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Ring Automorphisms that fix 1.

2

$begingroup$

This question is a follow - up to this question about Field Automorphisms of $mathbb{Q}[sqrt{2}]$.

Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:

$$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$

I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ but I don't know what property of ring isomorphisms would allow me to do this.

abstract-algebra ring-theory field-theory galois-theory

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asked 3 hours ago

Solarflare0Solarflare0

9813

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2

$begingroup$

This question is a follow - up to this question about Field Automorphisms of $mathbb{Q}[sqrt{2}]$.

Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:

$$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$

I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ but I don't know what property of ring isomorphisms would allow me to do this.

abstract-algebra ring-theory field-theory galois-theory

share|cite|improve this question

asked 3 hours ago

Solarflare0Solarflare0

9813

$endgroup$

add a comment | 

$begingroup$

This question is a follow - up to this question about Field Automorphisms of $mathbb{Q}[sqrt{2}]$.

Since $mathbb{Q}[sqrt{2}]$ is a vector space over $mathbb{Q}$ with basis ${1, sqrt{2}}$, I naively understand why it is the case that automorphisms $phi$ of $mathbb{Q}[sqrt{2}]$ are determined wholly by the image of $1$ and $sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt{2}) = sqrt{2}$, and I want to compute the value of $phi(frac{3}{2})$. I can do the following:

$$ phi(frac{3}{2}) = phi(3) phi(frac{1}{2}) = [phi(1) + phi(1) + phi(1)] phi(frac{1}{2}) = 3phi(frac{1}{2}).$$

I am unsure how to proceed from here. I would assume that it is true that $$phi(frac{1}{1 + 1}) = frac{phi(1)}{phi(1) + phi(1)} = frac{1}{2},$$ but I don't know what property of ring isomorphisms would allow me to do this.

abstract-algebra ring-theory field-theory galois-theory

share|cite|improve this question

asked 3 hours ago

Solarflare0Solarflare0

9813

$endgroup$

share|cite|improve this question

asked 3 hours ago

Solarflare0Solarflare0

9813

share|cite|improve this question

asked 3 hours ago

Solarflare0Solarflare0

9813

asked 3 hours ago

Solarflare0Solarflare0

9813

asked 3 hours ago

Solarflare0Solarflare0

9813

2 Answers
2

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3

$begingroup$

$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.

share|cite|improve this answer

answered 3 hours ago

lhflhf

168k11172405

$endgroup$

add a comment | 

1

$begingroup$

Every automorphism fixes $mathbb{Q}$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbb{Q}$.

For the proof, we assume WLOG that $mathbb{Q} subseteq K$. Then:

• $phi$ fixes $0$ and $1$, by definition.




• $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.




• $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.




• $phi$ fixes all rational numbers, since $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.

---

More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.

share|cite|improve this answer

answered 2 hours ago

60056005

37.1k752127

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3

$begingroup$

$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.

share|cite|improve this answer

answered 3 hours ago

lhflhf

168k11172405

$endgroup$

add a comment | 

3

$begingroup$

$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.

share|cite|improve this answer

answered 3 hours ago

lhflhf

168k11172405

$endgroup$

add a comment | 

$begingroup$

$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.

share|cite|improve this answer

answered 3 hours ago

lhflhf

168k11172405

$endgroup$

share|cite|improve this answer

answered 3 hours ago

lhflhf

168k11172405

answered 3 hours ago

lhflhf

168k11172405

answered 3 hours ago

lhflhf

168k11172405

1

$begingroup$

Every automorphism fixes $mathbb{Q}$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbb{Q}$.

For the proof, we assume WLOG that $mathbb{Q} subseteq K$. Then:

• $phi$ fixes $0$ and $1$, by definition.




• $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.




• $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.




• $phi$ fixes all rational numbers, since $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.

---

More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.

share|cite|improve this answer

answered 2 hours ago

60056005

37.1k752127

$endgroup$

add a comment | 

1

$begingroup$

Every automorphism fixes $mathbb{Q}$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbb{Q}$.

For the proof, we assume WLOG that $mathbb{Q} subseteq K$. Then:

• $phi$ fixes $0$ and $1$, by definition.




• $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.




• $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.




• $phi$ fixes all rational numbers, since $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.

---

More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.

share|cite|improve this answer

answered 2 hours ago

60056005

37.1k752127

$endgroup$

add a comment | 

$begingroup$

Every automorphism fixes $mathbb{Q}$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbb{Q}$.

For the proof, we assume WLOG that $mathbb{Q} subseteq K$. Then:

• $phi$ fixes $0$ and $1$, by definition.




• $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.




• $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.




• $phi$ fixes all rational numbers, since $n cdot phileft(frac{m}{n}right) = phi(m) = m$, so $phileft(frac{m}{n}right) = frac{m}{n}$.

---

More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbb{Q}$, since all automorphisms fix $mathbb{Q}$, such a restriction is unnecessary.

share|cite|improve this answer

answered 2 hours ago

60056005

37.1k752127

$endgroup$

share|cite|improve this answer

answered 2 hours ago

60056005

37.1k752127

answered 2 hours ago

60056005

37.1k752127

answered 2 hours ago

60056005

37.1k752127