Prove if two normal extensions over a field are isomorphic The Next CEO of Stack...
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Prove if two normal extensions over a field are isomorphic
The Next CEO of Stack OverflowNormal closure of field extension, axiom of choiceIs every element in a finite splitting field K over F a root in a polynomial?Why are separable and normal field extensions so called?Is normal extension algebraic?$E/F$ is a normal extension iff $E$ is a splitting field for some polynomial $fin F[X]$.Thoughts on normal extensions.$K subset E subset L$ finite field extensions and $L$ normal over $K$. Is $L$ normal over $E$, and is $E$ normal over $K$?Normal field extensionsKaplansky's vs common definition of normal field extensionIsomorphic field extensions
$begingroup$
Suppose $K_1,K_2$ are normal extensions over a field $F$ . Suppose the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$ . Then are $K_1,K_2$ isomorphic ?
My attempt:
I have proved the fact that an algebraic extension $K |F$ is normal iff $text{min}_F (alpha)$ splits in $K[X] , forall alpha in K$ . Combining this with the hypothesis that the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$, it seems to me that in fact $K_1$ must equal $K_2 $, but I do not have proof.
Thanks in advance for the help!
abstract-algebra field-theory normal-extension
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
asked Mar 17 at 5:56
reflexivereflexive
1,178625
$endgroup$
add a comment |
$begingroup$
Suppose $K_1,K_2$ are normal extensions over a field $F$ . Suppose the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$ . Then are $K_1,K_2$ isomorphic ?
My attempt:
I have proved the fact that an algebraic extension $K |F$ is normal iff $text{min}_F (alpha)$ splits in $K[X] , forall alpha in K$ . Combining this with the hypothesis that the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$, it seems to me that in fact $K_1$ must equal $K_2 $, but I do not have proof.
Thanks in advance for the help!
abstract-algebra field-theory normal-extension
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
asked Mar 17 at 5:56
reflexivereflexive
1,178625
$endgroup$
$begingroup$
If $ain K_1$, then the minimal polynomail of $a$ is the minimal polynomial of some element of $K_2$; thus, the minimal polynomial of $a$ must split in $K_2$ (because it has at least one root in $K_2$), and so $ain K_2$.
$endgroup$
– Arturo Magidin
Mar 17 at 6:00
$begingroup$
Yeah! So in particular $K_1 = K_2$ ? Right?
$endgroup$
– reflexive
Mar 17 at 6:04
$begingroup$
(Note that you could in principle have $K_1$ and $K_2$ contained in “different” algebraic closures, so they need not be equal; but you can always map $K_1$ into, say, the algebraic closure of $K_2$ over $F$ so that you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of $F$, so that you would get equality rather than isomorphism).
$endgroup$
– Arturo Magidin
Mar 17 at 6:04
$begingroup$
In case, $K_1$ and $K_2$ are contained in two different Algebraic closures, can you please provide more details regarding isomorphism (without considering 'you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of F$ ) ?
$endgroup$
– reflexive
Mar 17 at 6:10
1
$begingroup$
One of the properties of normal extensions is that if $K_1$ is a normal extension of $F$, and $L$ is an algebraic closure of $F$, then there is an embedding of $K_1$ into $L$ over $F$. So you can always map $K_1$ into the algebraic closure of $K_2$ without “disturbing” $F$, and work with that isomorphic copy of $K_1$. That’s why the question asks about isomorphism rather than equality.
$endgroup$
– Arturo Magidin
Mar 17 at 6:12
add a comment |
$begingroup$
Suppose $K_1,K_2$ are normal extensions over a field $F$ . Suppose the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$ . Then are $K_1,K_2$ isomorphic ?
My attempt:
I have proved the fact that an algebraic extension $K |F$ is normal iff $text{min}_F (alpha)$ splits in $K[X] , forall alpha in K$ . Combining this with the hypothesis that the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$, it seems to me that in fact $K_1$ must equal $K_2 $, but I do not have proof.
Thanks in advance for the help!
abstract-algebra field-theory normal-extension
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
asked Mar 17 at 5:56
reflexivereflexive
1,178625
$endgroup$
Suppose $K_1,K_2$ are normal extensions over a field $F$ . Suppose the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$ . Then are $K_1,K_2$ isomorphic ?
