How Did Galois Understand the Galois Group?
$begingroup$
We understand the Galois group of a polynomial as automorphisms of the polynomial's splitting field. How did Galois, who did not have a notion of "field," much less "automorphism" characterize Galois groups? I understand that it was as a subgroup of the permutation of the roots, but what determined whether a given permutation was in the "Galois group" or not? (I posted this in the Math group on FB, and figured out an answer, but realized that this is the logical place to post such questions. I'll put my answer below, but I'll wait before accepting it in case someone can give a better one.)
galois-theory
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
$endgroup$
add a comment |
$begingroup$
We understand the Galois group of a polynomial as automorphisms of the polynomial's splitting field. How did Galois, who did not have a notion of "field," much less "automorphism" characterize Galois groups? I understand that it was as a subgroup of the permutation of the roots, but what determined whether a given permutation was in the "Galois group" or not? (I posted this in the Math group on FB, and figured out an answer, but realized that this is the logical place to post such questions. I'll put my answer below, but I'll wait before accepting it in case someone can give a better one.)
galois-theory
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
$endgroup$
$begingroup$
Actually, a better place for you to post this question is here.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:00
$begingroup$
Thanks, next time I will know.
$endgroup$
– Nat Kuhn
Dec 23 '18 at 22:25
$begingroup$
According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated."
$endgroup$
– Somos
Dec 24 '18 at 1:15
add a comment |
$begingroup$
We understand the Galois group of a polynomial as automorphisms of the polynomial's splitting field. How did Galois, who did not have a notion of "field," much less "automorphism" characterize Galois groups? I understand that it was as a subgroup of the permutation of the roots, but what determined whether a given permutation was in the "Galois group" or not? (I posted this in the Math group on FB, and figured out an answer, but realized that this is the logical place to post such questions. I'll put my answer below, but I'll wait before accepting it in case someone can give a better one.)
galois-theory
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
$endgroup$
We understand the Galois group of a polynomial as automorphisms of the polynomial's splitting field. How did Galois, who did not have a notion of "field," much less "automorphism" characterize Galois groups? I understand that it was as a subgroup of the permutation of the roots, but what determined whether a given permutation was in the "Galois group" or not? (I posted this in the Math group on FB, and figured out an answer, but realized that this is the logical place to post such questions. I'll put my answer below, but I'll wait before accepting it in case someone can give a better one.)
galois-theory
galois-theory
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
asked Dec 23 '18 at 21:51
Nat KuhnNat Kuhn
1515
1515
$begingroup$
Actually, a better place for you to post this question is here.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:00
$begingroup$
Thanks, next time I will know.
$endgroup$
– Nat Kuhn
Dec 23 '18 at 22:25
$begingroup$
According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated."
$endgroup$
– Somos
Dec 24 '18 at 1:15
add a comment |
$begingroup$
Actually, a better place for you to post this question is here.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:00
$begingroup$
Thanks, next time I will know.
$endgroup$
– Nat Kuhn
Dec 23 '18 at 22:25
$begingroup$
According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated."
$endgroup$
– Somos
Dec 24 '18 at 1:15
$begingroup$
Actually, a better place for you to post this question is here.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:00
$begingroup$
Actually, a better place for you to post this question is here.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:00
$begingroup$
Thanks, next time I will know.
$endgroup$
– Nat Kuhn
Dec 23 '18 at 22:25
$begingroup$
Thanks, next time I will know.
$endgroup$
– Nat Kuhn
Dec 23 '18 at 22:25
$begingroup$
According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated."
$endgroup$
– Somos
Dec 24 '18 at 1:15
$begingroup$
According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated."
$endgroup$
– Somos
Dec 24 '18 at 1:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Michael Livshitz pointed me to this article by Harold M Edwards, according to which the answer is more-or-less this:
Let $A={a_1,...,a_n}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if):
(g) for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 iff g(pi(a_1),...,pi(a_n))=0$.
It is quite easy to show that this is equivalent to the modern definition. An automorphism of the splitting field of $f$ which leaves $k$ fixed gives a permutation of the roots, and (g) follows directly from the fact that it fixes $k$.
To see the converse, note that the set of all polynomials $g$ in $R$ with $g(a_1,....,a_n)=0$ forms an ideal $Isubset R$; it is just the kernel of the (surjective) map of $k[x_1,...,x_n]$ to $K=k(a_1,...,a_n)$ which takes $g$ to $g(a_1,...,a_n)$. As a result, $R/I$ is isomorphic to $K$. (In fact, $I$ is a maximal ideal but that plays no role in our proof.) If $pi$ is a permutation satisfying (g), the map from $R$ to $K$ which takes $g$ to $g(pi(a_1),...,pi(a_n))$ is also a surjective ring homomorphism, and (g) tells us that it has the same kernel, $I$. So it induces an isomorphism of $K=R/I$ to itself, obviously fixing $k$, and permuting the roots as specified by $pi$.
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3050732%2fhow-did-galois-understand-the-galois-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Michael Livshitz pointed me to this article by Harold M Edwards, according to which the answer is more-or-less this:
Let $A={a_1,...,a_n}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if):
(g) for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 iff g(pi(a_1),...,pi(a_n))=0$.
