Ring Automorphisms that fix 1. Announcing the arrival of Valued Associate #679: Cesar Manara ...
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Ring Automorphisms that fix 1.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Automorphisms of $mathbb Q(sqrt 2)$Automorphisms of $mathbb{R}^n$group of automorphisms of the ring $mathbb{Z}timesmathbb{Z}$Trying to understand a proof for the automorphisms of a polynomial ringAll automorphisms of splitting fieldsDetermining automorphisms of this extensionRing automorphisms of $mathbb{Q}[sqrt[3]{5}]$Automorphism of ring and isomorphism of quotient ringsThe automorphisms of the extension $mathbb{Q}(sqrt[4]{2})/mathbb{Q}$.Extension theorem for field automorphismsAre all verbal automorphisms inner power automorphisms?
$begingroup$
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
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
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
asked 3 hours ago
Solarflare0Solarflare0
9813
$endgroup$
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
abstract-algebra ring-theory field-theory galois-theory
asked 3 hours ago
Solarflare0Solarflare0
9813
asked 3 hours ago
Solarflare0Solarflare0
9813
asked 3 hours ago
Solarflare0Solarflare0
9813
asked 3 hours ago
Solarflare0Solarflare0
9813
asked 3 hours ago
Solarflare0Solarflare0
9813
9813
add a comment |
add a comment |
2 Answers
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$begingroup$
$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.
answered 3 hours ago
lhflhf
168k11172405
$endgroup$
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.
answered 3 hours ago
60056005
37.1k752127
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.
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 $q in mathbb Q$.
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 $q in mathbb Q$.
answered 3 hours ago
lhflhf
168k11172405
$endgroup$
$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.
answered 3 hours ago
lhflhf
168k11172405
answered 3 hours ago
lhflhf
168k11172405
answered 3 hours ago
lhflhf
168k11172405
answered 3 hours ago
lhflhf
168k11172405
168k11172405
add a comment |
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.
answered 3 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.
answered 3 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.
answered 3 hours ago
60056005
37.1k752127
$endgroup$
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.
answered 3 hours ago
60056005
37.1k752127
answered 3 hours ago
60056005
37.1k752127
answered 3 hours ago
60056005
37.1k752127
answered 3 hours ago
60056005
37.1k752127
37.1k752127
add a comment |
add a comment |
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