Mixing
There doesn't exists an epimorphism from $(mathbb{R}, +)$ to $(mathbb{Q}, +)$ proof verification. [duplicate]
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This question already has an answer here:
Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?
2 answers
Suppose that such epimorphism exists, and call it $f$.
For fixed $xinmathbb{R}$, take $N = text{Ker}(f)$ and $S = xmathbb{Q}$.
Apply the second theorem of isomorphisms, that is:
$$(SN)/Ncong S/(Scap N) $$
But $(SN)/N = {text{Ker}(f)}$, so that $S/(Scap N) = { x+Scap N: xin S} = { Scap N }$, hence $S = Scap N$. This means that $Ssubset N$, from which follows that $$mathbb{R} = text{Ker}(f)$$So that $f(x) = 0$ for all $xin mathbb{R}$, hence $f$ is not an epimorphism, and contradiction proofs that such epimorphism cannot exists.
Is this proof correct? Did I miss anything? Can it be done easier? Thank you.
Edit: Sorry, I see now that this proof isn't correct. Clearly, $NS$ doesn't have to equal $N$, which I thought is true because of additive and multiplicative notation confusion.
group-theory proof-verification
asked Jan 18 at 17:40
JakobianJakobian
2,650721
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marked as duplicate by Jakobian, Community♦ Jan 18 at 20:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?
2 answers
Suppose that such epimorphism exists, and call it $f$.
For fixed $xinmathbb{R}$, take $N = text{Ker}(f)$ and $S = xmathbb{Q}$.
Apply the second theorem of isomorphisms, that is:
$$(SN)/Ncong S/(Scap N) $$
But $(SN)/N = {text{Ker}(f)}$, so that $S/(Scap N) = { x+Scap N: xin S} = { Scap N }$, hence $S = Scap N$. This means that $Ssubset N$, from which follows that $$mathbb{R} = text{Ker}(f)$$So that $f(x) = 0$ for all $xin mathbb{R}$, hence $f$ is not an epimorphism, and contradiction proofs that such epimorphism cannot exists.
Is this proof correct? Did I miss anything? Can it be done easier? Thank you.
Edit: Sorry, I see now that this proof isn't correct. Clearly, $NS$ doesn't have to equal $N$, which I thought is true because of additive and multiplicative notation confusion.
group-theory proof-verification
asked Jan 18 at 17:40
JakobianJakobian
2,650721
$endgroup$
marked as duplicate by Jakobian, Community♦ Jan 18 at 20:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You seem to be mixing multiplicative and additive notation. Both groups in the headers are additive abelian groups, thus...what is $;S=xBbb Q;$ ? Is it really all the rational products of $;x;$ ? But then why $;SN;$ and not $;S+N;$ ? All this is confusing...
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– DonAntonio
Jan 18 at 17:54
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@DonAntonio: note that ${qxmid qinmathbb{Q}}$ is an additive subgroup of $mathbb{R}$.
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– Arturo Magidin
Jan 18 at 18:05
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@Jakobian: I cannot figure out why you conclude that $(SN)/N$ is a single element. That holds if and only if $xin N$, which is your desired conclusion, so it looks to me as if you are assuming what you are trying to prove. How do you justify that $(SN)/N$ consists only of $N$?
$endgroup$
– Arturo Magidin
Jan 18 at 18:06
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It's all because I've mixed up additive and multiplicative notation. Sorry for that, I was sligthly confused myself.
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– Jakobian
Jan 18 at 18:39
$begingroup$
@ArturoMagidin I know that, yet the the notation $;SN;$ seems off for additively written groups. That's why I asked.
$endgroup$
– DonAntonio
Jan 18 at 22:14
add a comment |
$begingroup$
This question already has an answer here:
Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?
2 answers
Suppose that such epimorphism exists, and call it $f$.
For fixed $xinmathbb{R}$, take $N = text{Ker}(f)$ and $S = xmathbb{Q}$.
Apply the second theorem of isomorphisms, that is:
$$(SN)/Ncong S/(Scap N) $$
But $(SN)/N = {text{Ker}(f)}$, so that $S/(Scap N) = { x+Scap N: xin S} = { Scap N }$, hence $S = Scap N$. This means that $Ssubset N$, from which follows that $$mathbb{R} = text{Ker}(f)$$So that $f(x) = 0$ for all $xin mathbb{R}$, hence $f$ is not an epimorphism, and contradiction proofs that such epimorphism cannot exists.
Is this proof correct? Did I miss anything? Can it be done easier? Thank you.
Edit: Sorry, I see now that this proof isn't correct. Clearly, $NS$ doesn't have to equal $N$, which I thought is true because of additive and multiplicative notation confusion.
group-theory proof-verification
asked Jan 18 at 17:40
JakobianJakobian
2,650721
$endgroup$
This question already has an answer here:
Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?
2 answers
Suppose that such epimorphism exists, and call it $f$.
For fixed $xinmathbb{R}$, take $N = text{Ker}(f)$ and $S = xmathbb{Q}$.
Apply the second theorem of isomorphisms, that is:
$$(SN)/Ncong S/(Scap N) $$
But $(SN)/N = {text{Ker}(f)}$, so that $S/(Scap N) = { x+Scap N: xin S} = { Scap N }$, hence $S = Scap N$. This means that $Ssubset N$, from which follows that $$mathbb{R} = text{Ker}(f)$$So that $f(x) = 0$ for all $xin mathbb{R}$, hence $f$ is not an epimorphism, and contradiction proofs that such epimorphism cannot exists.
Is this proof correct? Did I miss anything? Can it be done easier? Thank you.
Edit: Sorry, I see now that this proof isn't correct. Clearly, $NS$ doesn't have to equal $N$, which I thought is true because of additive and multiplicative notation confusion.
This question already has an answer here:
Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?
2 answers
group-theory proof-verification
group-theory proof-verification
asked Jan 18 at 17:40
JakobianJakobian
2,650721
asked Jan 18 at 17:40
JakobianJakobian
2,650721
edited Jan 18 at 17:50
Jakobian
asked Jan 18 at 17:40
JakobianJakobian
2,650721
asked Jan 18 at 17:40
JakobianJakobian
2,650721
asked Jan 18 at 17:40
JakobianJakobian
2,650721
2,650721
marked as duplicate by Jakobian, Community♦ Jan 18 at 20:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jakobian, Community♦ Jan 18 at 20:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You seem to be mixing multiplicative and additive notation. Both groups in the headers are additive abelian groups, thus...what is $;S=xBbb Q;$ ? Is it really all the rational products of $;x;$ ? But then why $;SN;$ and not $;S+N;$ ? All this is confusing...
$endgroup$
– DonAntonio
Jan 18 at 17:54
$begingroup$
@DonAntonio: note that ${qxmid qinmathbb{Q}}$ is an additive subgroup of $mathbb{R}$.
$endgroup$
– Arturo Magidin
Jan 18 at 18:05
$begingroup$
@Jakobian: I cannot figure out why you conclude that $(SN)/N$ is a single element. That holds if and only if $xin N$, which is your desired conclusion, so it looks to me as if you are assuming what you are trying to prove. How do you justify that $(SN)/N$ consists only of $N$?
$endgroup$
– Arturo Magidin
Jan 18 at 18:06
$begingroup$
It's all because I've mixed up additive and multiplicative notation. Sorry for that, I was sligthly confused myself.
$endgroup$
– Jakobian
Jan 18 at 18:39
$begingroup$
@ArturoMagidin I know that, yet the the notation $;SN;$ seems off for additively written groups. That's why I asked.
$endgroup$
– DonAntonio
Jan 18 at 22:14
add a comment |
$begingroup$
You seem to be mixing multiplicative and additive notation. Both groups in the headers are additive abelian groups, thus...what is $;S=xBbb Q;$ ? Is it really all the rational products of $;x;$ ? But then why $;SN;$ and not $;S+N;$ ? All this is confusing...
$endgroup$
– DonAntonio
Jan 18 at 17:54
$begingroup$
@DonAntonio: note that ${qxmid qinmathbb{Q}}$ is an additive subgroup of $mathbb{R}$.
$endgroup$
– Arturo Magidin
Jan 18 at 18:05
$begingroup$
@Jakobian: I cannot figure out why you conclude that $(SN)/N$ is a single element. That holds if and only if $xin N$, which is your desired conclusion, so it looks to me as if you are assuming what you are trying to prove. How do you justify that $(SN)/N$ consists only of $N$?
$endgroup$
– Arturo Magidin
Jan 18 at 18:06
$begingroup$
It's all because I've mixed up additive and multiplicative notation. Sorry for that, I was sligthly confused myself.
$endgroup$
– Jakobian
Jan 18 at 18:39
$begingroup$
@ArturoMagidin I know that, yet the the notation $;SN;$ seems off for additively written groups. That's why I asked.
$endgroup$
– DonAntonio
Jan 18 at 22:14
$begingroup$
You seem to be mixing multiplicative and additive notation. Both groups in the headers are additive abelian groups, thus...what is $;S=xBbb Q;$ ? Is it really all the rational products of $;x;$ ? But then why $;SN;$ and not $;S+N;$ ? All this is confusing...
$endgroup$
– DonAntonio
Jan 18 at 17:54
$begingroup$
You seem to be mixing multiplicative and additive notation. Both groups in the headers are additive abelian groups, thus...what is $;S=xBbb Q;$ ? Is it really all the rational products of $;x;$ ? But then why $;SN;$ and not $;S+N;$ ? All this is confusing...
$endgroup$
– DonAntonio
Jan 18 at 17:54
$begingroup$
@DonAntonio: note that ${qxmid qinmathbb{Q}}$ is an additive subgroup of $mathbb{R}$.
$endgroup$
– Arturo Magidin
Jan 18 at 18:05
$begingroup$
@DonAntonio: note that ${qxmid qinmathbb{Q}}$ is an additive subgroup of $mathbb{R}$.
$endgroup$
– Arturo Magidin
Jan 18 at 18:05
$begingroup$
@Jakobian: I cannot figure out why you conclude that $(SN)/N$ is a single element. That holds if and only if $xin N$, which is your desired conclusion, so it looks to me as if you are assuming what you are trying to prove. How do you justify that $(SN)/N$ consists only of $N$?