It is quite easy to show that this is equivalent to the modern definition. An automorphism of the splitting field of $f$ which leaves $k$ fixed gives a permutation of the roots, and (g) follows directly from the fact that it fixes $k$.
To see the converse, note that the set of all polynomials $g$ in $R$ with $g(a_1,....,a_n)=0$ forms an ideal $Isubset R$; it is just the kernel of the (surjective) map of $k[x_1,...,x_n]$ to $K=k(a_1,...,a_n)$ which takes $g$ to $g(a_1,...,a_n)$. As a result, $R/I$ is isomorphic to $K$. (In fact, $I$ is a maximal ideal but that plays no role in our proof.) If $pi$ is a permutation satisfying (g), the map from $R$ to $K$ which takes $g$ to $g(pi(a_1),...,pi(a_n))$ is also a surjective ring homomorphism, and (g) tells us that it has the same kernel, $I$. So it induces an isomorphism of $K=R/I$ to itself, obviously fixing $k$, and permuting the roots as specified by $pi$.
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
$endgroup$
add a comment |
$begingroup$
Michael Livshitz pointed me to this article by Harold M Edwards, according to which the answer is more-or-less this:
Let $A={a_1,...,a_n}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if):
(g) for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 iff g(pi(a_1),...,pi(a_n))=0$.
It is quite easy to show that this is equivalent to the modern definition. An automorphism of the splitting field of $f$ which leaves $k$ fixed gives a permutation of the roots, and (g) follows directly from the fact that it fixes $k$.
To see the converse, note that the set of all polynomials $g$ in $R$ with $g(a_1,....,a_n)=0$ forms an ideal $Isubset R$; it is just the kernel of the (surjective) map of $k[x_1,...,x_n]$ to $K=k(a_1,...,a_n)$ which takes $g$ to $g(a_1,...,a_n)$. As a result, $R/I$ is isomorphic to $K$. (In fact, $I$ is a maximal ideal but that plays no role in our proof.) If $pi$ is a permutation satisfying (g), the map from $R$ to $K$ which takes $g$ to $g(pi(a_1),...,pi(a_n))$ is also a surjective ring homomorphism, and (g) tells us that it has the same kernel, $I$. So it induces an isomorphism of $K=R/I$ to itself, obviously fixing $k$, and permuting the roots as specified by $pi$.
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
$endgroup$
add a comment |
$begingroup$
Michael Livshitz pointed me to this article by Harold M Edwards, according to which the answer is more-or-less this:
Let $A={a_1,...,a_n}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if):
(g) for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 iff g(pi(a_1),...,pi(a_n))=0$.
It is quite easy to show that this is equivalent to the modern definition. An automorphism of the splitting field of $f$ which leaves $k$ fixed gives a permutation of the roots, and (g) follows directly from the fact that it fixes $k$.
To see the converse, note that the set of all polynomials $g$ in $R$ with $g(a_1,....,a_n)=0$ forms an ideal $Isubset R$; it is just the kernel of the (surjective) map of $k[x_1,...,x_n]$ to $K=k(a_1,...,a_n)$ which takes $g$ to $g(a_1,...,a_n)$. As a result, $R/I$ is isomorphic to $K$. (In fact, $I$ is a maximal ideal but that plays no role in our proof.) If $pi$ is a permutation satisfying (g), the map from $R$ to $K$ which takes $g$ to $g(pi(a_1),...,pi(a_n))$ is also a surjective ring homomorphism, and (g) tells us that it has the same kernel, $I$. So it induces an isomorphism of $K=R/I$ to itself, obviously fixing $k$, and permuting the roots as specified by $pi$.
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
$endgroup$
Michael Livshitz pointed me to this article by Harold M Edwards, according to which the answer is more-or-less this:
Let $A={a_1,...,a_n}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if):
(g) for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 iff g(pi(a_1),...,pi(a_n))=0$.
It is quite easy to show that this is equivalent to the modern definition. An automorphism of the splitting field of $f$ which leaves $k$ fixed gives a permutation of the roots, and (g) follows directly from the fact that it fixes $k$.
To see the converse, note that the set of all polynomials $g$ in $R$ with $g(a_1,....,a_n)=0$ forms an ideal $Isubset R$; it is just the kernel of the (surjective) map of $k[x_1,...,x_n]$ to $K=k(a_1,...,a_n)$ which takes $g$ to $g(a_1,...,a_n)$. As a result, $R/I$ is isomorphic to $K$. (In fact, $I$ is a maximal ideal but that plays no role in our proof.) If $pi$ is a permutation satisfying (g), the map from $R$ to $K$ which takes $g$ to $g(pi(a_1),...,pi(a_n))$ is also a surjective ring homomorphism, and (g) tells us that it has the same kernel, $I$. So it induces an isomorphism of $K=R/I$ to itself, obviously fixing $k$, and permuting the roots as specified by $pi$.
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
answered Dec 23 '18 at 22:18
Nat KuhnNat Kuhn
1515
1515
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%2f3050732%2fhow-did-galois-understand-the-galois-group%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
$begingroup$
Actually, a better place for you to post this question is here.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:00
$begingroup$
Thanks, next time I will know.
$endgroup$
– Nat Kuhn
Dec 23 '18 at 22:25
$begingroup$
According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated."
$endgroup$
– Somos
Dec 24 '18 at 1:15