Determine all homomorphisms from $Q$ to $Q_{>0}^times$.











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This question is from a past year paper: Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



Let $Q$ denote the group of rationals under addition.
Let $Q_{>0}^times$ denote the group of positive rationals under multiplication.



(a) Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



(b) Determine all homomorphisms from$Q_{>0}^times$ to $Q$.



For part (a), I know that any homomorphism $f:Q to Q_{>0}^times$ is determined by $f(1)$, since any other $f(frac{m}{n}) = f(1)^{frac{m}{n}}$. Since the image of f is in rationals, then $f(1)$ must be $1$, otherwise we can find $frac{m}{n}$ such that $f(frac{m}{n})$ is irrational. Is this correct?



(b) I am not sure how to proceed for this part. I think $f$ would be determined by how it acts on this set: ${ p mid text{$p$ is prime} } cup { frac{1}{p} mid text{$p$ is prime} }$.










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  • 1




    Take a look at math.stackexchange.com/questions/229140/…
    – Anurag A
    Nov 17 at 3:11










  • Nice, thanks for the link!
    – eatfood
    Nov 17 at 6:42















up vote
1
down vote

favorite












This question is from a past year paper: Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



Let $Q$ denote the group of rationals under addition.
Let $Q_{>0}^times$ denote the group of positive rationals under multiplication.



(a) Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



(b) Determine all homomorphisms from$Q_{>0}^times$ to $Q$.



For part (a), I know that any homomorphism $f:Q to Q_{>0}^times$ is determined by $f(1)$, since any other $f(frac{m}{n}) = f(1)^{frac{m}{n}}$. Since the image of f is in rationals, then $f(1)$ must be $1$, otherwise we can find $frac{m}{n}$ such that $f(frac{m}{n})$ is irrational. Is this correct?



(b) I am not sure how to proceed for this part. I think $f$ would be determined by how it acts on this set: ${ p mid text{$p$ is prime} } cup { frac{1}{p} mid text{$p$ is prime} }$.










share|cite|improve this question


















  • 1




    Take a look at math.stackexchange.com/questions/229140/…
    – Anurag A
    Nov 17 at 3:11










  • Nice, thanks for the link!
    – eatfood
    Nov 17 at 6:42













up vote
1
down vote

favorite









up vote
1
down vote

favorite











This question is from a past year paper: Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



Let $Q$ denote the group of rationals under addition.
Let $Q_{>0}^times$ denote the group of positive rationals under multiplication.



(a) Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



(b) Determine all homomorphisms from$Q_{>0}^times$ to $Q$.



For part (a), I know that any homomorphism $f:Q to Q_{>0}^times$ is determined by $f(1)$, since any other $f(frac{m}{n}) = f(1)^{frac{m}{n}}$. Since the image of f is in rationals, then $f(1)$ must be $1$, otherwise we can find $frac{m}{n}$ such that $f(frac{m}{n})$ is irrational. Is this correct?



(b) I am not sure how to proceed for this part. I think $f$ would be determined by how it acts on this set: ${ p mid text{$p$ is prime} } cup { frac{1}{p} mid text{$p$ is prime} }$.










share|cite|improve this question













This question is from a past year paper: Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



Let $Q$ denote the group of rationals under addition.
Let $Q_{>0}^times$ denote the group of positive rationals under multiplication.



(a) Determine all homomorphisms from $Q$ to $Q_{>0}^times$.



(b) Determine all homomorphisms from$Q_{>0}^times$ to $Q$.



For part (a), I know that any homomorphism $f:Q to Q_{>0}^times$ is determined by $f(1)$, since any other $f(frac{m}{n}) = f(1)^{frac{m}{n}}$. Since the image of f is in rationals, then $f(1)$ must be $1$, otherwise we can find $frac{m}{n}$ such that $f(frac{m}{n})$ is irrational. Is this correct?



(b) I am not sure how to proceed for this part. I think $f$ would be determined by how it acts on this set: ${ p mid text{$p$ is prime} } cup { frac{1}{p} mid text{$p$ is prime} }$.







group-theory






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asked Nov 17 at 3:06









eatfood

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1827








  • 1




    Take a look at math.stackexchange.com/questions/229140/…
    – Anurag A
    Nov 17 at 3:11










  • Nice, thanks for the link!
    – eatfood
    Nov 17 at 6:42














  • 1




    Take a look at math.stackexchange.com/questions/229140/…
    – Anurag A
    Nov 17 at 3:11










  • Nice, thanks for the link!
    – eatfood
    Nov 17 at 6:42








1




1




Take a look at math.stackexchange.com/questions/229140/…
– Anurag A
Nov 17 at 3:11




Take a look at math.stackexchange.com/questions/229140/…
– Anurag A
Nov 17 at 3:11












Nice, thanks for the link!
– eatfood
Nov 17 at 6:42




Nice, thanks for the link!
– eatfood
Nov 17 at 6:42










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










The reasoning for the first question is correct : for $a^{frac mn}$ to remain rational for all $m,n$, we must have $a = 1$. Hence any such homomorphism is trivial.



