A question about the degree of an extension field












1












$begingroup$


Consider $f(x) := x^3+2x+2$ and the field $mathbb{Z_3}$. $f(x)$ is obviously irreducible over $mathbb{Z_3}$. Let $a$ be a root in an extension field of $mathbb{Z_3}$, then why is it that $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? What is the basis of $mathbb{Z_3}(a)$ over $mathbb{Z_3}$?



I know that $mathbb{Z_3}(a) simeq mathbb{Z_3}[x]/<f(x)>$ and since $f(x)$ is irreducible in $mathbb{Z_3}$, any polynomial in $mathbb{Z_3}[x]$ can have degree atmost 2. But I don't understand how that ties to $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? And how does that imply $mathbb{Z_3}(a)simeq GF(3^3)$? Thanks.










share|cite|improve this question









$endgroup$












  • $begingroup$
    See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15".
    $endgroup$
    – Lucas Corrêa
    Apr 21 at 21:44
















1












$begingroup$


Consider $f(x) := x^3+2x+2$ and the field $mathbb{Z_3}$. $f(x)$ is obviously irreducible over $mathbb{Z_3}$. Let $a$ be a root in an extension field of $mathbb{Z_3}$, then why is it that $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? What is the basis of $mathbb{Z_3}(a)$ over $mathbb{Z_3}$?



I know that $mathbb{Z_3}(a) simeq mathbb{Z_3}[x]/<f(x)>$ and since $f(x)$ is irreducible in $mathbb{Z_3}$, any polynomial in $mathbb{Z_3}[x]$ can have degree atmost 2. But I don't understand how that ties to $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? And how does that imply $mathbb{Z_3}(a)simeq GF(3^3)$? Thanks.










share|cite|improve this question









$endgroup$












  • $begingroup$
    See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15".
    $endgroup$
    – Lucas Corrêa
    Apr 21 at 21:44














1












1








1





$begingroup$


Consider $f(x) := x^3+2x+2$ and the field $mathbb{Z_3}$. $f(x)$ is obviously irreducible over $mathbb{Z_3}$. Let $a$ be a root in an extension field of $mathbb{Z_3}$, then why is it that $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? What is the basis of $mathbb{Z_3}(a)$ over $mathbb{Z_3}$?



I know that $mathbb{Z_3}(a) simeq mathbb{Z_3}[x]/<f(x)>$ and since $f(x)$ is irreducible in $mathbb{Z_3}$, any polynomial in $mathbb{Z_3}[x]$ can have degree atmost 2. But I don't understand how that ties to $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? And how does that imply $mathbb{Z_3}(a)simeq GF(3^3)$? Thanks.










share|cite|improve this question









$endgroup$




Consider $f(x) := x^3+2x+2$ and the field $mathbb{Z_3}$. $f(x)$ is obviously irreducible over $mathbb{Z_3}$. Let $a$ be a root in an extension field of $mathbb{Z_3}$, then why is it that $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? What is the basis of $mathbb{Z_3}(a)$ over $mathbb{Z_3}$?



I know that $mathbb{Z_3}(a) simeq mathbb{Z_3}[x]/<f(x)>$ and since $f(x)$ is irreducible in $mathbb{Z_3}$, any polynomial in $mathbb{Z_3}[x]$ can have degree atmost 2. But I don't understand how that ties to $[mathbb{Z_3}(a):mathbb{Z_3}] = 3$? And how does that imply $mathbb{Z_3}(a)simeq GF(3^3)$? Thanks.







abstract-algebra galois-theory finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 21 at 21:38









manifoldedmanifolded

53419




53419












  • $begingroup$
    See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15".
    $endgroup$
    – Lucas Corrêa
    Apr 21 at 21:44


















  • $begingroup$
    See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15".
    $endgroup$
    – Lucas Corrêa
    Apr 21 at 21:44
















$begingroup$
See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15".
$endgroup$
– Lucas Corrêa
Apr 21 at 21:44




$begingroup$
See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15".
$endgroup$
– Lucas Corrêa
Apr 21 at 21:44










2 Answers
2






active

oldest

votes


















2












$begingroup$

In general, the degree of $F(alpha)$ over $F$ is the degree of the minimal polynomial of $alpha$. In this case, the minimal polynomial is $f(x)=x^3+2x+2$ which has degree $3$. The basis is ${1,alpha,alpha^2}$.



