Mixing
Isomorphism between rationals and polynomials over $mathbb{Z}$
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For each of the following statements, prove it if it is true, and if it is false, give an explanation.
a) $mathbb{Q}^*$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^*=mathbb{Q}-{0}$.
b) $mathbb{Q}^+$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^+$ are positive rationals.
My approach: For the second part of the problem I have defined explicit mapping $$f:mathbb{Z}[x]to mathbb{Q}^+$$ defined by $f(a_1+a_2x+dots+a_{n+1}x^n)=p_1^{a_1}dots p_{n+1}^{a_{n+1}}$ where $p_n$ is $n$-th prime number.
I was able to show that this mapping is bijective homomorphism, i.e. isomorphism.
But I am not able to come up with something with the first part of the problem.
Would be very grateful for help.
P.S. More general question: How to solve such type of problems? Is there some technique?
P.P.S Also the second part of the problem was proven by constructing explicit isomorphism. Is it possible to prove it without counstructing mapping?
group-theory
asked 2 days ago
ZFR
4,88831337
add a comment |
up vote
1
down vote
favorite
For each of the following statements, prove it if it is true, and if it is false, give an explanation.
a) $mathbb{Q}^*$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^*=mathbb{Q}-{0}$.
b) $mathbb{Q}^+$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^+$ are positive rationals.
My approach: For the second part of the problem I have defined explicit mapping $$f:mathbb{Z}[x]to mathbb{Q}^+$$ defined by $f(a_1+a_2x+dots+a_{n+1}x^n)=p_1^{a_1}dots p_{n+1}^{a_{n+1}}$ where $p_n$ is $n$-th prime number.
I was able to show that this mapping is bijective homomorphism, i.e. isomorphism.
But I am not able to come up with something with the first part of the problem.
Would be very grateful for help.
P.S. More general question: How to solve such type of problems? Is there some technique?
P.P.S Also the second part of the problem was proven by constructing explicit isomorphism. Is it possible to prove it without counstructing mapping?
group-theory
asked 2 days ago
ZFR
4,88831337
HINT: $-1$ has finite order in $mathbb{Q}^*$. Are there elements of finite order in $mathbb{Z}[x]$?
– Arturo Magidin
2 days ago
1
It is possible to do Part 2 without an explicit map, but you'd need to do some work: prove that $mathbb{Z}[x]$ is a free abelian group of rank $aleph_0$, and then show that $mathbb{Q}^+$ is also a free abelian group of rank $aleph_0$, and conclude that they are isomorphic from that. But showing they are free of that rank will, implicitly construct isomorphisms between them...
– Arturo Magidin
2 days ago
@ArturoMagidin, because in lectures we dealing with free abelian groups. And honestly to say I think that these problems can be proven with the theory of free abelian groups. But i did not digest the theory quite well
– ZFR
2 days ago
yes, they can. You would want to show that $mathbb{Z}[x]$ is free on the set ${1,x,x^2,x^3,ldots}$, and that $mathbb{Q}^+$ is free on the basis ${p_1,p_2,p_3,ldots}$.
– Arturo Magidin
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For each of the following statements, prove it if it is true, and if it is false, give an explanation.
a) $mathbb{Q}^*$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^*=mathbb{Q}-{0}$.
b) $mathbb{Q}^+$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^+$ are positive rationals.
My approach: For the second part of the problem I have defined explicit mapping $$f:mathbb{Z}[x]to mathbb{Q}^+$$ defined by $f(a_1+a_2x+dots+a_{n+1}x^n)=p_1^{a_1}dots p_{n+1}^{a_{n+1}}$ where $p_n$ is $n$-th prime number.
I was able to show that this mapping is bijective homomorphism, i.e. isomorphism.
But I am not able to come up with something with the first part of the problem.
Would be very grateful for help.
P.S. More general question: How to solve such type of problems? Is there some technique?
P.P.S Also the second part of the problem was proven by constructing explicit isomorphism. Is it possible to prove it without counstructing mapping?
group-theory
asked 2 days ago
ZFR
4,88831337
For each of the following statements, prove it if it is true, and if it is false, give an explanation.
a) $mathbb{Q}^*$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^*=mathbb{Q}-{0}$.
b) $mathbb{Q}^+$ under multiplication is isomorphic to $mathbb{Z}[x]$ under $+$, where $mathbb{Q}^+$ are positive rationals.
