Prime field of a characteristic zero field is isomorphic to $mathbb{Q}$
I was attempting to prove that the prime field $P$ of a characteristic 0 field $K$ is isomorphic to $mathbb{Q}$. Here, I will use the notation
$nx=left{begin{array}{cc}underbrace{x+cdots+x}_{n text{ times}} & text{if }n>0\ 0 & text{if } n=0 \ underbrace{(-x)+cdots+(-x)}_{n text{ times}} & text{if }n<0 end{array}right.$
Firstly, I showed that $P={(m1)cdot(n1)^{-1}; m,ninmathbb{Z},nneq 0}$, where 1 is the mulitplicative identity of $K$. Then, I showed that $phi:mathbb{Q}to P$ such that $dfrac{m}{n}mapsto(m1)cdot(n1)^{-1}$ is the isomorphism that I needed to build.
I accessed the ProofWiki to check my answer and there I found this page with the proof of the same thing but, in their proof, they show that $phi$ is well-defined, i.e., given $dfrac{a}{b}=dfrac{c}{d}$, we have that $phileft(dfrac{a}{b}right)=phileft(dfrac{c}{d}right)$. Why is it necessary to show that $phi$ is well-defined?
abstract-algebra proof-explanation
asked Sep 11 '16 at 13:45
Edu W.
155
add a comment |
I was attempting to prove that the prime field $P$ of a characteristic 0 field $K$ is isomorphic to $mathbb{Q}$. Here, I will use the notation
$nx=left{begin{array}{cc}underbrace{x+cdots+x}_{n text{ times}} & text{if }n>0\ 0 & text{if } n=0 \ underbrace{(-x)+cdots+(-x)}_{n text{ times}} & text{if }n<0 end{array}right.$
Firstly, I showed that $P={(m1)cdot(n1)^{-1}; m,ninmathbb{Z},nneq 0}$, where 1 is the mulitplicative identity of $K$. Then, I showed that $phi:mathbb{Q}to P$ such that $dfrac{m}{n}mapsto(m1)cdot(n1)^{-1}$ is the isomorphism that I needed to build.
I accessed the ProofWiki to check my answer and there I found this page with the proof of the same thing but, in their proof, they show that $phi$ is well-defined, i.e., given $dfrac{a}{b}=dfrac{c}{d}$, we have that $phileft(dfrac{a}{b}right)=phileft(dfrac{c}{d}right)$. Why is it necessary to show that $phi$ is well-defined?
abstract-algebra proof-explanation
asked Sep 11 '16 at 13:45
Edu W.
155
Well, in principle it could be ill-defined. Rational numbers have no unique representation as fractions.
– Crostul
Sep 11 '16 at 13:57
@Crostul Ok, I see now. If, instead, I build the isomorphism $phi:Ptomathbb{Q}$ do I still need to show that this moprhism is well-defined?
– Edu W.
Sep 11 '16 at 14:29
add a comment |
I was attempting to prove that the prime field $P$ of a characteristic 0 field $K$ is isomorphic to $mathbb{Q}$. Here, I will use the notation
$nx=left{begin{array}{cc}underbrace{x+cdots+x}_{n text{ times}} & text{if }n>0\ 0 & text{if } n=0 \ underbrace{(-x)+cdots+(-x)}_{n text{ times}} & text{if }n<0 end{array}right.$
Firstly, I showed that $P={(m1)cdot(n1)^{-1}; m,ninmathbb{Z},nneq 0}$, where 1 is the mulitplicative identity of $K$. Then, I showed that $phi:mathbb{Q}to P$ such that $dfrac{m}{n}mapsto(m1)cdot(n1)^{-1}$ is the isomorphism that I needed to build.
I accessed the ProofWiki to check my answer and there I found this page with the proof of the same thing but, in their proof, they show that $phi$ is well-defined, i.e., given $dfrac{a}{b}=dfrac{c}{d}$, we have that $phileft(dfrac{a}{b}right)=phileft(dfrac{c}{d}right)$. Why is it necessary to show that $phi$ is well-defined?
abstract-algebra proof-explanation
asked Sep 11 '16 at 13:45
Edu W.
155
I was attempting to prove that the prime field $P$ of a characteristic 0 field $K$ is isomorphic to $mathbb{Q}$. Here, I will use the notation
$nx=left{begin{array}{cc}underbrace{x+cdots+x}_{n text{ times}} & text{if }n>0\ 0 & text{if } n=0 \ underbrace{(-x)+cdots+(-x)}_{n text{ times}} & text{if }n<0 end{array}right.$
Firstly, I showed that $P={(m1)cdot(n1)^{-1}; m,ninmathbb{Z},nneq 0}$, where 1 is the mulitplicative identity of $K$. Then, I showed that $phi:mathbb{Q}to P$ such that $dfrac{m}{n}mapsto(m1)cdot(n1)^{-1}$ is the isomorphism that I needed to build.
I accessed the ProofWiki to check my answer and there I found this page with the proof of the same thing but, in their proof, they show that $phi$ is well-defined, i.e., given $dfrac{a}{b}=dfrac{c}{d}$, we have that $phileft(dfrac{a}{b}right)=phileft(dfrac{c}{d}right)$. Why is it necessary to show that $phi$ is well-defined?
abstract-algebra proof-explanation
abstract-algebra proof-explanation
asked Sep 11 '16 at 13:45
Edu W.
155
asked Sep 11 '16 at 13:45
Edu W.
155
edited Feb 23 '17 at 15:52
asked Sep 11 '16 at 13:45
Edu W.
155
asked Sep 11 '16 at 13:45
Edu W.
155
asked Sep 11 '16 at 13:45
Edu W.