For the other direction, any element of $mathbb Q^+_{>0}$ is of the form $2^{n_1}3^{n_2}5^{n_3}7^{n_4}...$ where $n_i$ is an eventually zero sequence of integers. Therefore, $mathbb Q^+_{>0}$ is isomorphic to the group of "eventually zero integer sequences" under componentwise addition, under this isomorphism sending such a number to the corresponding power sequence.



Now, the sequence of eventually constant integer sequences has an integral basis given by $t_i$, where $t_i$ is the sequence having $1$ at the ith position and $0$ elsewhere. Every element is a finite integer linear combination of the $t_i$. Therefore, specifying a homomorphism to $mathbb Q$ is as good as specifying what it does on these $t_i$.



But this is easy : pick any sequence of rationals $q_i in mathbb Q$ and map $t_i to q_i$. This extends to a homomorphism via the map $sum s_it_i to sum s_iq_i$, where $s_i$ is an eventually zero sequence of integers, ensuring both sides are finite summations.





Via the identification of $mathbb Q^+_{>0}$ with the space of eventually zero integer sequences, this leads to :
$$
phi (2^{n_1}3^{n_2}5^{n_3}...) to sum q_in_i
$$



For any sequence of rationals $q_i$. Conversely, any homomorphism must be of this form, since the individual prime powers must map somewhere.



You can try to find conditions on $q_i$ which make this map injective/surjective.






share|cite|improve this answer





















  • Thank you for the detailed answer! this is what i need.
    – eatfood
    Nov 17 at 6:42










  • But of course, you are welcome!
    – астон вілла олоф мэллбэрг
    Nov 17 at 6:43











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1 Answer
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active

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1 Answer
1






active

oldest

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oldest

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oldest

votes








up vote
2
down vote



accepted










The reasoning for the first question is correct : for $a^{frac mn}$ to remain rational for all $m,n$, we must have $a = 1$. Hence any such homomorphism is trivial.



For the other direction, any element of $mathbb Q^+_{>0}$ is of the form $2^{n_1}3^{n_2}5^{n_3}7^{n_4}...$ where $n_i$ is an eventually zero sequence of integers. Therefore, $mathbb Q^+_{>0}$ is isomorphic to the group of "eventually zero integer sequences" under componentwise addition, under this isomorphism sending such a number to the corresponding power sequence.



Now, the sequence of eventually constant integer sequences has an integral basis given by $t_i$, where $t_i$ is the sequence having $1$ at the ith position and $0$ elsewhere. Every element is a finite integer linear combination of the $t_i$. Therefore, specifying a homomorphism to $mathbb Q$ is as good as specifying what it does on these $t_i$.



But this is easy : pick any sequence of rationals $q_i in mathbb Q$ and map $t_i to q_i$. This extends to a homomorphism via the map $sum s_it_i to sum s_iq_i$, where $s_i$ is an eventually zero sequence of integers, ensuring both sides are finite summations.





Via the identification of $mathbb Q^+_{>0}$ with the space of eventually zero integer sequences, this leads to :
$$
phi (2^{n_1}3^{n_2}5^{n_3}...) to sum q_in_i
$$



For any sequence of rationals $q_i$. Conversely, any homomorphism must be of this form, since the individual prime powers must map somewhere.



You can try to find conditions on $q_i$ which make this map injective/surjective.






share|cite|improve this answer





















  • Thank you for the detailed answer! this is what i need.
    – eatfood
    Nov 17 at 6:42










  • But of course, you are welcome!
    – астон вілла олоф мэллбэрг
    Nov 17 at 6:43















up vote
2
down vote



accepted










The reasoning for the first question is correct : for $a^{frac mn}$ to remain rational for all $m,n$, we must have $a = 1$. Hence any such homomorphism is trivial.



For the other direction, any element of $mathbb Q^+_{>0}$ is of the form $2^{n_1}3^{n_2}5^{n_3}7^{n_4}...$ where $n_i$ is an eventually zero sequence of integers. Therefore, $mathbb Q^+_{>0}$ is isomorphic to the group of "eventually zero integer sequences" under componentwise addition, under this isomorphism sending such a number to the corresponding power sequence.



Now, the sequence of eventually constant integer sequences has an integral basis given by $t_i$, where $t_i$ is the sequence having $1$ at the ith position and $0$ elsewhere. Every element is a finite integer linear combination of the $t_i$. Therefore, specifying a homomorphism to $mathbb Q$ is as good as specifying what it does on these $t_i$.



But this is easy : pick any sequence of rationals $q_i in mathbb Q$ and map $t_i to q_i$. This extends to a homomorphism via the map $sum s_it_i to sum s_iq_i$, where $s_i$ is an eventually zero sequence of integers, ensuring both sides are finite summations.





