Question about constructing fields












1












$begingroup$



Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10


















1












$begingroup$



Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10
















1












1








1





$begingroup$



Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$





Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials




field-theory irreducible-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 19:05









Leyla AlkanLeyla Alkan

1,5801724




1,5801724












  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10




















  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10


















$begingroup$
All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
$endgroup$
– Robert Israel
Dec 10 '18 at 19:10






$begingroup$
All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
$endgroup$
– Robert Israel
Dec 10 '18 at 19:10












1 Answer
1






active

oldest

votes


















0












$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034349%2fquestion-about-constructing-fields%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









0












$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56
















0












$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56














0












0








0





$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$



For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 19:22









Robert IsraelRobert Israel

324k23214468




324k23214468












  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56


















  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56
















$begingroup$
I didn't know Herstein's version of the term "isomorphism" was widely used.
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:04




$begingroup$
I didn't know Herstein's version of the term "isomorphism" was widely used.
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:04












$begingroup$
I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
$endgroup$
– Robert Israel
Dec 10 '18 at 20:56




$begingroup$
I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
$endgroup$
– Robert Israel
Dec 10 '18 at 20:56


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034349%2fquestion-about-constructing-fields%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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more