Extension of Splitting Fields over An Arbitrary FieldSplitting field implies Galois extensionWhat does it mean to take the splitting field of $f(x)in F[x]$ over $K$ where $K/F$ is a field extensionCalculating Splitting Field Degree of ExtensionDetermining whether or not an extension is a splitting fieldElementary Field Theory: Extension Field of Degree 2Splitting field of $x^3 - 2$ over $mathbbF_5$Normal field extension implies splitting fieldSplitting fields and their degreesWhat things we have to take care of while finding the degree of field extension, splitting fields for some polynomial?A question on the definition of splitting field
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Extension of Splitting Fields over An Arbitrary Field
Splitting field implies Galois extensionWhat does it mean to take the splitting field of $f(x)in F[x]$ over $K$ where $K/F$ is a field extensionCalculating Splitting Field Degree of ExtensionDetermining whether or not an extension is a splitting fieldElementary Field Theory: Extension Field of Degree 2Splitting field of $x^3 - 2$ over $mathbbF_5$Normal field extension implies splitting fieldSplitting fields and their degreesWhat things we have to take care of while finding the degree of field extension, splitting fields for some polynomial?A question on the definition of splitting 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 $mathbbQ$ 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 2 hours 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 $mathbbQ$ 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 2 hours 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
2 hours 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 $mathbbQ$ 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 2 hours 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 $mathbbQ$ 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 2 hours ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 2 hours ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 2 hours ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 2 hours ago
DevilofHell'sKitchenDevilofHell'sKitchen
405
asked 2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 2 hours ago
lhflhf
166k10171400
$endgroup$
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours ago
add a comment |
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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 2 hours ago
lhflhf
166k10171400
$endgroup$
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours 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 2 hours ago
lhflhf
166k10171400
$endgroup$
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours 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 2 hours 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 2 hours ago
lhflhf
166k10171400
answered 2 hours ago
lhflhf
166k10171400
answered 2 hours ago
lhflhf
166k10171400
answered 2 hours ago
lhflhf
166k10171400
166k10171400
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours ago
add a comment |
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours ago
2
2
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours ago
$begingroup$
And those powers of $theta$ are distinct elements of the field.
$endgroup$
– Gerry Myerson
2 hours ago
add a comment |
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l7qkS7b0,XaXO2z14bfhuDgMN0h1fTPJef,qQaubZpquw l,syt4x80ZwApnoYT,D
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
2 hours ago