Think of it this way: $F(alpha)$ should consist of elements of the form $p(alpha)/q(alpha)$, where $p,q$ are polynomials. But using the relation $alpha^3=-2alpha-2$, you can see that every polynomial in $alpha$ can be written as a linear combinations of $1,alpha,alpha^2$. And even $alpha^{-1}$ can be written as such. That means every element of $F(alpha)$ is a linear combination of $1,alpha,alpha^2$.



Let $K=mathbb{F}_3(alpha)$. To see why $Ksimeq mathbb{F}_9$, it's just a cardinality argument: since $K$ is a $3$-dimensional $mathbb{F}_3$-vector space, we know from linear algebra that $Ksimeq mathbb{F}_3^3$ as vector spaces. The right hand side has 27 elements. So $K$ is the field of 27 elements.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
    $endgroup$
    – manifolded
    Apr 21 at 21:49










  • $begingroup$
    Thanks @egreg, my arithmetic is suspect.
    $endgroup$
    – Ehsaan
    Apr 21 at 21:58












  • $begingroup$
    @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
    $endgroup$
    – Ehsaan
    Apr 21 at 21:59



















1












$begingroup$

Look at the situation from a more abstract point of view. Let $F$ be a field and $f(x)in F[x]$ an irreducible monic polynomial.



If $a$ is a root of $f(x)$ in some extension field $K$ of $F$, then, if $F(a)$ denotes the smallest subfield of $K$ containing $F$ and $a$, we have
$$
F(a)cong F[x]/langle f(x)rangle
$$

and moreover $F[a]$, the smallest subring of $K$ containing $F$ and $a$ is the same as $F(a)$. Therefore we can see $F(a)=F[a]={g(a):g(x)in F[x]}$.



On the other hand, as $f(a)=0$, given $g(x)in F[x]$, we can perform the division and write $g(x)=f(x)q(x)+r(x)$, where $r$ has degree less than the degree of $f$. Thus we also have
$$
F(a)=F[a]={g(a):g(x)in F[x],deg g<deg f} tag{*}
$$

which is probably what you refer to by saying “any polynomial in $mathbb{Z}_3[x]$ can have degree at most $2$” (which isn't a good way to express the fact).



Now, suppose $g(x)$ is a monic polynomial satisfying $g(a)=0$. Take $g$ of minimal degree. Since we can perform the division $f(x)=g(x)q(x)+r(x)$, the assumptions give us that $r(a)=0$; by minimality of $deg g$, we infer that $r(x)=0$. Therefore $g$ divides $f$. Since $f$ is irreducible, we deduce that $g(x)=f(x)$ (they can differ up to a nonzero multiplicative constant, but being both monic, the constant is $1$).



Hence $f(x)$ is the minimal polynomial of $a$.



Now we can see that the set ${1,a,a^2,dots,a^{n-1}}$ (where $n=deg f$) is a basis of $F[a]$ as a vector space over $F$. The fact it is a spanning set follows from (*); it is linearly independent because $f$ is the minimal polynomial and a linear combination of those elements is the value of a polynomial of lesser degree than $f$, so it cannot vanish unless all the coefficients are zero.