Via the identification of $mathbb Q^+_{>0}$ with the space of eventually zero integer sequences, this leads to :
$$
phi (2^{n_1}3^{n_2}5^{n_3}...) to sum q_in_i
$$



For any sequence of rationals $q_i$. Conversely, any homomorphism must be of this form, since the individual prime powers must map somewhere.



You can try to find conditions on $q_i$ which make this map injective/surjective.






share|cite|improve this answer





















  • Thank you for the detailed answer! this is what i need.
    – eatfood
    Nov 17 at 6:42










  • But of course, you are welcome!
    – астон вілла олоф мэллбэрг
    Nov 17 at 6:43













up vote
2
down vote



accepted







up vote
2
down vote



accepted






The reasoning for the first question is correct : for $a^{frac mn}$ to remain rational for all $m,n$, we must have $a = 1$. Hence any such homomorphism is trivial.



For the other direction, any element of $mathbb Q^+_{>0}$ is of the form $2^{n_1}3^{n_2}5^{n_3}7^{n_4}...$ where $n_i$ is an eventually zero sequence of integers. Therefore, $mathbb Q^+_{>0}$ is isomorphic to the group of "eventually zero integer sequences" under componentwise addition, under this isomorphism sending such a number to the corresponding power sequence.



Now, the sequence of eventually constant integer sequences has an integral basis given by $t_i$, where $t_i$ is the sequence having $1$ at the ith position and $0$ elsewhere. Every element is a finite integer linear combination of the $t_i$. Therefore, specifying a homomorphism to $mathbb Q$ is as good as specifying what it does on these $t_i$.



But this is easy : pick any sequence of rationals $q_i in mathbb Q$ and map $t_i to q_i$. This extends to a homomorphism via the map $sum s_it_i to sum s_iq_i$, where $s_i$ is an eventually zero sequence of integers, ensuring both sides are finite summations.





Via the identification of $mathbb Q^+_{>0}$ with the space of eventually zero integer sequences, this leads to :
$$
phi (2^{n_1}3^{n_2}5^{n_3}...) to sum q_in_i
$$



For any sequence of rationals $q_i$. Conversely, any homomorphism must be of this form, since the individual prime powers must map somewhere.



You can try to find conditions on $q_i$ which make this map injective/surjective.






share|cite|improve this answer












The reasoning for the first question is correct : for $a^{frac mn}$ to remain rational for all $m,n$, we must have $a = 1$. Hence any such homomorphism is trivial.



For the other direction, any element of $mathbb Q^+_{>0}$ is of the form $2^{n_1}3^{n_2}5^{n_3}7^{n_4}...$ where $n_i$ is an eventually zero sequence of integers. Therefore, $mathbb Q^+_{>0}$ is isomorphic to the group of "eventually zero integer sequences" under componentwise addition, under this isomorphism sending such a number to the corresponding power sequence.



Now, the sequence of eventually constant integer sequences has an integral basis given by $t_i$, where $t_i$ is the sequence having $1$ at the ith position and $0$ elsewhere. Every element is a finite integer linear combination of the $t_i$. Therefore, specifying a homomorphism to $mathbb Q$ is as good as specifying what it does on these $t_i$.



But this is easy : pick any sequence of rationals $q_i in mathbb Q$ and map $t_i to q_i$. This extends to a homomorphism via the map $sum s_it_i to sum s_iq_i$, where $s_i$ is an eventually zero sequence of integers, ensuring both sides are finite summations.





Via the identification of $mathbb Q^+_{>0}$ with the space of eventually zero integer sequences, this leads to :
$$
phi (2^{n_1}3^{n_2}5^{n_3}...) to sum q_in_i
$$



For any sequence of rationals $q_i$. Conversely, any homomorphism must be of this form, since the individual prime powers must map somewhere.



You can try to find conditions on $q_i$ which make this map injective/surjective.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 17 at 4:12









астон вілла олоф мэллбэрг

36.8k33376




36.8k33376












  • Thank you for the detailed answer! this is what i need.
    – eatfood
    Nov 17 at 6:42










  • But of course, you are welcome!
    – астон вілла олоф мэллбэрг
    Nov 17 at 6:43


















  • Thank you for the detailed answer! this is what i need.
    – eatfood
    Nov 17 at 6:42










  • But of course, you are welcome!
    – астон вілла олоф мэллбэрг
    Nov 17 at 6:43
















Thank you for the detailed answer! this is what i need.
– eatfood
Nov 17 at 6:42




Thank you for the detailed answer! this is what i need.
– eatfood
Nov 17 at 6:42












But of course, you are welcome!
– астон вілла олоф мэллбэрг
Nov 17 at 6:43




But of course, you are welcome!
– астон вілла олоф мэллбэрг
Nov 17 at 6:43


















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