# 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. 1.

additive: for $p,q\in\mathbb{Q}$, $\phi(p+q)=\phi(p)+\phi(q)$;

2. 2.

multiplicative: for $p,q\in\mathbb{Q}$, $\phi(pq)=\phi(p)\phi(q)$;

3. 3.

$\phi(1)=1_{F}$, and

4. 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 QIsThePrimeSubfieldOfAnyFieldOfCharacteristic0ProofThat 2013-03-22 15:39:57 2013-03-22 15:39:57 CWoo (3771) CWoo (3771) 16 CWoo (3771) Proof msc 15A99 msc 12F99 msc 12E99 msc 12E20 RationalNumbersAreRealNumbers prime field