Fast modular composition and irreducibility testing
Given a polynomials $fin F_q[t]$ we want an algorithm that says if $f$ is irreducible or not.
In the book "Modern Computer Algebra" by von zur Gathen and Jürgen Gerhard, an algorithm for that is proposed (algorithm $14.36$). The issue is not with the algorithm itself, but with the implementation advices.
The book recommends us to compute $t^qmod f$ as an auxiliary result, and then, use the fast modular composition algorithm to compute $t^{q^n}mod f$ (being $n:=deg(f)$), and, in general to compute $t^{q^{n/k}}mod f$ (being $k$ a prime divisor of $n$).
I understand this recommendation as follows: Let $n=2^k+sum_{i=0}^{k-1}n_i2^i$ be the binary representation of $n$ ($n_iin{0,1}$) and $s_i:=x^{q^i}mod f$.
It is clear that $$x^{q^n}mod f = x^{q^{2^k+sum_{i=0}^{k-1}n_i2^i}}mod f=s_{2^k}(s_{sum_{i=0}^{k-1}n_i2^i})mod f=s_{2^k}(s_{n_{k-1}2^{k-1}}(cdots(s_{n_0})cdots))mod f$$
So we will only need $mathcal{O}(log n)$ fast modular compositions to get the result.
I don't really get why this recommendations are useful, as we have to compute not only $x^qmod f$, but also $x^{q^{2^i}}mod f$ for some $iin{1,dots,k}$. Am I wrong?
Any idea?
abstract-algebra algorithms finite-fields irreducible-polynomials computer-algebra-systems
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
add a comment |
Given a polynomials $fin F_q[t]$ we want an algorithm that says if $f$ is irreducible or not.
In the book "Modern Computer Algebra" by von zur Gathen and Jürgen Gerhard, an algorithm for that is proposed (algorithm $14.36$). The issue is not with the algorithm itself, but with the implementation advices.
The book recommends us to compute $t^qmod f$ as an auxiliary result, and then, use the fast modular composition algorithm to compute $t^{q^n}mod f$ (being $n:=deg(f)$), and, in general to compute $t^{q^{n/k}}mod f$ (being $k$ a prime divisor of $n$).
I understand this recommendation as follows: Let $n=2^k+sum_{i=0}^{k-1}n_i2^i$ be the binary representation of $n$ ($n_iin{0,1}$) and $s_i:=x^{q^i}mod f$.
It is clear that $$x^{q^n}mod f = x^{q^{2^k+sum_{i=0}^{k-1}n_i2^i}}mod f=s_{2^k}(s_{sum_{i=0}^{k-1}n_i2^i})mod f=s_{2^k}(s_{n_{k-1}2^{k-1}}(cdots(s_{n_0})cdots))mod f$$
So we will only need $mathcal{O}(log n)$ fast modular compositions to get the result.
I don't really get why this recommendations are useful, as we have to compute not only $x^qmod f$, but also $x^{q^{2^i}}mod f$ for some $iin{1,dots,k}$. Am I wrong?
Any idea?
abstract-algebra algorithms finite-fields irreducible-polynomials computer-algebra-systems
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
1
Not my field at all, but isn't taking $q$-th powers very easy in $Bbb{F}_q$? For example, for any polynomial $ginBbb{F}_q[t]$ you have $(g(t))^q=g(t^q)$.
– Servaes
Dec 2 '18 at 18:37
@Servaes yes, what you said is true, but there are still some holes. Please, take a look to my edits
– Álvaro G. Tenorio
Dec 15 '18 at 14:48
add a comment |
Given a polynomials $fin F_q[t]$ we want an algorithm that says if $f$ is irreducible or not.
In the book "Modern Computer Algebra" by von zur Gathen and Jürgen Gerhard, an algorithm for that is proposed (algorithm $14.36$). The issue is not with the algorithm itself, but with the implementation advices.
The book recommends us to compute $t^qmod f$ as an auxiliary result, and then, use the fast modular composition algorithm to compute $t^{q^n}mod f$ (being $n:=deg(f)$), and, in general to compute $t^{q^{n/k}}mod f$ (being $k$ a prime divisor of $n$).
I understand this recommendation as follows: Let $n=2^k+sum_{i=0}^{k-1}n_i2^i$ be the binary representation of $n$ ($n_iin{0,1}$) and $s_i:=x^{q^i}mod f$.
