degree 3 Galois extension of $mathbbQ$ not radicalFinding a Galois extension of $Bbb Q$ of degree $3$Solvable but not radical.Examples of extension of $Bbb Q$ or prove there are no such examples.Galois group of a non-separable polynomial$f$ is solvable by radicals, but the splitting field $L:Q$ not radical extension.Is the splitting field of $x^3-2x-2$ a radical extension over $mathbbQ$?Proving an extension is radicalRadical extension and discriminant of cubicIf K $subset$ L, K Galois over F, L is a splitting field of f(x) over F, can we say L Galois over FHow do you prove that the Galois Group of a radical field extension is always soluble/solvable?Galois Extension.Radical field extension is soluble, **counter-example?**Any Sub-extension of a Radical Extension is Solvable?
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degree 3 Galois extension of $mathbbQ$ not radical
Finding a Galois extension of $Bbb Q$ of degree $3$Solvable but not radical.Examples of extension of $Bbb Q$ or prove there are no such examples.Galois group of a non-separable polynomial$f$ is solvable by radicals, but the splitting field $L:Q$ not radical extension.Is the splitting field of $x^3-2x-2$ a radical extension over $mathbbQ$?Proving an extension is radicalRadical extension and discriminant of cubicIf K $subset$ L, K Galois over F, L is a splitting field of f(x) over F, can we say L Galois over FHow do you prove that the Galois Group of a radical field extension is always soluble/solvable?Galois Extension.Radical field extension is soluble, **counter-example?**Any Sub-extension of a Radical Extension is Solvable?
$begingroup$
I have the following question.
I have the following result in Dummit and Foote abstract algebra (Theorem 39 p. 628) that says that over a field of $char=0$ then a polynomial $f(x)$ is soluble by radicals if and only if its Galois group is soluble group.
But I also have a result in one of my exams that says that a Galois extension of degree 3 over $mathbbQ$ is not a radical extension.
How is it that the two statements aren't contradicting one another? since I thought that if the Galois group was of size 3 then its cyclic so soluble, so if $G$ is the Galois group of some irreducible cubic then wouldn't the theorem say that it is soluble by radicals?
Or is there a difference between a polynomial being soluble by radicals and its splitting field being a radical extension?
Thank you
galois-theory
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
$endgroup$
add a comment |
$begingroup$
I have the following question.
I have the following result in Dummit and Foote abstract algebra (Theorem 39 p. 628) that says that over a field of $char=0$ then a polynomial $f(x)$ is soluble by radicals if and only if its Galois group is soluble group.
But I also have a result in one of my exams that says that a Galois extension of degree 3 over $mathbbQ$ is not a radical extension.
How is it that the two statements aren't contradicting one another? since I thought that if the Galois group was of size 3 then its cyclic so soluble, so if $G$ is the Galois group of some irreducible cubic then wouldn't the theorem say that it is soluble by radicals?
Or is there a difference between a polynomial being soluble by radicals and its splitting field being a radical extension?
Thank you
galois-theory
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
$endgroup$
add a comment |
$begingroup$
I have the following question.
I have the following result in Dummit and Foote abstract algebra (Theorem 39 p. 628) that says that over a field of $char=0$ then a polynomial $f(x)$ is soluble by radicals if and only if its Galois group is soluble group.
But I also have a result in one of my exams that says that a Galois extension of degree 3 over $mathbbQ$ is not a radical extension.
How is it that the two statements aren't contradicting one another? since I thought that if the Galois group was of size 3 then its cyclic so soluble, so if $G$ is the Galois group of some irreducible cubic then wouldn't the theorem say that it is soluble by radicals?
Or is there a difference between a polynomial being soluble by radicals and its splitting field being a radical extension?
Thank you
galois-theory
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
$endgroup$
I have the following question.
I have the following result in Dummit and Foote abstract algebra (Theorem 39 p. 628) that says that over a field of $char=0$ then a polynomial $f(x)$ is soluble by radicals if and only if its Galois group is soluble group.
But I also have a result in one of my exams that says that a Galois extension of degree 3 over $mathbbQ$ is not a radical extension.
How is it that the two statements aren't contradicting one another? since I thought that if the Galois group was of size 3 then its cyclic so soluble, so if $G$ is the Galois group of some irreducible cubic then wouldn't the theorem say that it is soluble by radicals?
Or is there a difference between a polynomial being soluble by radicals and its splitting field being a radical extension?
Thank you
galois-theory
galois-theory
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
edited Dec 8 '13 at 16:19
Chris Birkbeck
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
asked Dec 8 '13 at 16:11
Chris BirkbeckChris Birkbeck
758413
758413
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, there are some differences : an extension $L/K$ is solvable by radicals if there is an extension $L_1/L$ and a sequences of extensions $K subset K_1 subset ldots subset K_n subset L_1$ where each step is a radical extension.
If $L/K$ is cyclic of order $n$ and $K$ contains an $n$th root of unity, then we do have that $L/K$ is radical (there exists some $x in L$ such that $x^n in K$ and $L = K(x)$).
However, if $K$ doesn't contain those roots of unity, then no radical extension of order $n$ of $K$ is Galois. So we have to add them first, and then we obtain an extension $K subset K(zeta_n) subset L_1 = L(zeta_n)$ where the first step is soluble by radicals, and the second is radical. So $L$ is solvable by radicals, but can be very very far from being a radical extension.
