Mixing
A concept about normal extension and splitting field
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Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)
Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?
abstract-algebra extension-field
asked yesterday
Eric
1,737515
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up vote
2
down vote
favorite
Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)
Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?
abstract-algebra extension-field
asked yesterday
Eric
1,737515
Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday
Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday
$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)
Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?
abstract-algebra extension-field
asked yesterday
Eric
1,737515
Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)
Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?
abstract-algebra extension-field
abstract-algebra extension-field
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
edited yesterday
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
asked yesterday
Eric
1,737515
1,737515
Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday
Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday
$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday
add a comment |
Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday
Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday
$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday
Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday
Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday
Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday
Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday
$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday
$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday
add a comment |
2 Answers
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3
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No.
First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.
If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.
answered yesterday
Praneet Srivastava
762516
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up vote
0
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They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.
answered yesterday
Richard Martin
1,3238
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
No.
First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.
If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.
answered yesterday
Praneet Srivastava
762516
add a comment |
up vote
3
down vote
No.
First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.
If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.
answered yesterday
Praneet Srivastava
762516
add a comment |
up vote
3
down vote
up vote
3
down vote
No.
First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.
If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.
answered yesterday
Praneet Srivastava
762516
No.
First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.
If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.
answered yesterday
Praneet Srivastava
762516
edited yesterday
answered yesterday
Praneet Srivastava
762516
answered yesterday
Praneet Srivastava
762516
answered yesterday
Praneet Srivastava
762516
762516
add a comment |
add a comment |
up vote
0
down vote
They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
0
down vote
They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
0
down vote
up vote
0
down vote
They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.
answered yesterday
Richard Martin
1,3238
They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
1,3238
add a comment |
add a comment |
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Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday
Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday
$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday