Mixing
Galois group is determined by action on roots of polynomial [duplicate]
$begingroup$
This question is an exact duplicate of:
Galois Groups are isomorphic to subgroups of symmetric groups.
1 answer
This is a very simple question but I can't give a good answer to it. If we have a field $K$ and a Galois extension $L/K$ where $L=K(alpha_1,ldots,alpha_n)$ and $alpha_1,ldots,alpha_n$ are the roots of some separable polynomial in $K[x]$, then any automorphism of $L/K$ is uniquely determined by its action on the $alpha_i$. I suppose it is 'clear' because any element of $L$ is formed by combining elements of $K$ and the $alpha_i$ with field operations but this is not really rigorous enough for me. I'm sure there's a better explanation but I can't see it.
(Explanation in suggested problem didn't explain the specific point I did not understand, but I understand now.)
galois-theory galois-extensions
asked Jan 4 at 19:09
AlephNullAlephNull
2529
$endgroup$
marked as duplicate by Kenny Lau, amWhy, jgon, KReiser, Leucippus Jan 5 at 5:24
This question was marked as an exact duplicate of an existing question.
add a comment |
$begingroup$
This question is an exact duplicate of:
Galois Groups are isomorphic to subgroups of symmetric groups.
1 answer
This is a very simple question but I can't give a good answer to it. If we have a field $K$ and a Galois extension $L/K$ where $L=K(alpha_1,ldots,alpha_n)$ and $alpha_1,ldots,alpha_n$ are the roots of some separable polynomial in $K[x]$, then any automorphism of $L/K$ is uniquely determined by its action on the $alpha_i$. I suppose it is 'clear' because any element of $L$ is formed by combining elements of $K$ and the $alpha_i$ with field operations but this is not really rigorous enough for me. I'm sure there's a better explanation but I can't see it.
(Explanation in suggested problem didn't explain the specific point I did not understand, but I understand now.)
galois-theory galois-extensions
asked Jan 4 at 19:09
AlephNullAlephNull
2529
$endgroup$
marked as duplicate by Kenny Lau, amWhy, jgon, KReiser, Leucippus Jan 5 at 5:24
This question was marked as an exact duplicate of an existing question.
$begingroup$
A vector space map is determined by the image of a basis under such a map. A field extension $L/K$ is a $K$-vector space so it makes sense that any $K$-automorphism of $L$ is determined by the image of the $alpha_i$. This is what you are saying.
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 19:12
$begingroup$
@ÍgjøgnumMeg I'm aware of that, but you seem to be implying that the $alpha_i$ form a basis, or that they contain a basis, which I cannot explain.
$endgroup$
– AlephNull
Jan 4 at 19:16
$begingroup$
@KennyLau Is the explanation really that involved? It was just stated in passing in my notes. I suppose it's mainly that I don't understand why every $x in E$ can be expressed as a polynomial in the $alpha_i$ like that.
$endgroup$
– AlephNull
Jan 4 at 19:19
add a comment |
$begingroup$
This question is an exact duplicate of:
Galois Groups are isomorphic to subgroups of symmetric groups.
1 answer
This is a very simple question but I can't give a good answer to it. If we have a field $K$ and a Galois extension $L/K$ where $L=K(alpha_1,ldots,alpha_n)$ and $alpha_1,ldots,alpha_n$ are the roots of some separable polynomial in $K[x]$, then any automorphism of $L/K$ is uniquely determined by its action on the $alpha_i$. I suppose it is 'clear' because any element of $L$ is formed by combining elements of $K$ and the $alpha_i$ with field operations but this is not really rigorous enough for me. I'm sure there's a better explanation but I can't see it.
(Explanation in suggested problem didn't explain the specific point I did not understand, but I understand now.)
galois-theory galois-extensions
asked Jan 4 at 19:09
AlephNullAlephNull
2529
$endgroup$
This question is an exact duplicate of:
Galois Groups are isomorphic to subgroups of symmetric groups.
1 answer
This is a very simple question but I can't give a good answer to it. If we have a field $K$ and a Galois extension $L/K$ where $L=K(alpha_1,ldots,alpha_n)$ and $alpha_1,ldots,alpha_n$ are the roots of some separable polynomial in $K[x]$, then any automorphism of $L/K$ is uniquely determined by its action on the $alpha_i$. I suppose it is 'clear' because any element of $L$ is formed by combining elements of $K$ and the $alpha_i$ with field operations but this is not really rigorous enough for me. I'm sure there's a better explanation but I can't see it.
(Explanation in suggested problem didn't explain the specific point I did not understand, but I understand now.)
This question is an exact duplicate of:
Galois Groups are isomorphic to subgroups of symmetric groups.
1 answer
galois-theory galois-extensions
galois-theory galois-extensions
asked Jan 4 at 19:09
AlephNullAlephNull
2529
asked Jan 4 at 19:09
AlephNullAlephNull
2529
edited Jan 4 at 20:05
AlephNull
asked Jan 4 at 19:09
AlephNullAlephNull
2529
asked Jan 4 at 19:09
AlephNullAlephNull
2529
asked Jan 4 at 19:09
AlephNullAlephNull
2529
2529
marked as duplicate by Kenny Lau, amWhy, jgon, KReiser, Leucippus Jan 5 at 5:24
This question was marked as an exact duplicate of an existing question.
marked as duplicate by Kenny Lau, amWhy, jgon, KReiser, Leucippus Jan 5 at 5:24
This question was marked as an exact duplicate of an existing question.