The simplest example is $K = Bbb Q(2cos frac 2pi7) = Bbb Q[c]/(c^3+c^2-2c-1)$. It is a normal extension of $Bbb Q$ whose Galois group is cyclic of order $3$ (generated by $c mapsto c^2-2$). And in order to solve it by radicals and express $c$ with radicals, you will need to use $sqrt-3$ somewhere.
edited 2 days ago
Sam Nead
3,0141220
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
$endgroup$
1
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
add a comment |
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$begingroup$
Yes, there are some differences : an extension $L/K$ is solvable by radicals if there is an extension $L_1/L$ and a sequences of extensions $K subset K_1 subset ldots subset K_n subset L_1$ where each step is a radical extension.
If $L/K$ is cyclic of order $n$ and $K$ contains an $n$th root of unity, then we do have that $L/K$ is radical (there exists some $x in L$ such that $x^n in K$ and $L = K(x)$).
However, if $K$ doesn't contain those roots of unity, then no radical extension of order $n$ of $K$ is Galois. So we have to add them first, and then we obtain an extension $K subset K(zeta_n) subset L_1 = L(zeta_n)$ where the first step is soluble by radicals, and the second is radical. So $L$ is solvable by radicals, but can be very very far from being a radical extension.
The simplest example is $K = Bbb Q(2cos frac 2pi7) = Bbb Q[c]/(c^3+c^2-2c-1)$. It is a normal extension of $Bbb Q$ whose Galois group is cyclic of order $3$ (generated by $c mapsto c^2-2$). And in order to solve it by radicals and express $c$ with radicals, you will need to use $sqrt-3$ somewhere.
edited 2 days ago
Sam Nead
3,0141220
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
$endgroup$
1
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
add a comment |
$begingroup$
Yes, there are some differences : an extension $L/K$ is solvable by radicals if there is an extension $L_1/L$ and a sequences of extensions $K subset K_1 subset ldots subset K_n subset L_1$ where each step is a radical extension.
If $L/K$ is cyclic of order $n$ and $K$ contains an $n$th root of unity, then we do have that $L/K$ is radical (there exists some $x in L$ such that $x^n in K$ and $L = K(x)$).
However, if $K$ doesn't contain those roots of unity, then no radical extension of order $n$ of $K$ is Galois. So we have to add them first, and then we obtain an extension $K subset K(zeta_n) subset L_1 = L(zeta_n)$ where the first step is soluble by radicals, and the second is radical. So $L$ is solvable by radicals, but can be very very far from being a radical extension.
The simplest example is $K = Bbb Q(2cos frac 2pi7) = Bbb Q[c]/(c^3+c^2-2c-1)$. It is a normal extension of $Bbb Q$ whose Galois group is cyclic of order $3$ (generated by $c mapsto c^2-2$). And in order to solve it by radicals and express $c$ with radicals, you will need to use $sqrt-3$ somewhere.
edited 2 days ago
Sam Nead
3,0141220
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
$endgroup$
1
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
add a comment |
$begingroup$
Yes, there are some differences : an extension $L/K$ is solvable by radicals if there is an extension $L_1/L$ and a sequences of extensions $K subset K_1 subset ldots subset K_n subset L_1$ where each step is a radical extension.
If $L/K$ is cyclic of order $n$ and $K$ contains an $n$th root of unity, then we do have that $L/K$ is radical (there exists some $x in L$ such that $x^n in K$ and $L = K(x)$).
However, if $K$ doesn't contain those roots of unity, then no radical extension of order $n$ of $K$ is Galois. So we have to add them first, and then we obtain an extension $K subset K(zeta_n) subset L_1 = L(zeta_n)$ where the first step is soluble by radicals, and the second is radical. So $L$ is solvable by radicals, but can be very very far from being a radical extension.
The simplest example is $K = Bbb Q(2cos frac 2pi7) = Bbb Q[c]/(c^3+c^2-2c-1)$. It is a normal extension of $Bbb Q$ whose Galois group is cyclic of order $3$ (generated by $c mapsto c^2-2$). And in order to solve it by radicals and express $c$ with radicals, you will need to use $sqrt-3$ somewhere.
edited 2 days ago
Sam Nead
3,0141220
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
$endgroup$
Yes, there are some differences : an extension $L/K$ is solvable by radicals if there is an extension $L_1/L$ and a sequences of extensions $K subset K_1 subset ldots subset K_n subset L_1$ where each step is a radical extension.
If $L/K$ is cyclic of order $n$ and $K$ contains an $n$th root of unity, then we do have that $L/K$ is radical (there exists some $x in L$ such that $x^n in K$ and $L = K(x)$).
However, if $K$ doesn't contain those roots of unity, then no radical extension of order $n$ of $K$ is Galois. So we have to add them first, and then we obtain an extension $K subset K(zeta_n) subset L_1 = L(zeta_n)$ where the first step is soluble by radicals, and the second is radical. So $L$ is solvable by radicals, but can be very very far from being a radical extension.
The simplest example is $K = Bbb Q(2cos frac 2pi7) = Bbb Q[c]/(c^3+c^2-2c-1)$. It is a normal extension of $Bbb Q$ whose Galois group is cyclic of order $3$ (generated by $c mapsto c^2-2$). And in order to solve it by radicals and express $c$ with radicals, you will need to use $sqrt-3$ somewhere.
edited 2 days ago
Sam Nead
3,0141220
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
edited 2 days ago
Sam Nead
3,0141220
edited 2 days ago
Sam Nead
3,0141220
edited 2 days ago
Sam Nead
3,0141220
3,0141220
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
answered Dec 8 '13 at 16:44
merciomercio
44.7k258111
44.7k258111
1
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
add a comment |
1
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
1
1
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
$begingroup$
+1, fine explanation. I believe one might also mention en.wikipedia.org/wiki/Casus_irreducibilis
$endgroup$
– Andreas Caranti
Dec 8 '13 at 16:51
add a comment |
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