Finally apply this to your particular case: $mathbb{Z}_3[a]$ is a three-dimensional vector space over $mathbb{Z}_3$, so it has $3^3=27$ elements.






share|cite|improve this answer









$endgroup$














    Your Answer








    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%2f3196381%2fa-question-about-the-degree-of-an-extension-field%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









    2












    $begingroup$

    In general, the degree of $F(alpha)$ over $F$ is the degree of the minimal polynomial of $alpha$. In this case, the minimal polynomial is $f(x)=x^3+2x+2$ which has degree $3$. The basis is ${1,alpha,alpha^2}$.



    Think of it this way: $F(alpha)$ should consist of elements of the form $p(alpha)/q(alpha)$, where $p,q$ are polynomials. But using the relation $alpha^3=-2alpha-2$, you can see that every polynomial in $alpha$ can be written as a linear combinations of $1,alpha,alpha^2$. And even $alpha^{-1}$ can be written as such. That means every element of $F(alpha)$ is a linear combination of $1,alpha,alpha^2$.



    Let $K=mathbb{F}_3(alpha)$. To see why $Ksimeq mathbb{F}_9$, it's just a cardinality argument: since $K$ is a $3$-dimensional $mathbb{F}_3$-vector space, we know from linear algebra that $Ksimeq mathbb{F}_3^3$ as vector spaces. The right hand side has 27 elements. So $K$ is the field of 27 elements.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
      $endgroup$
      – manifolded
      Apr 21 at 21:49










    • $begingroup$
      Thanks @egreg, my arithmetic is suspect.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:58












    • $begingroup$
      @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:59
















    2












    $begingroup$

    In general, the degree of $F(alpha)$ over $F$ is the degree of the minimal polynomial of $alpha$. In this case, the minimal polynomial is $f(x)=x^3+2x+2$ which has degree $3$. The basis is ${1,alpha,alpha^2}$.



    Think of it this way: $F(alpha)$ should consist of elements of the form $p(alpha)/q(alpha)$, where $p,q$ are polynomials. But using the relation $alpha^3=-2alpha-2$, you can see that every polynomial in $alpha$ can be written as a linear combinations of $1,alpha,alpha^2$. And even $alpha^{-1}$ can be written as such. That means every element of $F(alpha)$ is a linear combination of $1,alpha,alpha^2$.



    Let $K=mathbb{F}_3(alpha)$. To see why $Ksimeq mathbb{F}_9$, it's just a cardinality argument: since $K$ is a $3$-dimensional $mathbb{F}_3$-vector space, we know from linear algebra that $Ksimeq mathbb{F}_3^3$ as vector spaces. The right hand side has 27 elements. So $K$ is the field of 27 elements.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
      $endgroup$
      – manifolded
      Apr 21 at 21:49










    • $begingroup$
      Thanks @egreg, my arithmetic is suspect.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:58












    • $begingroup$
      @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:59














    2












    2








    2





    $begingroup$

    In general, the degree of $F(alpha)$ over $F$ is the degree of the minimal polynomial of $alpha$. In this case, the minimal polynomial is $f(x)=x^3+2x+2$ which has degree $3$. The basis is ${1,alpha,alpha^2}$.



    Think of it this way: $F(alpha)$ should consist of elements of the form $p(alpha)/q(alpha)$, where $p,q$ are polynomials. But using the relation $alpha^3=-2alpha-2$, you can see that every polynomial in $alpha$ can be written as a linear combinations of $1,alpha,alpha^2$. And even $alpha^{-1}$ can be written as such. That means every element of $F(alpha)$ is a linear combination of $1,alpha,alpha^2$.



    Let $K=mathbb{F}_3(alpha)$. To see why $Ksimeq mathbb{F}_9$, it's just a cardinality argument: since $K$ is a $3$-dimensional $mathbb{F}_3$-vector space, we know from linear algebra that $Ksimeq mathbb{F}_3^3$ as vector spaces. The right hand side has 27 elements. So $K$ is the field of 27 elements.






    share|cite|improve this answer











    $endgroup$



    In general, the degree of $F(alpha)$ over $F$ is the degree of the minimal polynomial of $alpha$. In this case, the minimal polynomial is $f(x)=x^3+2x+2$ which has degree $3$. The basis is ${1,alpha,alpha^2}$.