My attempt:
I have proved the fact that an algebraic extension $K |F$ is normal iff $text{min}_F (alpha)$ splits in $K[X] , forall alpha in K$ . Combining this with the hypothesis that the set of minimal polynomials of elements in $K_1$ over $F$ = the set of minimal polynomials of elements in $K_2$ over $F$, it seems to me that in fact $K_1$ must equal $K_2 $, but I do not have proof.
Thanks in advance for the help!
abstract-algebra field-theory normal-extension
abstract-algebra field-theory normal-extension
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
asked Mar 17 at 5:56
reflexivereflexive
1,178625
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
asked Mar 17 at 5:56
reflexivereflexive
1,178625
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
edited Mar 17 at 7:23
YuiTo Cheng
2,1862937
2,1862937
asked Mar 17 at 5:56
reflexivereflexive
1,178625
asked Mar 17 at 5:56
reflexivereflexive
1,178625
asked Mar 17 at 5:56
reflexivereflexive
1,178625
1,178625
$begingroup$
If $ain K_1$, then the minimal polynomail of $a$ is the minimal polynomial of some element of $K_2$; thus, the minimal polynomial of $a$ must split in $K_2$ (because it has at least one root in $K_2$), and so $ain K_2$.
$endgroup$
– Arturo Magidin
Mar 17 at 6:00
$begingroup$
Yeah! So in particular $K_1 = K_2$ ? Right?
$endgroup$
– reflexive
Mar 17 at 6:04
$begingroup$
(Note that you could in principle have $K_1$ and $K_2$ contained in “different” algebraic closures, so they need not be equal; but you can always map $K_1$ into, say, the algebraic closure of $K_2$ over $F$ so that you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of $F$, so that you would get equality rather than isomorphism).
$endgroup$
– Arturo Magidin
Mar 17 at 6:04
$begingroup$
In case, $K_1$ and $K_2$ are contained in two different Algebraic closures, can you please provide more details regarding isomorphism (without considering 'you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of F$ ) ?
$endgroup$
– reflexive
Mar 17 at 6:10
1
$begingroup$
One of the properties of normal extensions is that if $K_1$ is a normal extension of $F$, and $L$ is an algebraic closure of $F$, then there is an embedding of $K_1$ into $L$ over $F$. So you can always map $K_1$ into the algebraic closure of $K_2$ without “disturbing” $F$, and work with that isomorphic copy of $K_1$. That’s why the question asks about isomorphism rather than equality.
$endgroup$
– Arturo Magidin
Mar 17 at 6:12
add a comment |
$begingroup$
If $ain K_1$, then the minimal polynomail of $a$ is the minimal polynomial of some element of $K_2$; thus, the minimal polynomial of $a$ must split in $K_2$ (because it has at least one root in $K_2$), and so $ain K_2$.
$endgroup$
– Arturo Magidin
Mar 17 at 6:00
$begingroup$
Yeah! So in particular $K_1 = K_2$ ? Right?
$endgroup$
– reflexive
Mar 17 at 6:04
$begingroup$
(Note that you could in principle have $K_1$ and $K_2$ contained in “different” algebraic closures, so they need not be equal; but you can always map $K_1$ into, say, the algebraic closure of $K_2$ over $F$ so that you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of $F$, so that you would get equality rather than isomorphism).
$endgroup$
– Arturo Magidin
Mar 17 at 6:04
$begingroup$
In case, $K_1$ and $K_2$ are contained in two different Algebraic closures, can you please provide more details regarding isomorphism (without considering 'you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of F$ ) ?
$endgroup$
– reflexive
Mar 17 at 6:10
1
$begingroup$
One of the properties of normal extensions is that if $K_1$ is a normal extension of $F$, and $L$ is an algebraic closure of $F$, then there is an embedding of $K_1$ into $L$ over $F$. So you can always map $K_1$ into the algebraic closure of $K_2$ without “disturbing” $F$, and work with that isomorphic copy of $K_1$. That’s why the question asks about isomorphism rather than equality.
$endgroup$
– Arturo Magidin
Mar 17 at 6:12
$begingroup$
If $ain K_1$, then the minimal polynomail of $a$ is the minimal polynomial of some element of $K_2$; thus, the minimal polynomial of $a$ must split in $K_2$ (because it has at least one root in $K_2$), and so $ain K_2$.