$endgroup$
– Arturo Magidin
Jan 18 at 18:06
$begingroup$
@Jakobian: I cannot figure out why you conclude that $(SN)/N$ is a single element. That holds if and only if $xin N$, which is your desired conclusion, so it looks to me as if you are assuming what you are trying to prove. How do you justify that $(SN)/N$ consists only of $N$?
$endgroup$
– Arturo Magidin
Jan 18 at 18:06
$begingroup$
It's all because I've mixed up additive and multiplicative notation. Sorry for that, I was sligthly confused myself.
$endgroup$
– Jakobian
Jan 18 at 18:39
$begingroup$
It's all because I've mixed up additive and multiplicative notation. Sorry for that, I was sligthly confused myself.
$endgroup$
– Jakobian
Jan 18 at 18:39
$begingroup$
@ArturoMagidin I know that, yet the the notation $;SN;$ seems off for additively written groups. That's why I asked.
$endgroup$
– DonAntonio
Jan 18 at 22:14
$begingroup$
@ArturoMagidin I know that, yet the the notation $;SN;$ seems off for additively written groups. That's why I asked.
$endgroup$
– DonAntonio
Jan 18 at 22:14
add a comment |
1 Answer
1
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$begingroup$
There is an epi $(mathbb R,+) rightarrow (mathbb Q ,+)$. See this question, especially the answer by N.S.
answered Jan 18 at 17:57
o.h.o.h.
4616
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$begingroup$
Thank you. I was convinced it doesn't exists.
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– Jakobian
Jan 18 at 18:40
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No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is an epi $(mathbb R,+) rightarrow (mathbb Q ,+)$. See this question, especially the answer by N.S.
answered Jan 18 at 17:57
o.h.o.h.
4616
$endgroup$
$begingroup$
Thank you. I was convinced it doesn't exists.
$endgroup$
– Jakobian
Jan 18 at 18:40
$begingroup$
No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
add a comment |
$begingroup$
There is an epi $(mathbb R,+) rightarrow (mathbb Q ,+)$. See this question, especially the answer by N.S.
answered Jan 18 at 17:57
o.h.o.h.
4616
$endgroup$
$begingroup$
Thank you. I was convinced it doesn't exists.
$endgroup$
– Jakobian
Jan 18 at 18:40
$begingroup$
No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
add a comment |
$begingroup$
There is an epi $(mathbb R,+) rightarrow (mathbb Q ,+)$. See this question, especially the answer by N.S.
answered Jan 18 at 17:57
o.h.o.h.
4616
$endgroup$
There is an epi $(mathbb R,+) rightarrow (mathbb Q ,+)$. See this question, especially the answer by N.S.
answered Jan 18 at 17:57
o.h.o.h.
4616
answered Jan 18 at 17:57
o.h.o.h.
4616
answered Jan 18 at 17:57
o.h.o.h.
4616
answered Jan 18 at 17:57
o.h.o.h.
4616
4616
$begingroup$
Thank you. I was convinced it doesn't exists.
$endgroup$
– Jakobian
Jan 18 at 18:40
$begingroup$
No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
add a comment |
$begingroup$
Thank you. I was convinced it doesn't exists.
$endgroup$
– Jakobian
Jan 18 at 18:40
$begingroup$
No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
$begingroup$
Thank you. I was convinced it doesn't exists.
$endgroup$
– Jakobian
Jan 18 at 18:40
$begingroup$
Thank you. I was convinced it doesn't exists.
$endgroup$
– Jakobian
Jan 18 at 18:40
$begingroup$
No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
$begingroup$
No problem. So was I when I at first glance :)
$endgroup$
– o.h.
Jan 18 at 18:51
add a comment |
$begingroup$
You seem to be mixing multiplicative and additive notation. Both groups in the headers are additive abelian groups, thus...what is $;S=xBbb Q;$ ? Is it really all the rational products of $;x;$ ? But then why $;SN;$ and not $;S+N;$ ? All this is confusing...
$endgroup$
– DonAntonio
Jan 18 at 17:54
$begingroup$
@DonAntonio: note that ${qxmid qinmathbb{Q}}$ is an additive subgroup of $mathbb{R}$.
$endgroup$
– Arturo Magidin
Jan 18 at 18:05
$begingroup$
@Jakobian: I cannot figure out why you conclude that $(SN)/N$ is a single element. That holds if and only if $xin N$, which is your desired conclusion, so it looks to me as if you are assuming what you are trying to prove. How do you justify that $(SN)/N$ consists only of $N$?
$endgroup$
– Arturo Magidin
Jan 18 at 18:06
$begingroup$
It's all because I've mixed up additive and multiplicative notation. Sorry for that, I was sligthly confused myself.
$endgroup$
– Jakobian
Jan 18 at 18:39
$begingroup$
@ArturoMagidin I know that, yet the the notation $;SN;$ seems off for additively written groups. That's why I asked.
$endgroup$
– DonAntonio
Jan 18 at 22:14