It is clear that $$x^{q^n}mod f = x^{q^{2^k+sum_{i=0}^{k-1}n_i2^i}}mod f=s_{2^k}(s_{sum_{i=0}^{k-1}n_i2^i})mod f=s_{2^k}(s_{n_{k-1}2^{k-1}}(cdots(s_{n_0})cdots))mod f$$
So we will only need $mathcal{O}(log n)$ fast modular compositions to get the result.
I don't really get why this recommendations are useful, as we have to compute not only $x^qmod f$, but also $x^{q^{2^i}}mod f$ for some $iin{1,dots,k}$. Am I wrong?
Any idea?
abstract-algebra algorithms finite-fields irreducible-polynomials computer-algebra-systems
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
Given a polynomials $fin F_q[t]$ we want an algorithm that says if $f$ is irreducible or not.
In the book "Modern Computer Algebra" by von zur Gathen and Jürgen Gerhard, an algorithm for that is proposed (algorithm $14.36$). The issue is not with the algorithm itself, but with the implementation advices.
The book recommends us to compute $t^qmod f$ as an auxiliary result, and then, use the fast modular composition algorithm to compute $t^{q^n}mod f$ (being $n:=deg(f)$), and, in general to compute $t^{q^{n/k}}mod f$ (being $k$ a prime divisor of $n$).
I understand this recommendation as follows: Let $n=2^k+sum_{i=0}^{k-1}n_i2^i$ be the binary representation of $n$ ($n_iin{0,1}$) and $s_i:=x^{q^i}mod f$.
It is clear that $$x^{q^n}mod f = x^{q^{2^k+sum_{i=0}^{k-1}n_i2^i}}mod f=s_{2^k}(s_{sum_{i=0}^{k-1}n_i2^i})mod f=s_{2^k}(s_{n_{k-1}2^{k-1}}(cdots(s_{n_0})cdots))mod f$$
So we will only need $mathcal{O}(log n)$ fast modular compositions to get the result.
I don't really get why this recommendations are useful, as we have to compute not only $x^qmod f$, but also $x^{q^{2^i}}mod f$ for some $iin{1,dots,k}$. Am I wrong?
Any idea?
abstract-algebra algorithms finite-fields irreducible-polynomials computer-algebra-systems
abstract-algebra algorithms finite-fields irreducible-polynomials computer-algebra-systems
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
edited Dec 15 '18 at 14:46
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
asked Dec 2 '18 at 18:23
Álvaro G. Tenorio
13710
13710
1
Not my field at all, but isn't taking $q$-th powers very easy in $Bbb{F}_q$? For example, for any polynomial $ginBbb{F}_q[t]$ you have $(g(t))^q=g(t^q)$.
– Servaes
Dec 2 '18 at 18:37
@Servaes yes, what you said is true, but there are still some holes. Please, take a look to my edits
– Álvaro G. Tenorio
Dec 15 '18 at 14:48
add a comment |
1
Not my field at all, but isn't taking $q$-th powers very easy in $Bbb{F}_q$? For example, for any polynomial $ginBbb{F}_q[t]$ you have $(g(t))^q=g(t^q)$.
– Servaes
Dec 2 '18 at 18:37
@Servaes yes, what you said is true, but there are still some holes. Please, take a look to my edits
– Álvaro G. Tenorio
Dec 15 '18 at 14:48
1
1
Not my field at all, but isn't taking $q$-th powers very easy in $Bbb{F}_q$? For example, for any polynomial $ginBbb{F}_q[t]$ you have $(g(t))^q=g(t^q)$.
– Servaes
Dec 2 '18 at 18:37
Not my field at all, but isn't taking $q$-th powers very easy in $Bbb{F}_q$? For example, for any polynomial $ginBbb{F}_q[t]$ you have $(g(t))^q=g(t^q)$.
– Servaes
Dec 2 '18 at 18:37
@Servaes yes, what you said is true, but there are still some holes. Please, take a look to my edits
– Álvaro G. Tenorio
Dec 15 '18 at 14:48
@Servaes yes, what you said is true, but there are still some holes. Please, take a look to my edits
– Álvaro G. Tenorio
Dec 15 '18 at 14:48
add a comment |
0
active
oldest
votes
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%2f3023011%2ffast-modular-composition-and-irreducibility-testing%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3023011%2ffast-modular-composition-and-irreducibility-testing%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
1
Not my field at all, but isn't taking $q$-th powers very easy in $Bbb{F}_q$? For example, for any polynomial $ginBbb{F}_q[t]$ you have $(g(t))^q=g(t^q)$.
– Servaes
Dec 2 '18 at 18:37
@Servaes yes, what you said is true, but there are still some holes. Please, take a look to my edits
– Álvaro G. Tenorio
Dec 15 '18 at 14:48