Determine whether a polynomial is solvable by radicals
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Let $f(x) = x^12+2x^6-2x^3+2$ and let $K$ be it's splitting field.
Is the Galois group of $K/Q$ solvable?
Since we're really only interested in the polynomial $x^4+2x^2-2x+2$, and we know that there exists a quartic formula. It is enough to conclude that the polynomial is solvable by radicals and therefore the Galois group is solvable. Is this solution correct?
The question also asks if this polynomial is irreducible. This is easy to show via Eisenstein's criterion. But is the irreducibility of the polynomial important in any way for showing it is solvable?
Feb 2016
abstract-algebra proof-verification galois-theory
edited Aug 12 at 9:34
Bernard
111k635103
asked Aug 12 at 8:31
iYOA
60549
add a comment |Â
up vote
1
down vote
favorite
Let $f(x) = x^12+2x^6-2x^3+2$ and let $K$ be it's splitting field.
Is the Galois group of $K/Q$ solvable?
Since we're really only interested in the polynomial $x^4+2x^2-2x+2$, and we know that there exists a quartic formula. It is enough to conclude that the polynomial is solvable by radicals and therefore the Galois group is solvable. Is this solution correct?
The question also asks if this polynomial is irreducible. This is easy to show via Eisenstein's criterion. But is the irreducibility of the polynomial important in any way for showing it is solvable?
Feb 2016
abstract-algebra proof-verification galois-theory
edited Aug 12 at 9:34
Bernard
111k635103
asked Aug 12 at 8:31
iYOA
60549
1
Irreducibility is not needed since when you compute the roots in radicals you compute them all.
– user583012
Aug 12 at 11:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f(x) = x^12+2x^6-2x^3+2$ and let $K$ be it's splitting field.
Is the Galois group of $K/Q$ solvable?
Since we're really only interested in the polynomial $x^4+2x^2-2x+2$, and we know that there exists a quartic formula. It is enough to conclude that the polynomial is solvable by radicals and therefore the Galois group is solvable. Is this solution correct?
The question also asks if this polynomial is irreducible. This is easy to show via Eisenstein's criterion. But is the irreducibility of the polynomial important in any way for showing it is solvable?
Feb 2016
abstract-algebra proof-verification galois-theory
edited Aug 12 at 9:34
Bernard
111k635103
asked Aug 12 at 8:31
iYOA
60549
Let $f(x) = x^12+2x^6-2x^3+2$ and let $K$ be it's splitting field.
Is the Galois group of $K/Q$ solvable?
Since we're really only interested in the polynomial $x^4+2x^2-2x+2$, and we know that there exists a quartic formula. It is enough to conclude that the polynomial is solvable by radicals and therefore the Galois group is solvable. Is this solution correct?
The question also asks if this polynomial is irreducible. This is easy to show via Eisenstein's criterion. But is the irreducibility of the polynomial important in any way for showing it is solvable?
Feb 2016
abstract-algebra proof-verification galois-theory
edited Aug 12 at 9:34
Bernard
111k635103
asked Aug 12 at 8:31
iYOA
60549
edited Aug 12 at 9:34
Bernard
111k635103
edited Aug 12 at 9:34
Bernard
111k635103
edited Aug 12 at 9:34
Bernard
111k635103
111k635103
asked Aug 12 at 8:31
iYOA
60549
asked Aug 12 at 8:31
iYOA
60549
asked Aug 12 at 8:31
iYOA
60549
60549
1
Irreducibility is not needed since when you compute the roots in radicals you compute them all.
– user583012
Aug 12 at 11:19
add a comment |Â
1
Irreducibility is not needed since when you compute the roots in radicals you compute them all.
– user583012
Aug 12 at 11:19
1
1
Irreducibility is not needed since when you compute the roots in radicals you compute them all.
– user583012
Aug 12 at 11:19
Irreducibility is not needed since when you compute the roots in radicals you compute them all.
– user583012
Aug 12 at 11:19
add a comment |Â
1 Answer
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The Galois group of $f(x)$ has order $3888=2^4cdot 3^5$. For the computation one can use Magma online here. This group is, by Burnside's $p^rq^s$-Theorem solvable, since the order has only two different prime divisors.
The polynomial $x^4+2x^2-2x+2$ has a solvable Galois group, see K. Conrad's article Galois group of cubics and quartics, namely one of the following solvable groups: $S_4,A_4,D_4,C_4,C_2times C_2$. However, the Galois group of $x^12+2x^6-2x^3+2$ is quite different from these groups.
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
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
The Galois group of $f(x)$ has order $3888=2^4cdot 3^5$. For the computation one can use Magma online here. This group is, by Burnside's $p^rq^s$-Theorem solvable, since the order has only two different prime divisors.
The polynomial $x^4+2x^2-2x+2$ has a solvable Galois group, see K. Conrad's article Galois group of cubics and quartics, namely one of the following solvable groups: $S_4,A_4,D_4,C_4,C_2times C_2$. However, the Galois group of $x^12+2x^6-2x^3+2$ is quite different from these groups.
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
add a comment |Â
up vote
0
down vote
The Galois group of $f(x)$ has order $3888=2^4cdot 3^5$. For the computation one can use Magma online here. This group is, by Burnside's $p^rq^s$-Theorem solvable, since the order has only two different prime divisors.
The polynomial $x^4+2x^2-2x+2$ has a solvable Galois group, see K. Conrad's article Galois group of cubics and quartics, namely one of the following solvable groups: $S_4,A_4,D_4,C_4,C_2times C_2$. However, the Galois group of $x^12+2x^6-2x^3+2$ is quite different from these groups.
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The Galois group of $f(x)$ has order $3888=2^4cdot 3^5$. For the computation one can use Magma online here. This group is, by Burnside's $p^rq^s$-Theorem solvable, since the order has only two different prime divisors.
The polynomial $x^4+2x^2-2x+2$ has a solvable Galois group, see K. Conrad's article Galois group of cubics and quartics, namely one of the following solvable groups: $S_4,A_4,D_4,C_4,C_2times C_2$. However, the Galois group of $x^12+2x^6-2x^3+2$ is quite different from these groups.
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
The Galois group of $f(x)$ has order $3888=2^4cdot 3^5$. For the computation one can use Magma online here. This group is, by Burnside's $p^rq^s$-Theorem solvable, since the order has only two different prime divisors.
The polynomial $x^4+2x^2-2x+2$ has a solvable Galois group, see K. Conrad's article Galois group of cubics and quartics, namely one of the following solvable groups: $S_4,A_4,D_4,C_4,C_2times C_2$. However, the Galois group of $x^12+2x^6-2x^3+2$ is quite different from these groups.
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
edited Aug 12 at 20:15
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
answered Aug 12 at 11:18
Dietrich Burde
74.8k64185
74.8k64185
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
add a comment |Â
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I was thinking more along the lines of "is the polynomial solvable by radicals?", in which case the galois group would also be solvable. And since every 4th degree polynomial is solvable by radicals, $y=x^3$ would imply $x$ is also a radical, so the original polynomial would also be solvable by radicals. The galois groups dont necessarily have to be the same though
– iYOA
Aug 12 at 19:44
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
I see. Looks like you are right.
– Dietrich Burde
Aug 12 at 20:17
add a comment |Â
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1
Irreducibility is not needed since when you compute the roots in radicals you compute them all.
– user583012
Aug 12 at 11:19