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Q is the prime subfield of any field of characteristic 0, proof that
The following two propositions show that $\mathbb{Q}$ can be embedded in any field of characteristic $0$, while $\mathbb{F}_{p}$ can be embedded in any field of characteristic $p$.
Proposition. $\mathbb{Q}$ is the prime subfield of any field of characteristic 0.
Proof.
Let $F$ be a field of characteristic $0$. We want to find a onetoone field homomorphism $\phi:\mathbb{Q}\to F$. For $\frac{m}{n}\in\mathbb{Q}$ with $m,\,n$ coprime, define the mapping $\phi$ that takes $\frac{m}{n}$ into $\frac{m1_{F}}{n1_{F}}\in F$. It is easy to check that $\phi$ is a welldefined function. Furthermore, it is elementary to show
1. additive: for $p,q\in\mathbb{Q}$, $\phi(p+q)=\phi(p)+\phi(q)$;
2. multiplicative: for $p,q\in\mathbb{Q}$, $\phi(pq)=\phi(p)\phi(q)$;
3. $\phi(1)=1_{F}$, and
4. $\phi(0)=0_{F}$.
This shows that $\phi$ is a field homomorphism. Finally, if $\phi(p)=0$ and $p\neq 0$, then $1=\phi(1)=\phi(pp^{{1}})=\phi(p)\phi(p^{{1}})=0\cdot\phi(p^{{1}})=0$, a contradiction. ∎
Proposition. $\mathbb{F}_{p}$ ($\cong\mathbb{Z}/p\mathbb{Z}$) is the prime subfield of any field of characteristic $p$.
Proof.
Let $F$ be a field of characteristic $p$. The idea again is to find an injective field homomorphism, this time, from $\mathbb{F}_{p}$ into $F$. Take $\phi$ to be the function that maps $m\in\mathbb{F}_{p}$ to $m\cdot 1_{F}$. It is welldefined, for if $m=n$ in $\mathbb{F}_{p}$, then $p\mid(mn)$, meaning $(mn)1_{F}=0$, or that $m\cdot 1_{F}=n\cdot 1_{F}$, (showing that one element in $\mathbb{F}_{p}$ does not get “mapped” to more than one element in $F$). Since the above argument is reversible, we see that $\phi$ is onetoone.
To complete the proof, we next show that $\phi$ is a field homomorphism. That $\phi(1)=1_{F}$ and $\phi(0)=0_{F}$ are clear from the definition of $\phi$. Additivity and multiplicativity of $\phi$ are readily verified, as follows:

$\phi(m+n)=(m+n)\cdot 1_{F}=m\cdot 1_{F}+n\cdot 1_{F}=\phi(m)+\phi(n)$;

$\phi(mn)=mn\cdot 1_{F}=mn\cdot 1_{F}\cdot 1_{F}=(m\cdot 1_{F})(n\cdot 1_{F})=% \phi(m)\phi(n)$.
This shows that $\phi$ is a field homomorphism. ∎
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Comments
A little sketchy
I think this entry confuses a couple of important points, stemming from the fact that you've tried to make the notion of a "ground field" halfway between an intuitive definition and a formal definition.
The formal counterpart to what you're talking about is the prime field, and your two theorems refer to those, Q and F_p (which, incidentally, is a better notation since Z_p can be confused with the padic integers).
The intuituve counterpart turns out to be not so intuitive after all..consider the scenario of considering the field extension \mathbb{C}(\sqrt{x}) over \mathbb{C}(x). One wouldn't really want to call Q the "ground field" in such a scenario, but further, nor is this notion really well defined. For example, \mathbb{C}(x) is just a degree 2 extension of \mathbb{C}(x^2), which in turn is a degree 2 extension of \mathbb{C}(x^4), etc. In this situation, there is no unique smallest subfield of the right type (i.e. a function field over \mathbb{C}) that embeds into all of the field in question.
I recommend either giving a formal definition of a prime field, or rewrite to be clear that "ground field" is not a formal term, and that the expression is used informally by mathematicians to refer to an obvious choice of an important field lying around somewhere.
Cam
Re: A little sketchy
NB, in some contexts one speaks also of "base field" (see e.g. the entry "extension field").
Jussi
Re: A little sketchy
Thank you for your comment! Sorry to have caused this confusion. I have changed the title and the content of the entry. I have also added a request for someone to define, at least informally, what a ground field is. Chi