Degree of a field extension of a minimal polynomial

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Suppose a polynomial $f(x)$ of degree $n$ over $mathbbQ$ is the minimal polynomial of an element $alpha$ in an extension field of $mathbbQ$. Is $[mathbbQ(alpha):mathbbQ]=n$? Please give reason(s) for your guess.




extension-field minimal-polynomials




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    That's perfectly correct.
    – Bernard
    Jul 29 at 10:08














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Suppose a polynomial $f(x)$ of degree $n$ over $mathbbQ$ is the minimal polynomial of an element $alpha$ in an extension field of $mathbbQ$. Is $[mathbbQ(alpha):mathbbQ]=n$? Please give reason(s) for your guess.




extension-field minimal-polynomials




share|cite|improve this question

















  • 1




    That's perfectly correct.
    – Bernard
    Jul 29 at 10:08












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Suppose a polynomial $f(x)$ of degree $n$ over $mathbbQ$ is the minimal polynomial of an element $alpha$ in an extension field of $mathbbQ$. Is $[mathbbQ(alpha):mathbbQ]=n$? Please give reason(s) for your guess.




extension-field minimal-polynomials




share|cite|improve this question













Suppose a polynomial $f(x)$ of degree $n$ over $mathbbQ$ is the minimal polynomial of an element $alpha$ in an extension field of $mathbbQ$. Is $[mathbbQ(alpha):mathbbQ]=n$? Please give reason(s) for your guess.





extension-field minimal-polynomials





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edited Jul 29 at 10:11
























asked Jul 29 at 10:07









user319994

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385







  • 1




    That's perfectly correct.
    – Bernard
    Jul 29 at 10:08












  • 1




    That's perfectly correct.
    – Bernard
    Jul 29 at 10:08







1




1




That's perfectly correct.
– Bernard
Jul 29 at 10:08




That's perfectly correct.
– Bernard
Jul 29 at 10:08










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First $mathbf Q[α]=p(α)mid deg p<n$.



Indeed,we can divide any polynomial $p(x)$ by $f(x)$ by Euclidean division:
beginalign
p(x)&=q(x)f(x)+r(x),quad& r&=0;textor;deg r<deg f=n.
endalign



Thus $p(α)=r(α)$ and $mathbf Q[α]$ is a finite-dimensional $bf Q$-vectorspace, of dimension $n$ since $f$ is the minimal polynomial of $α$.



Next, note that $mathbf Q(α)=mathbf Q[α]$.



Indeed, since $mathbf Q[α]$ is a finite dimensional $mathbf Q$-vector space, so that multiplication by a non-zero element of $mathbf Q[α]$, which is injective, is also surjective, i.e. $1$ is attained – in other words this non-zero element has in inverse $mathbf Q[α]$, which is therefore a field.






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    First $mathbf Q[α]=p(α)mid deg p<n$.



    Indeed,we can divide any polynomial $p(x)$ by $f(x)$ by Euclidean division:
    beginalign
    p(x)&=q(x)f(x)+r(x),quad& r&=0;textor;deg r<deg f=n.
    endalign



    Thus $p(α)=r(α)$ and $mathbf Q[α]$ is a finite-dimensional $bf Q$-vectorspace, of dimension $n$ since $f$ is the minimal polynomial of $α$.



    Next, note that $mathbf Q(α)=mathbf Q[α]$.



    Indeed, since $mathbf Q[α]$ is a finite dimensional $mathbf Q$-vector space, so that multiplication by a non-zero element of $mathbf Q[α]$, which is injective, is also surjective, i.e. $1$ is attained – in other words this non-zero element has in inverse $mathbf Q[α]$, which is therefore a field.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      First $mathbf Q[α]=p(α)mid deg p<n$.



      Indeed,we can divide any polynomial $p(x)$ by $f(x)$ by Euclidean division:
      beginalign
      p(x)&=q(x)f(x)+r(x),quad& r&=0;textor;deg r<deg f=n.
      endalign



      Thus $p(α)=r(α)$ and $mathbf Q[α]$ is a finite-dimensional $bf Q$-vectorspace, of dimension $n$ since $f$ is the minimal polynomial of $α$.



      Next, note that $mathbf Q(α)=mathbf Q[α]$.



      Indeed, since $mathbf Q[α]$ is a finite dimensional $mathbf Q$-vector space, so that multiplication by a non-zero element of $mathbf Q[α]$, which is injective, is also surjective, i.e. $1$ is attained – in other words this non-zero element has in inverse $mathbf Q[α]$, which is therefore a field.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        First $mathbf Q[α]=p(α)mid deg p<n$.



        Indeed,we can divide any polynomial $p(x)$ by $f(x)$ by Euclidean division:
        beginalign
        p(x)&=q(x)f(x)+r(x),quad& r&=0;textor;deg r<deg f=n.
        endalign



        Thus $p(α)=r(α)$ and $mathbf Q[α]$ is a finite-dimensional $bf Q$-vectorspace, of dimension $n$ since $f$ is the minimal polynomial of $α$.



        Next, note that $mathbf Q(α)=mathbf Q[α]$.



        Indeed, since $mathbf Q[α]$ is a finite dimensional $mathbf Q$-vector space, so that multiplication by a non-zero element of $mathbf Q[α]$, which is injective, is also surjective, i.e. $1$ is attained – in other words this non-zero element has in inverse $mathbf Q[α]$, which is therefore a field.






        share|cite|improve this answer













        First $mathbf Q[α]=p(α)mid deg p<n$.



        Indeed,we can divide any polynomial $p(x)$ by $f(x)$ by Euclidean division:
        beginalign
        p(x)&=q(x)f(x)+r(x),quad& r&=0;textor;deg r<deg f=n.
        endalign



        Thus $p(α)=r(α)$ and $mathbf Q[α]$ is a finite-dimensional $bf Q$-vectorspace, of dimension $n$ since $f$ is the minimal polynomial of $α$.



        Next, note that $mathbf Q(α)=mathbf Q[α]$.



        Indeed, since $mathbf Q[α]$ is a finite dimensional $mathbf Q$-vector space, so that multiplication by a non-zero element of $mathbf Q[α]$, which is injective, is also surjective, i.e. $1$ is attained – in other words this non-zero element has in inverse $mathbf Q[α]$, which is therefore a field.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 29 at 10:48









        Bernard

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