Mixing
Field homomorphism to algebraic closure and Galois group
Clash Royale CLAN TAG#URR8PPP
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I read from my notes that let $alpha$ be algebraic over $K$. Consider $K(alpha)$ which has extension degree the same as $p=textirr(alpha,K)$ the irreducible polynomial of $alpha$ over $K$. Now, let $L$ be an algebraic closure of $K$. We claim that there are exactly as many homomorphisms $K(alpha)to L$ that fix $K$ as the number of roots of $p$ in $L$.
But this sounds weird to me because if we restrict to the image of those homomorphisms, the homomorphisms can be regarded as automorphisms from $K(alpha)$ to it. Then it is the same as asking its Galois group which is sometimes not the same as the number as the roots, usually not trivial.
For example, if we consider $X^3-2$ having 3 roots, whose Galois group is $S_6$ rather than a one with just 3 elements.
(Possibly, I got a fake note.)
abstract-algebra galois-theory
asked Jul 28 at 21:06
Xavier Yang
440314
add a comment |Â
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I read from my notes that let $alpha$ be algebraic over $K$. Consider $K(alpha)$ which has extension degree the same as $p=textirr(alpha,K)$ the irreducible polynomial of $alpha$ over $K$. Now, let $L$ be an algebraic closure of $K$. We claim that there are exactly as many homomorphisms $K(alpha)to L$ that fix $K$ as the number of roots of $p$ in $L$.
But this sounds weird to me because if we restrict to the image of those homomorphisms, the homomorphisms can be regarded as automorphisms from $K(alpha)$ to it. Then it is the same as asking its Galois group which is sometimes not the same as the number as the roots, usually not trivial.
For example, if we consider $X^3-2$ having 3 roots, whose Galois group is $S_6$ rather than a one with just 3 elements.
(Possibly, I got a fake note.)
abstract-algebra galois-theory
asked Jul 28 at 21:06
Xavier Yang
440314
You're second paragraph is wrong. The image of these homomorphisms are not necessarily $K(alpha)$. They are $K(beta)$ where $beta$ is a root of $p$. I am not really sure what your example is demonstrating.
– Dionel Jaime
Jul 28 at 21:23
add a comment |Â
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I read from my notes that let $alpha$ be algebraic over $K$. Consider $K(alpha)$ which has extension degree the same as $p=textirr(alpha,K)$ the irreducible polynomial of $alpha$ over $K$. Now, let $L$ be an algebraic closure of $K$. We claim that there are exactly as many homomorphisms $K(alpha)to L$ that fix $K$ as the number of roots of $p$ in $L$.
But this sounds weird to me because if we restrict to the image of those homomorphisms, the homomorphisms can be regarded as automorphisms from $K(alpha)$ to it. Then it is the same as asking its Galois group which is sometimes not the same as the number as the roots, usually not trivial.
For example, if we consider $X^3-2$ having 3 roots, whose Galois group is $S_6$ rather than a one with just 3 elements.
(Possibly, I got a fake note.)
abstract-algebra galois-theory
asked Jul 28 at 21:06
Xavier Yang
440314
I read from my notes that let $alpha$ be algebraic over $K$. Consider $K(alpha)$ which has extension degree the same as $p=textirr(alpha,K)$ the irreducible polynomial of $alpha$ over $K$. Now, let $L$ be an algebraic closure of $K$. We claim that there are exactly as many homomorphisms $K(alpha)to L$ that fix $K$ as the number of roots of $p$ in $L$.
But this sounds weird to me because if we restrict to the image of those homomorphisms, the homomorphisms can be regarded as automorphisms from $K(alpha)$ to it. Then it is the same as asking its Galois group which is sometimes not the same as the number as the roots, usually not trivial.
For example, if we consider $X^3-2$ having 3 roots, whose Galois group is $S_6$ rather than a one with just 3 elements.
(Possibly, I got a fake note.)
abstract-algebra galois-theory
asked Jul 28 at 21:06
Xavier Yang
440314
asked Jul 28 at 21:06
Xavier Yang
440314
asked Jul 28 at 21:06
Xavier Yang
440314
asked Jul 28 at 21:06
Xavier Yang
440314
440314
You're second paragraph is wrong. The image of these homomorphisms are not necessarily $K(alpha)$. They are $K(beta)$ where $beta$ is a root of $p$. I am not really sure what your example is demonstrating.
– Dionel Jaime
Jul 28 at 21:23
add a comment |Â
You're second paragraph is wrong. The image of these homomorphisms are not necessarily $K(alpha)$. They are $K(beta)$ where $beta$ is a root of $p$. I am not really sure what your example is demonstrating.