My approach: For the second part of the problem I have defined explicit mapping $$f:mathbb{Z}[x]to mathbb{Q}^+$$ defined by $f(a_1+a_2x+dots+a_{n+1}x^n)=p_1^{a_1}dots p_{n+1}^{a_{n+1}}$ where $p_n$ is $n$-th prime number.
I was able to show that this mapping is bijective homomorphism, i.e. isomorphism.
But I am not able to come up with something with the first part of the problem.
Would be very grateful for help.
P.S. More general question: How to solve such type of problems? Is there some technique?
P.P.S Also the second part of the problem was proven by constructing explicit isomorphism. Is it possible to prove it without counstructing mapping?
group-theory
group-theory
asked 2 days ago
ZFR
4,88831337
asked 2 days ago
ZFR
4,88831337
asked 2 days ago
ZFR
4,88831337
asked 2 days ago
ZFR
4,88831337
asked 2 days ago
ZFR
4,88831337
4,88831337
HINT: $-1$ has finite order in $mathbb{Q}^*$. Are there elements of finite order in $mathbb{Z}[x]$?
– Arturo Magidin
2 days ago
1
It is possible to do Part 2 without an explicit map, but you'd need to do some work: prove that $mathbb{Z}[x]$ is a free abelian group of rank $aleph_0$, and then show that $mathbb{Q}^+$ is also a free abelian group of rank $aleph_0$, and conclude that they are isomorphic from that. But showing they are free of that rank will, implicitly construct isomorphisms between them...
– Arturo Magidin
2 days ago
@ArturoMagidin, because in lectures we dealing with free abelian groups. And honestly to say I think that these problems can be proven with the theory of free abelian groups. But i did not digest the theory quite well
– ZFR
2 days ago
yes, they can. You would want to show that $mathbb{Z}[x]$ is free on the set ${1,x,x^2,x^3,ldots}$, and that $mathbb{Q}^+$ is free on the basis ${p_1,p_2,p_3,ldots}$.
– Arturo Magidin
2 days ago
add a comment |
HINT: $-1$ has finite order in $mathbb{Q}^*$. Are there elements of finite order in $mathbb{Z}[x]$?
– Arturo Magidin
2 days ago
1
It is possible to do Part 2 without an explicit map, but you'd need to do some work: prove that $mathbb{Z}[x]$ is a free abelian group of rank $aleph_0$, and then show that $mathbb{Q}^+$ is also a free abelian group of rank $aleph_0$, and conclude that they are isomorphic from that. But showing they are free of that rank will, implicitly construct isomorphisms between them...
– Arturo Magidin
2 days ago
@ArturoMagidin, because in lectures we dealing with free abelian groups. And honestly to say I think that these problems can be proven with the theory of free abelian groups. But i did not digest the theory quite well
– ZFR
2 days ago
yes, they can. You would want to show that $mathbb{Z}[x]$ is free on the set ${1,x,x^2,x^3,ldots}$, and that $mathbb{Q}^+$ is free on the basis ${p_1,p_2,p_3,ldots}$.
– Arturo Magidin
2 days ago
HINT: $-1$ has finite order in $mathbb{Q}^*$. Are there elements of finite order in $mathbb{Z}[x]$?
– Arturo Magidin
2 days ago
HINT: $-1$ has finite order in $mathbb{Q}^*$. Are there elements of finite order in $mathbb{Z}[x]$?
– Arturo Magidin
2 days ago
1
1
It is possible to do Part 2 without an explicit map, but you'd need to do some work: prove that $mathbb{Z}[x]$ is a free abelian group of rank $aleph_0$, and then show that $mathbb{Q}^+$ is also a free abelian group of rank $aleph_0$, and conclude that they are isomorphic from that. But showing they are free of that rank will, implicitly construct isomorphisms between them...
– Arturo Magidin
2 days ago
It is possible to do Part 2 without an explicit map, but you'd need to do some work: prove that $mathbb{Z}[x]$ is a free abelian group of rank $aleph_0$, and then show that $mathbb{Q}^+$ is also a free abelian group of rank $aleph_0$, and conclude that they are isomorphic from that. But showing they are free of that rank will, implicitly construct isomorphisms between them...
– Arturo Magidin
2 days ago
@ArturoMagidin, because in lectures we dealing with free abelian groups. And honestly to say I think that these problems can be proven with the theory of free abelian groups. But i did not digest the theory quite well
– ZFR
2 days ago
@ArturoMagidin, because in lectures we dealing with free abelian groups. And honestly to say I think that these problems can be proven with the theory of free abelian groups. But i did not digest the theory quite well
– ZFR
2 days ago
yes, they can. You would want to show that $mathbb{Z}[x]$ is free on the set ${1,x,x^2,x^3,ldots}$, and that $mathbb{Q}^+$ is free on the basis ${p_1,p_2,p_3,ldots}$.
