Extension of Splitting Fields over An Arbitrary Field












4












$begingroup$


Let $F$ be a field in which $0 neq2$ in $F$, and consider $f=x^4+1$. If $E$ is the splitting field for $f$ over $F$, it turns out that $E$ is a simple extension of $F$. How does one realize this fact? I'm not so sure as to what field element I can adjoin to $F$ to allow $f$ to split into linear factors. Finding the splitting field over something like $mathbb{Q}$ is straight forward and easy in comparison, but I'm having trouble working with any general field $F$.



Also, if we indeed did have that $0=2$ in our field $F$, then $f=x^4+1=(x+1)^4$, so $F$ is its own splitting field, is this correct reasoning?






abstract-algebra field-theory extension-field splitting-field







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$endgroup$








  • 3




    $begingroup$
    Letting $alpha$ be any root, then $f$ splits as $(x-alpha)(x+alpha)(x-alpha^3)(x+alpha^3)$ in $F[alpha]$.
    $endgroup$
    – Mike Earnest
    yesterday


















4












$begingroup$


Let $F$ be a field in which $0 neq2$ in $F$, and consider $f=x^4+1$. If $E$ is the splitting field for $f$ over $F$, it turns out that $E$ is a simple extension of $F$. How does one realize this fact? I'm not so sure as to what field element I can adjoin to $F$ to allow $f$ to split into linear factors. Finding the splitting field over something like $mathbb{Q}$ is straight forward and easy in comparison, but I'm having trouble working with any general field $F$.



Also, if we indeed did have that $0=2$ in our field $F$, then $f=x^4+1=(x+1)^4$, so $F$ is its own splitting field, is this correct reasoning?






abstract-algebra field-theory extension-field splitting-field







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Letting $alpha$ be any root, then $f$ splits as $(x-alpha)(x+alpha)(x-alpha^3)(x+alpha^3)$ in $F[alpha]$.
    $endgroup$
    – Mike Earnest
    yesterday
















4












4








4


0



$begingroup$


Let $F$ be a field in which $0 neq2$ in $F$, and consider $f=x^4+1$. If $E$ is the splitting field for $f$ over $F$, it turns out that $E$ is a simple extension of $F$. How does one realize this fact? I'm not so sure as to what field element I can adjoin to $F$ to allow $f$ to split into linear factors. Finding the splitting field over something like $mathbb{Q}$ is straight forward and easy in comparison, but I'm having trouble working with any general field $F$.



Also, if we indeed did have that $0=2$ in our field $F$, then $f=x^4+1=(x+1)^4$, so $F$ is its own splitting field, is this correct reasoning?






abstract-algebra field-theory extension-field splitting-field







share|cite|improve this question









$endgroup$




Let $F$ be a field in which $0 neq2$ in $F$, and consider $f=x^4+1$. If $E$ is the splitting field for $f$ over $F$, it turns out that $E$ is a simple extension of $F$. How does one realize this fact? I'm not so sure as to what field element I can adjoin to $F$ to allow $f$ to split into linear factors. Finding the splitting field over something like $mathbb{Q}$ is straight forward and easy in comparison, but I'm having trouble working with any general field $F$.



Also, if we indeed did have that $0=2$ in our field $F$, then $f=x^4+1=(x+1)^4$, so $F$ is its own splitting field, is this correct reasoning?






abstract-algebra field-theory extension-field splitting-field




abstract-algebra field-theory extension-field splitting-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









DevilofHell'sKitchenDevilofHell'sKitchen

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  • 3




    $begingroup$
    Letting $alpha$ be any root, then $f$ splits as $(x-alpha)(x+alpha)(x-alpha^3)(x+alpha^3)$ in $F[alpha]$.
    $endgroup$
    – Mike Earnest
    yesterday
















  • 3




    $begingroup$
    Letting $alpha$ be any root, then $f$ splits as $(x-alpha)(x+alpha)(x-alpha^3)(x+alpha^3)$ in $F[alpha]$.
    $endgroup$
    – Mike Earnest
    yesterday










3




3




$begingroup$
Letting $alpha$ be any root, then $f$ splits as $(x-alpha)(x+alpha)(x-alpha^3)(x+alpha^3)$ in $F[alpha]$.
$endgroup$
– Mike Earnest
yesterday






$begingroup$
Letting $alpha$ be any root, then $f$ splits as $(x-alpha)(x+alpha)(x-alpha^3)(x+alpha^3)$ in $F[alpha]$.
$endgroup$
– Mike Earnest
yesterday












1 Answer
1






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4












$begingroup$

If $theta$ is a root of $x^4+1$, then so are $theta,theta^3,theta^5,theta^7$.
These are all different because $theta$ has order $8$ in $E^times$, since $theta^4=-1ne1$.
Therefore, $x^4+1$ splits in $F(theta)$.



If $operatorname{char}(F)=2$, then $x^4+1=(x+1)^4$ and $E=F=F(1)$, also a simple extension.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    And those powers of $theta$ are distinct elements of the field.
    $endgroup$
    – Gerry Myerson
    yesterday






  • 1




    $begingroup$
    @GerryMyerson, thanks, fixed.
    $endgroup$
    – lhf
    18 hours ago











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4












$begingroup$

If $theta$ is a root of $x^4+1$, then so are $theta,theta^3,theta^5,theta^7$.
These are all different because $theta$ has order $8$ in $E^times$, since $theta^4=-1ne1$.
Therefore, $x^4+1$ splits in $F(theta)$.



If $operatorname{char}(F)=2$, then $x^4+1=(x+1)^4$ and $E=F=F(1)$, also a simple extension.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    And those powers of $theta$ are distinct elements of the field.
    $endgroup$
    – Gerry Myerson
    yesterday






  • 1




    $begingroup$
    @GerryMyerson, thanks, fixed.
    $endgroup$
    – lhf
    18 hours ago
















4












$begingroup$

If $theta$ is a root of $x^4+1$, then so are $theta,theta^3,theta^5,theta^7$.
These are all different because $theta$ has order $8$ in $E^times$, since $theta^4=-1ne1$.
Therefore, $x^4+1$ splits in $F(theta)$.



If $operatorname{char}(F)=2$, then $x^4+1=(x+1)^4$ and $E=F=F(1)$, also a simple extension.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    And those powers of $theta$ are distinct elements of the field.
    $endgroup$
    – Gerry Myerson
    yesterday






  • 1




    $begingroup$
    @GerryMyerson, thanks, fixed.
    $endgroup$
    – lhf
    18 hours ago














4












4








4





$begingroup$

If $theta$ is a root of $x^4+1$, then so are $theta,theta^3,theta^5,theta^7$.
These are all different because $theta$ has order $8$ in $E^times$, since $theta^4=-1ne1$.
Therefore, $x^4+1$ splits in $F(theta)$.



If $operatorname{char}(F)=2$, then $x^4+1=(x+1)^4$ and $E=F=F(1)$, also a simple extension.






share|cite|improve this answer











$endgroup$



If $theta$ is a root of $x^4+1$, then so are $theta,theta^3,theta^5,theta^7$.
These are all different because $theta$ has order $8$ in $E^times$, since $theta^4=-1ne1$.
Therefore, $x^4+1$ splits in $F(theta)$.



If $operatorname{char}(F)=2$, then $x^4+1=(x+1)^4$ and $E=F=F(1)$, also a simple extension.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 18 hours ago

























answered yesterday









lhflhf

166k10171400




166k10171400








  • 3




    $begingroup$
    And those powers of $theta$ are distinct elements of the field.
    $endgroup$
    – Gerry Myerson
    yesterday






  • 1




    $begingroup$
    @GerryMyerson, thanks, fixed.
    $endgroup$
    – lhf
    18 hours ago














  • 3




    $begingroup$
    And those powers of $theta$ are distinct elements of the field.
    $endgroup$
    – Gerry Myerson
    yesterday






  • 1




    $begingroup$
    @GerryMyerson, thanks, fixed.
    $endgroup$
    – lhf
    18 hours ago








3




3




$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
yesterday




$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
yesterday




1




1




$begingroup$
@GerryMyerson, thanks, fixed.
$endgroup$
– lhf
18 hours ago




$begingroup$
@GerryMyerson, thanks, fixed.
$endgroup$
– lhf
18 hours ago


















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