$endgroup$
– Arturo Magidin
Mar 17 at 6:00
$begingroup$
If $ain K_1$, then the minimal polynomail of $a$ is the minimal polynomial of some element of $K_2$; thus, the minimal polynomial of $a$ must split in $K_2$ (because it has at least one root in $K_2$), and so $ain K_2$.
$endgroup$
– Arturo Magidin
Mar 17 at 6:00
$begingroup$
Yeah! So in particular $K_1 = K_2$ ? Right?
$endgroup$
– reflexive
Mar 17 at 6:04
$begingroup$
Yeah! So in particular $K_1 = K_2$ ? Right?
$endgroup$
– reflexive
Mar 17 at 6:04
$begingroup$
(Note that you could in principle have $K_1$ and $K_2$ contained in “different” algebraic closures, so they need not be equal; but you can always map $K_1$ into, say, the algebraic closure of $K_2$ over $F$ so that you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of $F$, so that you would get equality rather than isomorphism).
$endgroup$
– Arturo Magidin
Mar 17 at 6:04
$begingroup$
(Note that you could in principle have $K_1$ and $K_2$ contained in “different” algebraic closures, so they need not be equal; but you can always map $K_1$ into, say, the algebraic closure of $K_2$ over $F$ so that you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of $F$, so that you would get equality rather than isomorphism).
$endgroup$
– Arturo Magidin
Mar 17 at 6:04
$begingroup$
In case, $K_1$ and $K_2$ are contained in two different Algebraic closures, can you please provide more details regarding isomorphism (without considering 'you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of F$ ) ?
$endgroup$
– reflexive
Mar 17 at 6:10
$begingroup$
In case, $K_1$ and $K_2$ are contained in two different Algebraic closures, can you please provide more details regarding isomorphism (without considering 'you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of F$ ) ?
$endgroup$
– reflexive
Mar 17 at 6:10
1
1
$begingroup$
One of the properties of normal extensions is that if $K_1$ is a normal extension of $F$, and $L$ is an algebraic closure of $F$, then there is an embedding of $K_1$ into $L$ over $F$. So you can always map $K_1$ into the algebraic closure of $K_2$ without “disturbing” $F$, and work with that isomorphic copy of $K_1$. That’s why the question asks about isomorphism rather than equality.
$endgroup$
– Arturo Magidin
Mar 17 at 6:12
$begingroup$
One of the properties of normal extensions is that if $K_1$ is a normal extension of $F$, and $L$ is an algebraic closure of $F$, then there is an embedding of $K_1$ into $L$ over $F$. So you can always map $K_1$ into the algebraic closure of $K_2$ without “disturbing” $F$, and work with that isomorphic copy of $K_1$. That’s why the question asks about isomorphism rather than equality.
$endgroup$
– Arturo Magidin
Mar 17 at 6:12
add a comment |
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$begingroup$
If $ain K_1$, then the minimal polynomail of $a$ is the minimal polynomial of some element of $K_2$; thus, the minimal polynomial of $a$ must split in $K_2$ (because it has at least one root in $K_2$), and so $ain K_2$.
$endgroup$
– Arturo Magidin
Mar 17 at 6:00
$begingroup$
Yeah! So in particular $K_1 = K_2$ ? Right?
$endgroup$
– reflexive
Mar 17 at 6:04
$begingroup$
(Note that you could in principle have $K_1$ and $K_2$ contained in “different” algebraic closures, so they need not be equal; but you can always map $K_1$ into, say, the algebraic closure of $K_2$ over $F$ so that you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of $F$, so that you would get equality rather than isomorphism).
$endgroup$
– Arturo Magidin
Mar 17 at 6:04
$begingroup$
In case, $K_1$ and $K_2$ are contained in two different Algebraic closures, can you please provide more details regarding isomorphism (without considering 'you may assume that $K_1$ and $K_2$ are in fact contained in a common overfield of F$ ) ?
$endgroup$
– reflexive
Mar 17 at 6:10
1
$begingroup$
One of the properties of normal extensions is that if $K_1$ is a normal extension of $F$, and $L$ is an algebraic closure of $F$, then there is an embedding of $K_1$ into $L$ over $F$. So you can always map $K_1$ into the algebraic closure of $K_2$ without “disturbing” $F$, and work with that isomorphic copy of $K_1$. That’s why the question asks about isomorphism rather than equality.
$endgroup$
– Arturo Magidin
Mar 17 at 6:12