– Arturo Magidin
2 days ago
yes, they can. You would want to show that $mathbb{Z}[x]$ is free on the set ${1,x,x^2,x^3,ldots}$, and that $mathbb{Q}^+$ is free on the basis ${p_1,p_2,p_3,ldots}$.
– Arturo Magidin
2 days ago
add a comment |
1 Answer
1
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2
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You've done a great job with the b part! In general, constructing an isomorphism is the cleanest way to show it exists. To prove that no isomorphism exists, you can show that one group possesses some property that the other does not.
For the first question, try to find an element in $mathbb Q^*$ of order 2 ($xneq 1$ such that $x^2=1$), and show that no element of $mathbb Z[x]$ under addition has order 2 (i.e. that no non-zero polynomial $p inmathbb Z[x]$ satisfies $p+p=0$).
answered 2 days ago
lisyarus
10.3k21433
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You've done a great job with the b part! In general, constructing an isomorphism is the cleanest way to show it exists. To prove that no isomorphism exists, you can show that one group possesses some property that the other does not.
For the first question, try to find an element in $mathbb Q^*$ of order 2 ($xneq 1$ such that $x^2=1$), and show that no element of $mathbb Z[x]$ under addition has order 2 (i.e. that no non-zero polynomial $p inmathbb Z[x]$ satisfies $p+p=0$).
answered 2 days ago
lisyarus
10.3k21433
add a comment |
up vote
2
down vote
You've done a great job with the b part! In general, constructing an isomorphism is the cleanest way to show it exists. To prove that no isomorphism exists, you can show that one group possesses some property that the other does not.
For the first question, try to find an element in $mathbb Q^*$ of order 2 ($xneq 1$ such that $x^2=1$), and show that no element of $mathbb Z[x]$ under addition has order 2 (i.e. that no non-zero polynomial $p inmathbb Z[x]$ satisfies $p+p=0$).
answered 2 days ago
lisyarus
10.3k21433
add a comment |
up vote
2
down vote
up vote
2
down vote
You've done a great job with the b part! In general, constructing an isomorphism is the cleanest way to show it exists. To prove that no isomorphism exists, you can show that one group possesses some property that the other does not.
For the first question, try to find an element in $mathbb Q^*$ of order 2 ($xneq 1$ such that $x^2=1$), and show that no element of $mathbb Z[x]$ under addition has order 2 (i.e. that no non-zero polynomial $p inmathbb Z[x]$ satisfies $p+p=0$).
answered 2 days ago
lisyarus
10.3k21433
You've done a great job with the b part! In general, constructing an isomorphism is the cleanest way to show it exists. To prove that no isomorphism exists, you can show that one group possesses some property that the other does not.
For the first question, try to find an element in $mathbb Q^*$ of order 2 ($xneq 1$ such that $x^2=1$), and show that no element of $mathbb Z[x]$ under addition has order 2 (i.e. that no non-zero polynomial $p inmathbb Z[x]$ satisfies $p+p=0$).
answered 2 days ago
lisyarus
10.3k21433
answered 2 days ago
lisyarus
10.3k21433
answered 2 days ago
lisyarus
10.3k21433
answered 2 days ago
lisyarus
10.3k21433
10.3k21433
add a comment |
add a comment |
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HINT: $-1$ has finite order in $mathbb{Q}^*$. Are there elements of finite order in $mathbb{Z}[x]$?
– Arturo Magidin
2 days ago
1
It is possible to do Part 2 without an explicit map, but you'd need to do some work: prove that $mathbb{Z}[x]$ is a free abelian group of rank $aleph_0$, and then show that $mathbb{Q}^+$ is also a free abelian group of rank $aleph_0$, and conclude that they are isomorphic from that. But showing they are free of that rank will, implicitly construct isomorphisms between them...
– Arturo Magidin
2 days ago
@ArturoMagidin, because in lectures we dealing with free abelian groups. And honestly to say I think that these problems can be proven with the theory of free abelian groups. But i did not digest the theory quite well
– ZFR
2 days ago
yes, they can. You would want to show that $mathbb{Z}[x]$ is free on the set ${1,x,x^2,x^3,ldots}$, and that $mathbb{Q}^+$ is free on the basis ${p_1,p_2,p_3,ldots}$.
– Arturo Magidin
2 days ago