Irreducibility of $x^4 + x^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$
I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.
So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$
And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.
Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.
Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$
Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!
finite-fields irreducible-polynomials
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
add a comment |
I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.
So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$
And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.
Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.
Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$
Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!
finite-fields irreducible-polynomials
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53
Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55
add a comment |
I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.
So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$
And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.
Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.
Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$
Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!
finite-fields irreducible-polynomials
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.
So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$
And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.
Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.
Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$
Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!
finite-fields irreducible-polynomials
finite-fields irreducible-polynomials
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
edited Nov 19 at 1:21
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
asked Nov 19 at 0:50
Naweed G. Seldon
1,299419
1,299419
"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53
Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55
add a comment |
"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53
Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55
"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53
"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53
Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55
Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55
add a comment |
2 Answers
2
active
oldest
votes
$P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).
answered Nov 19 at 1:54
Chris Custer
10.8k3724
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
1
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
add a comment |
$Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?
answered Nov 19 at 1:53
sharding4
4,0961823
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%2f3004358%2firreducibility-of-x4-x3-1-over-finite-field-mathbbf-2a-1-le%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
$P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).
answered Nov 19 at 1:54
Chris Custer
10.8k3724
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
1
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
add a comment |
$P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).
answered Nov 19 at 1:54
Chris Custer
10.8k3724
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
1
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
add a comment |
$P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).
answered Nov 19 at 1:54
Chris Custer
10.8k3724
$P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).
answered Nov 19 at 1:54
Chris Custer
10.8k3724
answered Nov 19 at 1:54
Chris Custer
10.8k3724
answered Nov 19 at 1:54
Chris Custer
10.8k3724
answered Nov 19 at 1:54
Chris Custer
10.8k3724
10.8k3724
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
1
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
add a comment |
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
1
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
– Naweed G. Seldon
Nov 19 at 2:05
1
1
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
– Chris Custer
Nov 19 at 2:28
add a comment |
$Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?
answered Nov 19 at 1:53
sharding4
4,0961823
add a comment |
$Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?
answered Nov 19 at 1:53
sharding4
4,0961823
add a comment |
$Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?
answered Nov 19 at 1:53
sharding4
4,0961823
$Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?
answered Nov 19 at 1:53
sharding4
4,0961823
edited Nov 19 at 2:49
answered Nov 19 at 1:53
sharding4
4,0961823
answered Nov 19 at 1:53
sharding4
4,0961823
answered Nov 19 at 1:53
sharding4
4,0961823
4,0961823
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%2f3004358%2firreducibility-of-x4-x3-1-over-finite-field-mathbbf-2a-1-le%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
"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53
Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55