155
155
Well, in principle it could be ill-defined. Rational numbers have no unique representation as fractions.
– Crostul
Sep 11 '16 at 13:57
@Crostul Ok, I see now. If, instead, I build the isomorphism $phi:Ptomathbb{Q}$ do I still need to show that this moprhism is well-defined?
– Edu W.
Sep 11 '16 at 14:29
add a comment |
Well, in principle it could be ill-defined. Rational numbers have no unique representation as fractions.
– Crostul
Sep 11 '16 at 13:57
@Crostul Ok, I see now. If, instead, I build the isomorphism $phi:Ptomathbb{Q}$ do I still need to show that this moprhism is well-defined?
– Edu W.
Sep 11 '16 at 14:29
Well, in principle it could be ill-defined. Rational numbers have no unique representation as fractions.
– Crostul
Sep 11 '16 at 13:57
Well, in principle it could be ill-defined. Rational numbers have no unique representation as fractions.
– Crostul
Sep 11 '16 at 13:57
@Crostul Ok, I see now. If, instead, I build the isomorphism $phi:Ptomathbb{Q}$ do I still need to show that this moprhism is well-defined?
– Edu W.
Sep 11 '16 at 14:29
@Crostul Ok, I see now. If, instead, I build the isomorphism $phi:Ptomathbb{Q}$ do I still need to show that this moprhism is well-defined?
– Edu W.
Sep 11 '16 at 14:29
add a comment |
1 Answer
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If you define a map $phi : Sto T$, you must always check that $phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $phi$ based on a presentation of an element $xin S$, you must check that if $x'$ is a different presentation of $x$, then $phi(x) = phi(x')$. If $phi$ is to be a function, then $phi(x)$ must have one and only one value. So, since $x = x'$, $phi(x)$ must equal $phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = dots$. So, if you define a map $phi : Bbb Qto S$, and you define $phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $phi : Bbb QtoBbb Z$ given by $phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = phi(8/4) = phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $phi : PtoBbb Q$ gives the same value $phi(x)$ no matter which way you write $x$.
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
add a comment |
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If you define a map $phi : Sto T$, you must always check that $phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $phi$ based on a presentation of an element $xin S$, you must check that if $x'$ is a different presentation of $x$, then $phi(x) = phi(x')$. If $phi$ is to be a function, then $phi(x)$ must have one and only one value. So, since $x = x'$, $phi(x)$ must equal $phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = dots$. So, if you define a map $phi : Bbb Qto S$, and you define $phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $phi : Bbb QtoBbb Z$ given by $phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = phi(8/4) = phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $phi : PtoBbb Q$ gives the same value $phi(x)$ no matter which way you write $x$.
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
add a comment |
If you define a map $phi : Sto T$, you must always check that $phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $phi$ based on a presentation of an element $xin S$, you must check that if $x'$ is a different presentation of $x$, then $phi(x) = phi(x')$. If $phi$ is to be a function, then $phi(x)$ must have one and only one value. So, since $x = x'$, $phi(x)$ must equal $phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = dots$. So, if you define a map $phi : Bbb Qto S$, and you define $phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $phi : Bbb QtoBbb Z$ given by $phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = phi(8/4) = phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $phi : PtoBbb Q$ gives the same value $phi(x)$ no matter which way you write $x$.
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
add a comment |
If you define a map $phi : Sto T$, you must always check that $phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $phi$ based on a presentation of an element $xin S$, you must check that if $x'$ is a different presentation of $x$, then $phi(x) = phi(x')$. If $phi$ is to be a function, then $phi(x)$ must have one and only one value. So, since $x = x'$, $phi(x)$ must equal $phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = dots$. So, if you define a map $phi : Bbb Qto S$, and you define $phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $phi : Bbb QtoBbb Z$ given by $phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = phi(8/4) = phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $phi : PtoBbb Q$ gives the same value $phi(x)$ no matter which way you write $x$.
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
If you define a map $phi : Sto T$, you must always check that $phi$ is well defined; if it isn't well defined, it's not a function at all! More accurately, whenever you define $phi$ based on a presentation of an element $xin S$, you must check that if $x'$ is a different presentation of $x$, then $phi(x) = phi(x')$. If $phi$ is to be a function, then $phi(x)$ must have one and only one value. So, since $x = x'$, $phi(x)$ must equal $phi(x')$.
In this case, a rational number may be written in more than one way: $1/2 = 2/4 = 4/8 = dots$. So, if you define a map $phi : Bbb Qto S$, and you define $phi(a/b)$ by some formula that you plug $a$ and $b$ into, then you need to check that if $a/b = c/d$, then the formula gives the same values when you put $c$ and $d$ into it instead of $a$ and $b$. As an example: $phi : Bbb QtoBbb Z$ given by $phi(a/b) = a - b$ is not well defined, since it would imply that $4 = 8 - 4 = phi(8/4) = phi(2/1) = 2 - 1 = 1$.
If an element $x$ of your $P$ can be written in more than one way, you must check that your $phi : PtoBbb Q$ gives the same value $phi(x)$ no matter which way you write $x$.
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
edited Nov 26 at 1:20
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
answered Sep 11 '16 at 20:51
Stahl
16.5k43355
16.5k43355
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
add a comment |
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
Ok, now it's clear. Thanks.
– Edu W.
Sep 12 '16 at 2:15
add a comment |
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Well, in principle it could be ill-defined. Rational numbers have no unique representation as fractions.
– Crostul
Sep 11 '16 at 13:57
@Crostul Ok, I see now. If, instead, I build the isomorphism $phi:Ptomathbb{Q}$ do I still need to show that this moprhism is well-defined?
– Edu W.
Sep 11 '16 at 14:29