Algebra - endomorphisms of field












-1














Find all endomorphisms of $mathbb{Q}$. ($mathbb{Q}$ is the field.)



When finding isomorphism of $2mathbb{Z}$ and $3mathbb{Z}$, we define the mapping like $φ: 2mathbb{Z} → 3mathbb{Z}$ and then we say $φ(2) = 3k, k∈Z$.



How to start in this case of $mathbb{Q}$? Can you just tell me how does the mapping look like? Thanks for your time!






ring-homomorphism







share|cite|improve this question




















  • 1




    It looks just like the identity map.
    – José Carlos Santos
    Dec 6 at 17:45










  • Like φ (k) = k ?
    – Haus
    Dec 6 at 17:46






  • 1




    Yes, just like that.
    – José Carlos Santos
    Dec 6 at 17:47
















-1














Find all endomorphisms of $mathbb{Q}$. ($mathbb{Q}$ is the field.)



When finding isomorphism of $2mathbb{Z}$ and $3mathbb{Z}$, we define the mapping like $φ: 2mathbb{Z} → 3mathbb{Z}$ and then we say $φ(2) = 3k, k∈Z$.



How to start in this case of $mathbb{Q}$? Can you just tell me how does the mapping look like? Thanks for your time!






ring-homomorphism







share|cite|improve this question




















  • 1




    It looks just like the identity map.
    – José Carlos Santos
    Dec 6 at 17:45










  • Like φ (k) = k ?
    – Haus
    Dec 6 at 17:46






  • 1




    Yes, just like that.
    – José Carlos Santos
    Dec 6 at 17:47














-1












-1








-1







Find all endomorphisms of $mathbb{Q}$. ($mathbb{Q}$ is the field.)



When finding isomorphism of $2mathbb{Z}$ and $3mathbb{Z}$, we define the mapping like $φ: 2mathbb{Z} → 3mathbb{Z}$ and then we say $φ(2) = 3k, k∈Z$.



How to start in this case of $mathbb{Q}$? Can you just tell me how does the mapping look like? Thanks for your time!






ring-homomorphism







share|cite|improve this question















Find all endomorphisms of $mathbb{Q}$. ($mathbb{Q}$ is the field.)



When finding isomorphism of $2mathbb{Z}$ and $3mathbb{Z}$, we define the mapping like $φ: 2mathbb{Z} → 3mathbb{Z}$ and then we say $φ(2) = 3k, k∈Z$.



How to start in this case of $mathbb{Q}$? Can you just tell me how does the mapping look like? Thanks for your time!






ring-homomorphism




ring-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 at 17:49









ODF

79938




79938










asked Dec 6 at 17:44









Haus

307




307








  • 1




    It looks just like the identity map.
    – José Carlos Santos
    Dec 6 at 17:45










  • Like φ (k) = k ?
    – Haus
    Dec 6 at 17:46






  • 1




    Yes, just like that.
    – José Carlos Santos
    Dec 6 at 17:47














  • 1




    It looks just like the identity map.
    – José Carlos Santos
    Dec 6 at 17:45










  • Like φ (k) = k ?
    – Haus
    Dec 6 at 17:46






  • 1




    Yes, just like that.
    – José Carlos Santos
    Dec 6 at 17:47








1




1




It looks just like the identity map.
– José Carlos Santos
Dec 6 at 17:45




It looks just like the identity map.
– José Carlos Santos
Dec 6 at 17:45












Like φ (k) = k ?
– Haus
Dec 6 at 17:46




Like φ (k) = k ?
– Haus
Dec 6 at 17:46




1




1




Yes, just like that.
– José Carlos Santos
Dec 6 at 17:47




Yes, just like that.
– José Carlos Santos
Dec 6 at 17:47










1 Answer
1






active

oldest

votes


















1














This is different if we are looking for ring endomorphisms or group endomorphisms.



Let's start with ring endomorphisms, as they are easier. Suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a ring homomorphism. Then $$phi(q) = qphi(1) = q$$ for any $qinmathbb{Q}$ and hence $phi$ is just the identity. Therefore ${textrm{ring endomorphisms of } mathbb{Q}} cong {textrm{id}}$.



Now suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a group homorphism. For any $n in mathbb{Z}$ and $qinmathbb{Q}$, we have $$phi(nq) = phi(q + cdots + q) = phi(q) + cdots +phi(q) = nphi(q)$$ Therefore if $q = frac{m}{n}$ is a fraction, $nphi(q) = phi(m)$ so $$phi(q) = frac{phi(m)}{n} = frac{m}{n}phi(1)$$ Hence, $phi$ is determined by $phi(1)$. Since $phi(1)$ can take any value in $mathbb{Q}$, and each homomorphism is distinct, we obtain a (ring) isomorphism:



$$ begin{align} {textrm{group endomorphisms of } mathbb{Q}} & rightarrow mathbb{Q} \ phi & mapsto phi(1)end{align}$$



and so ${textrm{group endomorphisms of } mathbb{Q}}cong mathbb{Q}$.



This example shows it is always important to be precise which category we are working over!






share|cite|improve this answer





















  • Thank you so much!
    – Haus
    Dec 6 at 18:06











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This is different if we are looking for ring endomorphisms or group endomorphisms.



Let's start with ring endomorphisms, as they are easier. Suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a ring homomorphism. Then $$phi(q) = qphi(1) = q$$ for any $qinmathbb{Q}$ and hence $phi$ is just the identity. Therefore ${textrm{ring endomorphisms of } mathbb{Q}} cong {textrm{id}}$.



