Ring Automorphisms that fix 1.
$begingroup$
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
2
active
oldest
votes
$begingroup$
$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
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 |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190546%2fring-automorphisms-that-fix-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
2phi(frac{3}{2}) = phi(3) = 3phi(1) = 3
implies
phi(frac{3}{2}) =frac{3}{2}
$$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190546%2fring-automorphisms-that-fix-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