    Think of it this way: $F(alpha)$ should consist of elements of the form $p(alpha)/q(alpha)$, where $p,q$ are polynomials. But using the relation $alpha^3=-2alpha-2$, you can see that every polynomial in $alpha$ can be written as a linear combinations of $1,alpha,alpha^2$. And even $alpha^{-1}$ can be written as such. That means every element of $F(alpha)$ is a linear combination of $1,alpha,alpha^2$.



    Let $K=mathbb{F}_3(alpha)$. To see why $Ksimeq mathbb{F}_9$, it's just a cardinality argument: since $K$ is a $3$-dimensional $mathbb{F}_3$-vector space, we know from linear algebra that $Ksimeq mathbb{F}_3^3$ as vector spaces. The right hand side has 27 elements. So $K$ is the field of 27 elements.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 21 at 21:56

























    answered Apr 21 at 21:42









    EhsaanEhsaan

    1,120514




    1,120514












    • $begingroup$
      I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
      $endgroup$
      – manifolded
      Apr 21 at 21:49










    • $begingroup$
      Thanks @egreg, my arithmetic is suspect.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:58












    • $begingroup$
      @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:59


















    • $begingroup$
      I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
      $endgroup$
      – manifolded
      Apr 21 at 21:49










    • $begingroup$
      Thanks @egreg, my arithmetic is suspect.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:58












    • $begingroup$
      @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
      $endgroup$
      – Ehsaan
      Apr 21 at 21:59
















    $begingroup$
    I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
    $endgroup$
    – manifolded
    Apr 21 at 21:49




    $begingroup$
    I see, thanks. Why is $f(x)$ the minimal polynomial for $alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $alpha$?
    $endgroup$
    – manifolded
    Apr 21 at 21:49












    $begingroup$
    Thanks @egreg, my arithmetic is suspect.
    $endgroup$
    – Ehsaan
    Apr 21 at 21:58






    $begingroup$
    Thanks @egreg, my arithmetic is suspect.
    $endgroup$
    – Ehsaan
    Apr 21 at 21:58














    $begingroup$
    @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
    $endgroup$
    – Ehsaan
    Apr 21 at 21:59




    $begingroup$
    @manifolded: The minimal polynomial of $alpha$ has the property that it generates the ideal of all polynomials which vanish at $alpha$. It is the unique (monic) irreducible polynomial with $alpha$ as a root.
    $endgroup$
    – Ehsaan
    Apr 21 at 21:59











    1












    $begingroup$

    Look at the situation from a more abstract point of view. Let $F$ be a field and $f(x)in F[x]$ an irreducible monic polynomial.



    If $a$ is a root of $f(x)$ in some extension field $K$ of $F$, then, if $F(a)$ denotes the smallest subfield of $K$ containing $F$ and $a$, we have
    $$
    F(a)cong F[x]/langle f(x)rangle
    $$

    and moreover $F[a]$, the smallest subring of $K$ containing $F$ and $a$ is the same as $F(a)$. Therefore we can see $F(a)=F[a]={g(a):g(x)in F[x]}$.



    On the other hand, as $f(a)=0$, given $g(x)in F[x]$, we can perform the division and write $g(x)=f(x)q(x)+r(x)$, where $r$ has degree less than the degree of $f$. Thus we also have
    $$
    F(a)=F[a]={g(a):g(x)in F[x],deg g<deg f} tag{*}
    $$

    which is probably what you refer to by saying “any polynomial in $mathbb{Z}_3[x]$ can have degree at most $2$” (which isn't a good way to express the fact).