$begingroup$
A vector space map is determined by the image of a basis under such a map. A field extension $L/K$ is a $K$-vector space so it makes sense that any $K$-automorphism of $L$ is determined by the image of the $alpha_i$. This is what you are saying.
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 19:12
$begingroup$
@ÍgjøgnumMeg I'm aware of that, but you seem to be implying that the $alpha_i$ form a basis, or that they contain a basis, which I cannot explain.
$endgroup$
– AlephNull
Jan 4 at 19:16
$begingroup$
@KennyLau Is the explanation really that involved? It was just stated in passing in my notes. I suppose it's mainly that I don't understand why every $x in E$ can be expressed as a polynomial in the $alpha_i$ like that.
$endgroup$
– AlephNull
Jan 4 at 19:19
add a comment |
$begingroup$
A vector space map is determined by the image of a basis under such a map. A field extension $L/K$ is a $K$-vector space so it makes sense that any $K$-automorphism of $L$ is determined by the image of the $alpha_i$. This is what you are saying.
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 19:12
$begingroup$
@ÍgjøgnumMeg I'm aware of that, but you seem to be implying that the $alpha_i$ form a basis, or that they contain a basis, which I cannot explain.
$endgroup$
– AlephNull
Jan 4 at 19:16
$begingroup$
@KennyLau Is the explanation really that involved? It was just stated in passing in my notes. I suppose it's mainly that I don't understand why every $x in E$ can be expressed as a polynomial in the $alpha_i$ like that.
$endgroup$
– AlephNull
Jan 4 at 19:19
$begingroup$
A vector space map is determined by the image of a basis under such a map. A field extension $L/K$ is a $K$-vector space so it makes sense that any $K$-automorphism of $L$ is determined by the image of the $alpha_i$. This is what you are saying.
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 19:12
$begingroup$
A vector space map is determined by the image of a basis under such a map. A field extension $L/K$ is a $K$-vector space so it makes sense that any $K$-automorphism of $L$ is determined by the image of the $alpha_i$. This is what you are saying.
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 19:12
$begingroup$
@ÍgjøgnumMeg I'm aware of that, but you seem to be implying that the $alpha_i$ form a basis, or that they contain a basis, which I cannot explain.
$endgroup$
– AlephNull
Jan 4 at 19:16
$begingroup$
@ÍgjøgnumMeg I'm aware of that, but you seem to be implying that the $alpha_i$ form a basis, or that they contain a basis, which I cannot explain.
$endgroup$
– AlephNull
Jan 4 at 19:16
$begingroup$
@KennyLau Is the explanation really that involved? It was just stated in passing in my notes. I suppose it's mainly that I don't understand why every $x in E$ can be expressed as a polynomial in the $alpha_i$ like that.
$endgroup$
– AlephNull
Jan 4 at 19:19
$begingroup$
@KennyLau Is the explanation really that involved? It was just stated in passing in my notes. I suppose it's mainly that I don't understand why every $x in E$ can be expressed as a polynomial in the $alpha_i$ like that.
$endgroup$
– AlephNull
Jan 4 at 19:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you have a base field $F$ and $K/F$ is an extension, say $K = F(alpha)$ then $K cong F[X]/(p(X))$ where $p(X) = X^n + a_{n-1}X^{n-1} + dots + a_1X + a_0$ is an irreducible polynomial having $alpha$ as a root. Now in $K$ you have $alpha^n = -(a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0)$ so $lbrace 1, alpha, dots, alpha^{n-1}rbrace$ definitely span $K$ as an $F$-vector space.
If you had some non-trivial linear dependence $a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0 = 0$ then this would imply
$$a_{n-1}X^{n-1} + dots + a_1X + a_0 equiv 0 bmod p(X)$$
so $p(X)$ divides $a_{n-1}X^{n-1} + dots + a_1 X + a_0$. But $deg p(X) = n < n-1$ so this can't happen unless $a_i = 0$ for all $i$, and this translates into linear independence of the $alpha_i$ in $K$.
You can just continue this process inductively and use the tower law for field extensions to do this for any number of generators.
Alternatively, instead of continuing this process, you can note that a finite separable extension has a primitive element.
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you have a base field $F$ and $K/F$ is an extension, say $K = F(alpha)$ then $K cong F[X]/(p(X))$ where $p(X) = X^n + a_{n-1}X^{n-1} + dots + a_1X + a_0$ is an irreducible polynomial having $alpha$ as a root. Now in $K$ you have $alpha^n = -(a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0)$ so $lbrace 1, alpha, dots, alpha^{n-1}rbrace$ definitely span $K$ as an $F$-vector space.
