Factorising $X^{16}- X$ over $mathbb F_4$.











up vote
2
down vote

favorite












I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring







share|cite|improve this question




















  • 1




    For reference.
    – Git Gud
    Dec 4 at 1:25










  • Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    – Jyrki Lahtonen
    Dec 4 at 4:25















up vote
2
down vote

favorite












I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring







share|cite|improve this question




















  • 1




    For reference.
    – Git Gud
    Dec 4 at 1:25










  • Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    – Jyrki Lahtonen
    Dec 4 at 4:25













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring







share|cite|improve this question















I need to factorise $X^{16}- X$ over $mathbb F_4$. How might I go about this? I have factorised over $mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?



Here is the factorisation over $mathbb F_2$:



$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$






polynomials finite-fields factoring




polynomials finite-fields factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 1:26









Git Gud

28.6k104899




28.6k104899










asked Dec 4 at 1:16









Devilo

11718




11718








  • 1




    For reference.
    – Git Gud
    Dec 4 at 1:25










  • Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    – Jyrki Lahtonen
    Dec 4 at 4:25














  • 1




    For reference.
    – Git Gud
    Dec 4 at 1:25










  • Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
    – Jyrki Lahtonen
    Dec 4 at 4:25








1




1




For reference.
– Git Gud
Dec 4 at 1:25




For reference.
– Git Gud
Dec 4 at 1:25












Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
– Jyrki Lahtonen
Dec 4 at 4:25




Your factorization over $Bbb{F}_2$ is incomplete. The octic is a product of two quartics. See here.
– Jyrki Lahtonen
Dec 4 at 4:25










1 Answer
1






active

oldest

votes

















up vote
3
down vote













Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer























  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 at 2:36






  • 1




    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    – Batominovski
    Dec 4 at 2:43











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024981%2ffactorising-x16-x-over-mathbb-f-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








up vote
3
down vote













Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer























  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 at 2:36






  • 1




    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    – Batominovski
    Dec 4 at 2:43















up vote
3
down vote













Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer























  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 at 2:36






  • 1




    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    – Batominovski
    Dec 4 at 2:43













up vote
3
down vote










up vote
3
down vote









Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$






share|cite|improve this answer














Let $q:=4$. You are factorizing $X^{q^2}-X$ over $mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X),f_1(X),f_2(X),cdots,f_k(X)$$
where $f_1(X),f_2(X),ldots,f_k(X)inmathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $mathbb{F}_q$ (whence $k=dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $mathbb{F}_q=mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1),,$$
where $t$ is an element of $mathbb{F}_4setminusmathbb{F}_2$.



Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b,,$$
with $a,binmathbb{F}_4={0,1,t,t+1}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $cinmathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $aneq 0$.



If $a=1$, then $bin{t,t+1}$, and both choices work. If $a=t$, then $bin{1,t}$ and both choices work. If $a=t+1$, then $bin{1,t+1}$ and both choices work. This shows that
$$begin{align}X^{16}-X&=X(X+1)(X+t)big(X+(t+1)big),(X^2+X+t),big(X^2+X+(t+1)big)\&phantom{abcf}(X^2+tX+1),(X^2+tX+t),big(X^2+(t+1),X+1big),big(X^2+(t+1),X+(t+1)big),.end{align}$$



I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+dfrac{1}{X}$. Then,
$$begin{align}X^4+X^3+X^2+X+1&=X^2,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\&=(X^2+tX+1),big(X^2+(t+1)X+1big),.end{align}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 at 1:58

























answered Dec 4 at 1:37









Batominovski

33.2k23292




33.2k23292












  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 at 2:36






  • 1




    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    – Batominovski
    Dec 4 at 2:43


















  • How do you know in advance that they should factorise into quadratics?
    – Devilo
    Dec 4 at 2:36






  • 1




    @Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
    – Batominovski
    Dec 4 at 2:43
















How do you know in advance that they should factorise into quadratics?
– Devilo
Dec 4 at 2:36




How do you know in advance that they should factorise into quadratics?
– Devilo
Dec 4 at 2:36




1




1




@Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
– Batominovski
Dec 4 at 2:43




@Devilo It follows from a theorem which states that $mathbb{F}_{q}$ is the splitting field of $X^{q}-X$, where $q$ is a prime power. Therefore, $mathbb{F}_{16}$ is the splitting field of $X^{16}-X$. Since $[mathbb{F}_{16}:mathbb{F}_4]=2$, it follows that the irreducible factors of $X^{16}-X$ over $mathbb{F}_4$ are either linear or quadratic.
– Batominovski
Dec 4 at 2:43


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024981%2ffactorising-x16-x-over-mathbb-f-4%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more