    Now, suppose $g(x)$ is a monic polynomial satisfying $g(a)=0$. Take $g$ of minimal degree. Since we can perform the division $f(x)=g(x)q(x)+r(x)$, the assumptions give us that $r(a)=0$; by minimality of $deg g$, we infer that $r(x)=0$. Therefore $g$ divides $f$. Since $f$ is irreducible, we deduce that $g(x)=f(x)$ (they can differ up to a nonzero multiplicative constant, but being both monic, the constant is $1$).



    Hence $f(x)$ is the minimal polynomial of $a$.



    Now we can see that the set ${1,a,a^2,dots,a^{n-1}}$ (where $n=deg f$) is a basis of $F[a]$ as a vector space over $F$. The fact it is a spanning set follows from (*); it is linearly independent because $f$ is the minimal polynomial and a linear combination of those elements is the value of a polynomial of lesser degree than $f$, so it cannot vanish unless all the coefficients are zero.



    Finally apply this to your particular case: $mathbb{Z}_3[a]$ is a three-dimensional vector space over $mathbb{Z}_3$, so it has $3^3=27$ elements.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Look at the situation from a more abstract point of view. Let $F$ be a field and $f(x)in F[x]$ an irreducible monic polynomial.



      If $a$ is a root of $f(x)$ in some extension field $K$ of $F$, then, if $F(a)$ denotes the smallest subfield of $K$ containing $F$ and $a$, we have
      $$
      F(a)cong F[x]/langle f(x)rangle
      $$

      and moreover $F[a]$, the smallest subring of $K$ containing $F$ and $a$ is the same as $F(a)$. Therefore we can see $F(a)=F[a]={g(a):g(x)in F[x]}$.



      On the other hand, as $f(a)=0$, given $g(x)in F[x]$, we can perform the division and write $g(x)=f(x)q(x)+r(x)$, where $r$ has degree less than the degree of $f$. Thus we also have
      $$
      F(a)=F[a]={g(a):g(x)in F[x],deg g<deg f} tag{*}
      $$

      which is probably what you refer to by saying “any polynomial in $mathbb{Z}_3[x]$ can have degree at most $2$” (which isn't a good way to express the fact).



      Now, suppose $g(x)$ is a monic polynomial satisfying $g(a)=0$. Take $g$ of minimal degree. Since we can perform the division $f(x)=g(x)q(x)+r(x)$, the assumptions give us that $r(a)=0$; by minimality of $deg g$, we infer that $r(x)=0$. Therefore $g$ divides $f$. Since $f$ is irreducible, we deduce that $g(x)=f(x)$ (they can differ up to a nonzero multiplicative constant, but being both monic, the constant is $1$).



      Hence $f(x)$ is the minimal polynomial of $a$.



      Now we can see that the set ${1,a,a^2,dots,a^{n-1}}$ (where $n=deg f$) is a basis of $F[a]$ as a vector space over $F$. The fact it is a spanning set follows from (*); it is linearly independent because $f$ is the minimal polynomial and a linear combination of those elements is the value of a polynomial of lesser degree than $f$, so it cannot vanish unless all the coefficients are zero.



      Finally apply this to your particular case: $mathbb{Z}_3[a]$ is a three-dimensional vector space over $mathbb{Z}_3$, so it has $3^3=27$ elements.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Look at the situation from a more abstract point of view. Let $F$ be a field and $f(x)in F[x]$ an irreducible monic polynomial.



        If $a$ is a root of $f(x)$ in some extension field $K$ of $F$, then, if $F(a)$ denotes the smallest subfield of $K$ containing $F$ and $a$, we have
        $$
        F(a)cong F[x]/langle f(x)rangle
        $$

        and moreover $F[a]$, the smallest subring of $K$ containing $F$ and $a$ is the same as $F(a)$. Therefore we can see $F(a)=F[a]={g(a):g(x)in F[x]}$.



        On the other hand, as $f(a)=0$, given $g(x)in F[x]$, we can perform the division and write $g(x)=f(x)q(x)+r(x)$, where $r$ has degree less than the degree of $f$. Thus we also have
        $$
        F(a)=F[a]={g(a):g(x)in F[x],deg g<deg f} tag{*}
        $$

        which is probably what you refer to by saying “any polynomial in $mathbb{Z}_3[x]$ can have degree at most $2$” (which isn't a good way to express the fact).



        Now, suppose $g(x)$ is a monic polynomial satisfying $g(a)=0$. Take $g$ of minimal degree. Since we can perform the division $f(x)=g(x)q(x)+r(x)$, the assumptions give us that $r(a)=0$; by minimality of $deg g$, we infer that $r(x)=0$. Therefore $g$ divides $f$. Since $f$ is irreducible, we deduce that $g(x)=f(x)$ (they can differ up to a nonzero multiplicative constant, but being both monic, the constant is $1$).



        Hence $f(x)$ is the minimal polynomial of $a$.



        Now we can see that the set ${1,a,a^2,dots,a^{n-1}}$ (where $n=deg f$) is a basis of $F[a]$ as a vector space over $F$. The fact it is a spanning set follows from (*); it is linearly independent because $f$ is the minimal polynomial and a linear combination of those elements is the value of a polynomial of lesser degree than $f$, so it cannot vanish unless all the coefficients are zero.



        Finally apply this to your particular case: $mathbb{Z}_3[a]$ is a three-dimensional vector space over $mathbb{Z}_3$, so it has $3^3=27$ elements.






        share|cite|improve this answer









        $endgroup$



        Look at the situation from a more abstract point of view. Let $F$ be a field and $f(x)in F[x]$ an irreducible monic polynomial.



        If $a$ is a root of $f(x)$ in some extension field $K$ of $F$, then, if $F(a)$ denotes the smallest subfield of $K$ containing $F$ and $a$, we have
        $$
        F(a)cong F[x]/langle f(x)rangle
        $$

        and moreover $F[a]$, the smallest subring of $K$ containing $F$ and $a$ is the same as $F(a)$. Therefore we can see $F(a)=F[a]={g(a):g(x)in F[x]}$.



        On the other hand, as $f(a)=0$, given $g(x)in F[x]$, we can perform the division and write $g(x)=f(x)q(x)+r(x)$, where $r$ has degree less than the degree of $f$. Thus we also have
        $$
        F(a)=F[a]={g(a):g(x)in F[x],deg g<deg f} tag{*}
        $$

        which is probably what you refer to by saying “any polynomial in $mathbb{Z}_3[x]$ can have degree at most $2$” (which isn't a good way to express the fact).



        Now, suppose $g(x)$ is a monic polynomial satisfying $g(a)=0$. Take $g$ of minimal degree. Since we can perform the division $f(x)=g(x)q(x)+r(x)$, the assumptions give us that $r(a)=0$; by minimality of $deg g$, we infer that $r(x)=0$. Therefore $g$ divides $f$. Since $f$ is irreducible, we deduce that $g(x)=f(x)$ (they can differ up to a nonzero multiplicative constant, but being both monic, the constant is $1$).



        Hence $f(x)$ is the minimal polynomial of $a$.



        Now we can see that the set ${1,a,a^2,dots,a^{n-1}}$ (where $n=deg f$) is a basis of $F[a]$ as a vector space over $F$. The fact it is a spanning set follows from (*); it is linearly independent because $f$ is the minimal polynomial and a linear combination of those elements is the value of a polynomial of lesser degree than $f$, so it cannot vanish unless all the coefficients are zero.



        Finally apply this to your particular case: $mathbb{Z}_3[a]$ is a three-dimensional vector space over $mathbb{Z}_3$, so it has $3^3=27$ elements.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 21 at 22:11









        egregegreg

        186k1486209




        186k1486209






























            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%2f3196381%2fa-question-about-the-degree-of-an-extension-field%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

            Puebla de Zaragoza

            Musa