If you had some non-trivial linear dependence $a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0 = 0$ then this would imply
$$a_{n-1}X^{n-1} + dots + a_1X + a_0 equiv 0 bmod p(X)$$
so $p(X)$ divides $a_{n-1}X^{n-1} + dots + a_1 X + a_0$. But $deg p(X) = n < n-1$ so this can't happen unless $a_i = 0$ for all $i$, and this translates into linear independence of the $alpha_i$ in $K$.
You can just continue this process inductively and use the tower law for field extensions to do this for any number of generators.
Alternatively, instead of continuing this process, you can note that a finite separable extension has a primitive element.
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
add a comment |
$begingroup$
If you have a base field $F$ and $K/F$ is an extension, say $K = F(alpha)$ then $K cong F[X]/(p(X))$ where $p(X) = X^n + a_{n-1}X^{n-1} + dots + a_1X + a_0$ is an irreducible polynomial having $alpha$ as a root. Now in $K$ you have $alpha^n = -(a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0)$ so $lbrace 1, alpha, dots, alpha^{n-1}rbrace$ definitely span $K$ as an $F$-vector space.
If you had some non-trivial linear dependence $a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0 = 0$ then this would imply
$$a_{n-1}X^{n-1} + dots + a_1X + a_0 equiv 0 bmod p(X)$$
so $p(X)$ divides $a_{n-1}X^{n-1} + dots + a_1 X + a_0$. But $deg p(X) = n < n-1$ so this can't happen unless $a_i = 0$ for all $i$, and this translates into linear independence of the $alpha_i$ in $K$.
You can just continue this process inductively and use the tower law for field extensions to do this for any number of generators.
Alternatively, instead of continuing this process, you can note that a finite separable extension has a primitive element.
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
add a comment |
$begingroup$
If you have a base field $F$ and $K/F$ is an extension, say $K = F(alpha)$ then $K cong F[X]/(p(X))$ where $p(X) = X^n + a_{n-1}X^{n-1} + dots + a_1X + a_0$ is an irreducible polynomial having $alpha$ as a root. Now in $K$ you have $alpha^n = -(a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0)$ so $lbrace 1, alpha, dots, alpha^{n-1}rbrace$ definitely span $K$ as an $F$-vector space.
If you had some non-trivial linear dependence $a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0 = 0$ then this would imply
$$a_{n-1}X^{n-1} + dots + a_1X + a_0 equiv 0 bmod p(X)$$
so $p(X)$ divides $a_{n-1}X^{n-1} + dots + a_1 X + a_0$. But $deg p(X) = n < n-1$ so this can't happen unless $a_i = 0$ for all $i$, and this translates into linear independence of the $alpha_i$ in $K$.
You can just continue this process inductively and use the tower law for field extensions to do this for any number of generators.
Alternatively, instead of continuing this process, you can note that a finite separable extension has a primitive element.
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
$endgroup$
If you have a base field $F$ and $K/F$ is an extension, say $K = F(alpha)$ then $K cong F[X]/(p(X))$ where $p(X) = X^n + a_{n-1}X^{n-1} + dots + a_1X + a_0$ is an irreducible polynomial having $alpha$ as a root. Now in $K$ you have $alpha^n = -(a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0)$ so $lbrace 1, alpha, dots, alpha^{n-1}rbrace$ definitely span $K$ as an $F$-vector space.
If you had some non-trivial linear dependence $a_{n-1}alpha^{n-1} + dots + a_1alpha + a_0 = 0$ then this would imply
$$a_{n-1}X^{n-1} + dots + a_1X + a_0 equiv 0 bmod p(X)$$
so $p(X)$ divides $a_{n-1}X^{n-1} + dots + a_1 X + a_0$. But $deg p(X) = n < n-1$ so this can't happen unless $a_i = 0$ for all $i$, and this translates into linear independence of the $alpha_i$ in $K$.
You can just continue this process inductively and use the tower law for field extensions to do this for any number of generators.
Alternatively, instead of continuing this process, you can note that a finite separable extension has a primitive element.
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
edited Jan 4 at 20:39
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
answered Jan 4 at 19:25
ÍgjøgnumMegÍgjøgnumMeg
2,83511029
2,83511029
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
add a comment |
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
$begingroup$
Ah, I knew the characterisation of simple algebraic extensions but for some reason didn't see how it easily extended inductively to show that everything can be expressed in terms of the $alpha_i$.
$endgroup$
– AlephNull
Jan 4 at 19:32
add a comment |
$begingroup$
A vector space map is determined by the image of a basis under such a map. A field extension $L/K$ is a $K$-vector space so it makes sense that any $K$-automorphism of $L$ is determined by the image of the $alpha_i$. This is what you are saying.
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 19:12
$begingroup$
@ÍgjøgnumMeg I'm aware of that, but you seem to be implying that the $alpha_i$ form a basis, or that they contain a basis, which I cannot explain.
$endgroup$
– AlephNull
Jan 4 at 19:16
$begingroup$
@KennyLau Is the explanation really that involved? It was just stated in passing in my notes. I suppose it's mainly that I don't understand why every $x in E$ can be expressed as a polynomial in the $alpha_i$ like that.
$endgroup$
– AlephNull
Jan 4 at 19:19