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[parent] proof that Q is the prime subfield of any field of characteristic 0 (Proof)

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 one-to-one 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 well-defined 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\ne 0$, then $ 1=\phi(1)=\phi(pp^{-1})=\phi(p)\phi(p^{-1})=0\cdot\phi(p^{-1})=0$, a contradiction. $ \qedsymbol$

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 well-defined, for if $ m=n$ in $ \mathbb{F}_p$, then $ p\mid (m-n)$, meaning $ (m-n)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 one-to-one.

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. $ \qedsymbol$



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"proof that Q is the prime subfield of any field of characteristic 0" is owned by CWoo. [ full author list (3) ]
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See Also: rational numbers are real numbers

Also defines:  prime field

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Cross-references: additivity, clear, complete, argument, injective, contradiction, multiplicative, additive, function, well-defined, mapping, coprime, field homomorphism, one-to-one, prime subfield, characteristic, field, propositions
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This is version 13 of proof that Q is the prime subfield of any field of characteristic 0, born on 2006-02-06, modified 2006-03-13.
Object id is 7600, canonical name is GroundField.
Accessed 2727 times total.

Classification:
AMS MSC: 12E20 (Field theory and polynomials :: General field theory :: Finite fields )
 12E99 (Field theory and polynomials :: General field theory :: Miscellaneous)
 12F99 (Field theory and polynomials :: Field extensions :: Miscellaneous)
 15A99 (Linear and multilinear algebra; matrix theory :: Miscellaneous topics)
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A little sketchy by mathcam on 2006-02-09 12:55:20
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 p-adic 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 re-write 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
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