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 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 $p\neq 0$ , 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 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. ∎

Title	Q is the prime subfield of any field of characteristic 0, proof that
Canonical name	QIsThePrimeSubfieldOfAnyFieldOfCharacteristic0ProofThat
Date of creation	2013-03-22 15:39:57
Last modified on	2013-03-22 15:39:57
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	16
Author	CWoo (3771)
Entry type	Proof
Classification	msc 15A99
Classification	msc 12F99
Classification	msc 12E99
Classification	msc 12E20
Related topic	RationalNumbersAreRealNumbers
Defines	prime field