Galois group of $x^3-x^2-4$
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichado
6,2551929
add a comment |
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichado
6,2551929
4
Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59
1
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
– jgon
Jan 3 at 2:00
1
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
– dezdichado
Jan 3 at 2:03
It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24
@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago
add a comment |
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichado
6,2551929
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichado
6,2551929
asked Jan 3 at 1:42
dezdichado
6,2551929
asked Jan 3 at 1:42
dezdichado
6,2551929
asked Jan 3 at 1:42
dezdichado
6,2551929
asked Jan 3 at 1:42
dezdichado
6,2551929
6,2551929
4
Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59
1
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
– jgon
Jan 3 at 2:00
1
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
– dezdichado
Jan 3 at 2:03
It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24
@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago
add a comment |
4
Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59
1
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
– jgon
Jan 3 at 2:00
1
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
– dezdichado
Jan 3 at 2:03
It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24
@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago
4
4
Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59
Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59
1
1
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
– jgon
Jan 3 at 2:00
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
– jgon
Jan 3 at 2:00
1
1
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
– dezdichado
Jan 3 at 2:03
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
– dezdichado
Jan 3 at 2:03
It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24
It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24
@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago
@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered 2 days ago
Cheerful Parsnip
20.9k23396
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%2f3060179%2fgalois-group-of-x3-x2-4%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
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered 2 days ago
Cheerful Parsnip
20.9k23396
add a comment |
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered 2 days ago
Cheerful Parsnip
20.9k23396
add a comment |
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered 2 days ago
Cheerful Parsnip
20.9k23396
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered 2 days ago
Cheerful Parsnip
20.9k23396
answered 2 days ago
Cheerful Parsnip
20.9k23396
answered 2 days ago
Cheerful Parsnip
20.9k23396
answered 2 days ago
Cheerful Parsnip
20.9k23396
20.9k23396
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3060179%2fgalois-group-of-x3-x2-4%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
4
Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59
1
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
– jgon
Jan 3 at 2:00
1
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
– dezdichado
Jan 3 at 2:03
It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24
@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago