Extension of Splitting Fields over An Arbitrary Field
$begingroup$
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
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
$endgroup$
2
$begingroup$
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
58 mins ago
add a comment |
$begingroup$
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
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
$endgroup$
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
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 1 hour ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
405
2
$begingroup$
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
58 mins ago
add a comment |
2
$begingroup$
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
58 mins ago
2
2
$begingroup$
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
58 mins ago
$begingroup$
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
58 mins ago
add a comment |
1 Answer
1
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$begingroup$
If $theta$ is a root of $x^4+1$, then so are $theta^k$ for $k=1,3,5,7$, and so $x^4+1$ splits in $F(theta)$.
answered 1 hour ago
lhflhf
166k10171400
$endgroup$
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $theta$ is a root of $x^4+1$, then so are $theta^k$ for $k=1,3,5,7$, and so $x^4+1$ splits in $F(theta)$.
answered 1 hour ago
lhflhf
166k10171400
$endgroup$
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
add a comment |
$begingroup$
If $theta$ is a root of $x^4+1$, then so are $theta^k$ for $k=1,3,5,7$, and so $x^4+1$ splits in $F(theta)$.
answered 1 hour ago
lhflhf
166k10171400
$endgroup$
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
add a comment |
$begingroup$
If $theta$ is a root of $x^4+1$, then so are $theta^k$ for $k=1,3,5,7$, and so $x^4+1$ splits in $F(theta)$.
answered 1 hour ago
lhflhf
166k10171400
$endgroup$
If $theta$ is a root of $x^4+1$, then so are $theta^k$ for $k=1,3,5,7$, and so $x^4+1$ splits in $F(theta)$.
answered 1 hour ago
lhflhf
166k10171400
answered 1 hour ago
lhflhf
166k10171400
answered 1 hour ago
lhflhf
166k10171400
answered 1 hour ago
lhflhf
166k10171400
166k10171400
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
add a comment |
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
2
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
50 mins ago
add a comment |
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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
58 mins ago