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 $p\eq 0$ , 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.∎

单词	QIsThePrimeSubfieldOfAnyFieldOfCharacteristic0ProofThat
释义	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 $p\eq 0$ , 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.∎
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