Compute the splitting field of $X^4 + 5 X^3 + 10 X^2 + 10 X + 5$ over $Bbb Q$

up vote
0
down vote

favorite

I'm trying compute the splitting field of $P(X) := X^4 + 5 X^3 + 10 X^2 + 10 X + 5$ over $mathbb{Q}$.

This is what I thought: I tried find the roots of $P$ observing that

$$P(X-1) = X^4 + X^3 + X^2 + X + 11$$

but I'm stuck here.

Can anyone give me a direction in order to find the splitting field of $P(X)$?

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

asked Nov 14 at 3:52

Math enthusiast

1176

1

You made a mistake in $P(X-1)$. Hint: $XP(X)=(1+X)^5-1$
– user10354138
Nov 14 at 4:00

$P(x)=frac{(x+1)^5-1}{x}$
– Thomas Andrews
Nov 14 at 4:00

add a comment |

up vote
0
down vote

favorite

I'm trying compute the splitting field of $P(X) := X^4 + 5 X^3 + 10 X^2 + 10 X + 5$ over $mathbb{Q}$.

This is what I thought: I tried find the roots of $P$ observing that

$$P(X-1) = X^4 + X^3 + X^2 + X + 11$$

but I'm stuck here.

Can anyone give me a direction in order to find the splitting field of $P(X)$?

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

asked Nov 14 at 3:52

Math enthusiast

1176

1

You made a mistake in $P(X-1)$. Hint: $XP(X)=(1+X)^5-1$
– user10354138
Nov 14 at 4:00

$P(x)=frac{(x+1)^5-1}{x}$
– Thomas Andrews
Nov 14 at 4:00

add a comment |

up vote
0
down vote

favorite

I'm trying compute the splitting field of $P(X) := X^4 + 5 X^3 + 10 X^2 + 10 X + 5$ over $mathbb{Q}$.

This is what I thought: I tried find the roots of $P$ observing that

$$P(X-1) = X^4 + X^3 + X^2 + X + 11$$

but I'm stuck here.

Can anyone give me a direction in order to find the splitting field of $P(X)$?

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

asked Nov 14 at 3:52

Math enthusiast

1176

I'm trying compute the splitting field of $P(X) := X^4 + 5 X^3 + 10 X^2 + 10 X + 5$ over $mathbb{Q}$.

This is what I thought: I tried find the roots of $P$ observing that

$$P(X-1) = X^4 + X^3 + X^2 + X + 11$$

but I'm stuck here.

Can anyone give me a direction in order to find the splitting field of $P(X)$?

abstract-algebra splitting-field

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

asked Nov 14 at 3:52

Math enthusiast

1176

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

asked Nov 14 at 3:52

Math enthusiast

1176

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

edited Nov 14 at 4:31

Chinnapparaj R

4,6081725

asked Nov 14 at 3:52

Math enthusiast

1176

asked Nov 14 at 3:52

Math enthusiast

1176

asked Nov 14 at 3:52

Math enthusiast

1176

1

You made a mistake in $P(X-1)$. Hint: $XP(X)=(1+X)^5-1$
– user10354138
Nov 14 at 4:00

$P(x)=frac{(x+1)^5-1}{x}$
– Thomas Andrews
Nov 14 at 4:00

add a comment |

1

You made a mistake in $P(X-1)$. Hint: $XP(X)=(1+X)^5-1$
– user10354138
Nov 14 at 4:00

$P(x)=frac{(x+1)^5-1}{x}$
– Thomas Andrews
Nov 14 at 4:00

You made a mistake in $P(X-1)$. Hint: $XP(X)=(1+X)^5-1$
– user10354138
Nov 14 at 4:00

$P(x)=frac{(x+1)^5-1}{x}$
– Thomas Andrews
Nov 14 at 4:00

add a comment |

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

Finding roots of this polynomial is difficult so we use the following result:

Result: Let $f (x) in F[x]$ and let $a in F$. Then $f(x)$ and $f(x+ a)$ have the
same splitting field over $F$.

Here $-1 in Bbb{Q}$ and $f(x+(-1))=1+x+x^2+x^3+x^4$ not $11$ in the constant term

But the last polynomial is a Cyclotomic polynomial. So comparing to original polynomial, this one is easy to find!

edited Nov 14 at 4:30

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

Thanks! I thought that $f(x-1)$ would be a cyclotomic polynomial, but I was not identifying my mistake.
– Math enthusiast
Nov 14 at 17:46

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',
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
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997755%2fcompute-the-splitting-field-of-x4-5-x3-10-x2-10-x-5-over-bbb-q%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

Finding roots of this polynomial is difficult so we use the following result:

Result: Let $f (x) in F[x]$ and let $a in F$. Then $f(x)$ and $f(x+ a)$ have the
same splitting field over $F$.

Here $-1 in Bbb{Q}$ and $f(x+(-1))=1+x+x^2+x^3+x^4$ not $11$ in the constant term

But the last polynomial is a Cyclotomic polynomial. So comparing to original polynomial, this one is easy to find!

edited Nov 14 at 4:30

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

Thanks! I thought that $f(x-1)$ would be a cyclotomic polynomial, but I was not identifying my mistake.
– Math enthusiast
Nov 14 at 17:46

add a comment |

up vote
2
down vote

accepted

Finding roots of this polynomial is difficult so we use the following result:

Result: Let $f (x) in F[x]$ and let $a in F$. Then $f(x)$ and $f(x+ a)$ have the
same splitting field over $F$.

Here $-1 in Bbb{Q}$ and $f(x+(-1))=1+x+x^2+x^3+x^4$ not $11$ in the constant term

But the last polynomial is a Cyclotomic polynomial. So comparing to original polynomial, this one is easy to find!

edited Nov 14 at 4:30

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

Thanks! I thought that $f(x-1)$ would be a cyclotomic polynomial, but I was not identifying my mistake.
– Math enthusiast
Nov 14 at 17:46

add a comment |

up vote
2
down vote

accepted

Finding roots of this polynomial is difficult so we use the following result:

Result: Let $f (x) in F[x]$ and let $a in F$. Then $f(x)$ and $f(x+ a)$ have the
same splitting field over $F$.

Here $-1 in Bbb{Q}$ and $f(x+(-1))=1+x+x^2+x^3+x^4$ not $11$ in the constant term

But the last polynomial is a Cyclotomic polynomial. So comparing to original polynomial, this one is easy to find!

edited Nov 14 at 4:30

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

Finding roots of this polynomial is difficult so we use the following result:

Result: Let $f (x) in F[x]$ and let $a in F$. Then $f(x)$ and $f(x+ a)$ have the
same splitting field over $F$.

Here $-1 in Bbb{Q}$ and $f(x+(-1))=1+x+x^2+x^3+x^4$ not $11$ in the constant term

But the last polynomial is a Cyclotomic polynomial. So comparing to original polynomial, this one is easy to find!

edited Nov 14 at 4:30

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

edited Nov 14 at 4:30

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

answered Nov 14 at 4:12

Chinnapparaj R

4,6081725

Thanks! I thought that $f(x-1)$ would be a cyclotomic polynomial, but I was not identifying my mistake.
– Math enthusiast
Nov 14 at 17:46

add a comment |

Thanks! I thought that $f(x-1)$ would be a cyclotomic polynomial, but I was not identifying my mistake.
– Math enthusiast
Nov 14 at 17:46

Thanks! I thought that $f(x-1)$ would be a cyclotomic polynomial, but I was not identifying my mistake.
– Math enthusiast
Nov 14 at 17:46

add a comment |

draft saved

draft discarded

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Cfrgtkky