding $\displaystyle \mathbb{Z}$ it must contain by an embedding a field of quotients of $\displaystyle \mathbb{Z}$ i.e. $\displaystyle \mathbb{Q}$.
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