Algebra - endomorphisms of field
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
add a comment |
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
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
add a comment |
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
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
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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!
Thank you so much!
– Haus
Dec 6 at 18:06
add a comment |
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1 Answer
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active
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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!
Thank you so much!
– Haus
Dec 6 at 18:06
add a comment |
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!
Thank you so much!
– Haus
Dec 6 at 18:06
add a comment |
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!
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!
answered Dec 6 at 18:03
ODF
79938
79938
Thank you so much!
– Haus
Dec 6 at 18:06
add a comment |
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
add a comment |
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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