– Dionel Jaime
Jul 28 at 21:23
You're second paragraph is wrong. The image of these homomorphisms are not necessarily $K(alpha)$. They are $K(beta)$ where $beta$ is a root of $p$. I am not really sure what your example is demonstrating.
– Dionel Jaime
Jul 28 at 21:23
You're second paragraph is wrong. The image of these homomorphisms are not necessarily $K(alpha)$. They are $K(beta)$ where $beta$ is a root of $p$. I am not really sure what your example is demonstrating.
– Dionel Jaime
Jul 28 at 21:23
add a comment |Â
1 Answer
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I guess you got things mixed up.
First of all the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2)$ is the trivial group of just one element. You are confusing this with the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, i.e. the splitting field of $x^3-2$.
So in particular the statement above doesn't say anything about the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, at least explicitly. In fact it says that $mathbbQ(sqrt[3]2)$ can be embedded in three different way into $L$. And indeed the images of it are $mathbbQ(sqrt[3]2)$ ,$mathbbQ(zeta_3sqrt[3]2)$,$mathbbQ(zeta_3^2sqrt[3]2)$
answered Jul 28 at 21:25
Stefan4024
28k52974
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I guess you got things mixed up.
First of all the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2)$ is the trivial group of just one element. You are confusing this with the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, i.e. the splitting field of $x^3-2$.
So in particular the statement above doesn't say anything about the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, at least explicitly. In fact it says that $mathbbQ(sqrt[3]2)$ can be embedded in three different way into $L$. And indeed the images of it are $mathbbQ(sqrt[3]2)$ ,$mathbbQ(zeta_3sqrt[3]2)$,$mathbbQ(zeta_3^2sqrt[3]2)$
answered Jul 28 at 21:25
Stefan4024
28k52974
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
add a comment |Â
up vote
0
down vote
I guess you got things mixed up.
First of all the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2)$ is the trivial group of just one element. You are confusing this with the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, i.e. the splitting field of $x^3-2$.
So in particular the statement above doesn't say anything about the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, at least explicitly. In fact it says that $mathbbQ(sqrt[3]2)$ can be embedded in three different way into $L$. And indeed the images of it are $mathbbQ(sqrt[3]2)$ ,$mathbbQ(zeta_3sqrt[3]2)$,$mathbbQ(zeta_3^2sqrt[3]2)$
answered Jul 28 at 21:25
Stefan4024
28k52974
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I guess you got things mixed up.
First of all the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2)$ is the trivial group of just one element. You are confusing this with the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, i.e. the splitting field of $x^3-2$.
So in particular the statement above doesn't say anything about the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, at least explicitly. In fact it says that $mathbbQ(sqrt[3]2)$ can be embedded in three different way into $L$. And indeed the images of it are $mathbbQ(sqrt[3]2)$ ,$mathbbQ(zeta_3sqrt[3]2)$,$mathbbQ(zeta_3^2sqrt[3]2)$
answered Jul 28 at 21:25
Stefan4024
28k52974
I guess you got things mixed up.
First of all the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2)$ is the trivial group of just one element. You are confusing this with the Galois group of the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, i.e. the splitting field of $x^3-2$.
So in particular the statement above doesn't say anything about the extension $mathbbQ subseteq mathbbQ(sqrt[3]2,zeta_3)$, at least explicitly. In fact it says that $mathbbQ(sqrt[3]2)$ can be embedded in three different way into $L$. And indeed the images of it are $mathbbQ(sqrt[3]2)$ ,$mathbbQ(zeta_3sqrt[3]2)$,$mathbbQ(zeta_3^2sqrt[3]2)$
answered Jul 28 at 21:25
Stefan4024
28k52974
answered Jul 28 at 21:25
Stefan4024
28k52974
answered Jul 28 at 21:25
Stefan4024
28k52974
answered Jul 28 at 21:25
Stefan4024
28k52974
28k52974
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
add a comment |Â
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@Stefan4020 If we have two isomorphisms splitting fields of $f$ over $K$, is the number of isomorphisms between them (fixing $K$) the same as the size of galois group?
– Xavier Yang
Jul 29 at 12:31
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
@XavierYang Yeah. Composing any such isomorphism with the isomorphism (used to conclude that the splitting fields are isomorphic) yields an automorphism of the splitting field.
– Stefan4024
Jul 29 at 16:32
add a comment |Â
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You're second paragraph is wrong. The image of these homomorphisms are not necessarily $K(alpha)$. They are $K(beta)$ where $beta$ is a root of $p$. I am not really sure what your example is demonstrating.
– Dionel Jaime
Jul 28 at 21:23