Now suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a group homorphism. For any $n in mathbb{Z}$ and $qinmathbb{Q}$, we have $$phi(nq) = phi(q + cdots + q) = phi(q) + cdots +phi(q) = nphi(q)$$ Therefore if $q = frac{m}{n}$ is a fraction, $nphi(q) = phi(m)$ so $$phi(q) = frac{phi(m)}{n} = frac{m}{n}phi(1)$$ Hence, $phi$ is determined by $phi(1)$. Since $phi(1)$ can take any value in $mathbb{Q}$, and each homomorphism is distinct, we obtain a (ring) isomorphism:



$$ begin{align} {textrm{group endomorphisms of } mathbb{Q}} & rightarrow mathbb{Q} \ phi & mapsto phi(1)end{align}$$



and so ${textrm{group endomorphisms of } mathbb{Q}}cong mathbb{Q}$.



This example shows it is always important to be precise which category we are working over!






share|cite|improve this answer





















  • Thank you so much!
    – Haus
    Dec 6 at 18:06
















1














This is different if we are looking for ring endomorphisms or group endomorphisms.



Let's start with ring endomorphisms, as they are easier. Suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a ring homomorphism. Then $$phi(q) = qphi(1) = q$$ for any $qinmathbb{Q}$ and hence $phi$ is just the identity. Therefore ${textrm{ring endomorphisms of } mathbb{Q}} cong {textrm{id}}$.



Now suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a group homorphism. For any $n in mathbb{Z}$ and $qinmathbb{Q}$, we have $$phi(nq) = phi(q + cdots + q) = phi(q) + cdots +phi(q) = nphi(q)$$ Therefore if $q = frac{m}{n}$ is a fraction, $nphi(q) = phi(m)$ so $$phi(q) = frac{phi(m)}{n} = frac{m}{n}phi(1)$$ Hence, $phi$ is determined by $phi(1)$. Since $phi(1)$ can take any value in $mathbb{Q}$, and each homomorphism is distinct, we obtain a (ring) isomorphism:



$$ begin{align} {textrm{group endomorphisms of } mathbb{Q}} & rightarrow mathbb{Q} \ phi & mapsto phi(1)end{align}$$



and so ${textrm{group endomorphisms of } mathbb{Q}}cong mathbb{Q}$.



This example shows it is always important to be precise which category we are working over!






share|cite|improve this answer





















  • Thank you so much!
    – Haus
    Dec 6 at 18:06














1












1








1






This is different if we are looking for ring endomorphisms or group endomorphisms.



Let's start with ring endomorphisms, as they are easier. Suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a ring homomorphism. Then $$phi(q) = qphi(1) = q$$ for any $qinmathbb{Q}$ and hence $phi$ is just the identity. Therefore ${textrm{ring endomorphisms of } mathbb{Q}} cong {textrm{id}}$.



Now suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a group homorphism. For any $n in mathbb{Z}$ and $qinmathbb{Q}$, we have $$phi(nq) = phi(q + cdots + q) = phi(q) + cdots +phi(q) = nphi(q)$$ Therefore if $q = frac{m}{n}$ is a fraction, $nphi(q) = phi(m)$ so $$phi(q) = frac{phi(m)}{n} = frac{m}{n}phi(1)$$ Hence, $phi$ is determined by $phi(1)$. Since $phi(1)$ can take any value in $mathbb{Q}$, and each homomorphism is distinct, we obtain a (ring) isomorphism:



$$ begin{align} {textrm{group endomorphisms of } mathbb{Q}} & rightarrow mathbb{Q} \ phi & mapsto phi(1)end{align}$$



and so ${textrm{group endomorphisms of } mathbb{Q}}cong mathbb{Q}$.



This example shows it is always important to be precise which category we are working over!






share|cite|improve this answer












This is different if we are looking for ring endomorphisms or group endomorphisms.



Let's start with ring endomorphisms, as they are easier. Suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a ring homomorphism. Then $$phi(q) = qphi(1) = q$$ for any $qinmathbb{Q}$ and hence $phi$ is just the identity. Therefore ${textrm{ring endomorphisms of } mathbb{Q}} cong {textrm{id}}$.



Now suppose $phi : mathbb{Q} rightarrow mathbb{Q}$ is a group homorphism. For any $n in mathbb{Z}$ and $qinmathbb{Q}$, we have $$phi(nq) = phi(q + cdots + q) = phi(q) + cdots +phi(q) = nphi(q)$$ Therefore if $q = frac{m}{n}$ is a fraction, $nphi(q) = phi(m)$ so $$phi(q) = frac{phi(m)}{n} = frac{m}{n}phi(1)$$ Hence, $phi$ is determined by $phi(1)$. Since $phi(1)$ can take any value in $mathbb{Q}$, and each homomorphism is distinct, we obtain a (ring) isomorphism:



$$ begin{align} {textrm{group endomorphisms of } mathbb{Q}} & rightarrow mathbb{Q} \ phi & mapsto phi(1)end{align}$$



and so ${textrm{group endomorphisms of } mathbb{Q}}cong mathbb{Q}$.



This example shows it is always important to be precise which category we are working over!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 at 18:03









ODF

79938




79938












  • Thank you so much!
    – Haus
    Dec 6 at 18:06


















  • Thank you so much!
    – Haus
    Dec 6 at 18:06
















Thank you so much!
– Haus
Dec 6 at 18:06




Thank you so much!
– Haus
Dec 6 at 18:06


















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