Galois group of ($x^2$ + $2x$ − $1$)($x^3$ + $3x^2$ + $3x$ − 6) over $Bbb Q$ Announcing...
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Galois group of ($x^2$ + $2x$ − $1$)($x^3$ + $3x^2$ + $3x$ − 6) over $Bbb Q$
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$begingroup$
What is the Galois group of $(x^2 + 2x − 1)(x^3 + 3x^2 + 3x − 6)$ over $mathbb Q$?
I noticed that this polynomial can also be written as $((x+1)^2-2)((x+1)^3-7)$, can this be of any help?
Can I treat the polynomial as $((x')^2-2)((x')^3-7)$ for $x'=x+1$ and would that change the answer?
abstract-algebra polynomials ring-theory galois-theory
edited Mar 23 at 12:14
Andrews
1,3012423
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
$endgroup$
add a comment |
$begingroup$
What is the Galois group of $(x^2 + 2x − 1)(x^3 + 3x^2 + 3x − 6)$ over $mathbb Q$?
I noticed that this polynomial can also be written as $((x+1)^2-2)((x+1)^3-7)$, can this be of any help?
Can I treat the polynomial as $((x')^2-2)((x')^3-7)$ for $x'=x+1$ and would that change the answer?
abstract-algebra polynomials ring-theory galois-theory
edited Mar 23 at 12:14
Andrews
1,3012423
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
$endgroup$
add a comment |
$begingroup$
What is the Galois group of $(x^2 + 2x − 1)(x^3 + 3x^2 + 3x − 6)$ over $mathbb Q$?
I noticed that this polynomial can also be written as $((x+1)^2-2)((x+1)^3-7)$, can this be of any help?
Can I treat the polynomial as $((x')^2-2)((x')^3-7)$ for $x'=x+1$ and would that change the answer?
abstract-algebra polynomials ring-theory galois-theory
edited Mar 23 at 12:14
Andrews
1,3012423
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
$endgroup$
What is the Galois group of $(x^2 + 2x − 1)(x^3 + 3x^2 + 3x − 6)$ over $mathbb Q$?
I noticed that this polynomial can also be written as $((x+1)^2-2)((x+1)^3-7)$, can this be of any help?
Can I treat the polynomial as $((x')^2-2)((x')^3-7)$ for $x'=x+1$ and would that change the answer?
abstract-algebra polynomials ring-theory galois-theory
abstract-algebra polynomials ring-theory galois-theory
edited Mar 23 at 12:14
Andrews
1,3012423
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
edited Mar 23 at 12:14
Andrews
1,3012423
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
edited Mar 23 at 12:14
Andrews
1,3012423
edited Mar 23 at 12:14
Andrews
1,3012423
edited Mar 23 at 12:14
Andrews
1,3012423
1,3012423
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
asked Mar 23 at 12:01
bensimmonsisarookiebensimmonsisarookie
84
84
add a comment |
add a comment |
1 Answer
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$begingroup$
Both things you've noticed is of help. The second note just shows that the roots of $(x+1)^{2}-2$ are $x=pmsqrt{2}-1$ because $x'=x+1$ and $x'=pmsqrt{2}$ are the roots of $(x')^{2}-2$ and you can show that $K_{1}=mathbb{Q}(sqrt{2})=mathbb{Q}(sqrt{2}-1)$. Likewise for the second factor (where the splitting field is $K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3})$ where $zeta_{3}$ is a primitive third root of unity). Now just show that $$mathbb{Q}(sqrt[3]{7},zeta_{3})capmathbb{Q}(sqrt{2})=mathbb{Q},$$ and consider the compositum $L=K_{1}K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3},sqrt{2})$ as an extension over $mathbb{Q}$. $L/mathbb{Q}$ is Galois and we have the following isomorphism of Galois groups
$$operatorname{Gal}(L/mathbb{Q})stackrel{simeq}{longrightarrow}operatorname{Gal}(K_{1}/mathbb{Q})timesoperatorname{Gal}(K_{2}/mathbb{Q}),$$
which you can use to deduce the answer.
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
$endgroup$
add a comment |
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$begingroup$
Both things you've noticed is of help. The second note just shows that the roots of $(x+1)^{2}-2$ are $x=pmsqrt{2}-1$ because $x'=x+1$ and $x'=pmsqrt{2}$ are the roots of $(x')^{2}-2$ and you can show that $K_{1}=mathbb{Q}(sqrt{2})=mathbb{Q}(sqrt{2}-1)$. Likewise for the second factor (where the splitting field is $K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3})$ where $zeta_{3}$ is a primitive third root of unity). Now just show that $$mathbb{Q}(sqrt[3]{7},zeta_{3})capmathbb{Q}(sqrt{2})=mathbb{Q},$$ and consider the compositum $L=K_{1}K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3},sqrt{2})$ as an extension over $mathbb{Q}$. $L/mathbb{Q}$ is Galois and we have the following isomorphism of Galois groups
$$operatorname{Gal}(L/mathbb{Q})stackrel{simeq}{longrightarrow}operatorname{Gal}(K_{1}/mathbb{Q})timesoperatorname{Gal}(K_{2}/mathbb{Q}),$$
which you can use to deduce the answer.
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
$endgroup$
add a comment |
$begingroup$
Both things you've noticed is of help. The second note just shows that the roots of $(x+1)^{2}-2$ are $x=pmsqrt{2}-1$ because $x'=x+1$ and $x'=pmsqrt{2}$ are the roots of $(x')^{2}-2$ and you can show that $K_{1}=mathbb{Q}(sqrt{2})=mathbb{Q}(sqrt{2}-1)$. Likewise for the second factor (where the splitting field is $K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3})$ where $zeta_{3}$ is a primitive third root of unity). Now just show that $$mathbb{Q}(sqrt[3]{7},zeta_{3})capmathbb{Q}(sqrt{2})=mathbb{Q},$$ and consider the compositum $L=K_{1}K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3},sqrt{2})$ as an extension over $mathbb{Q}$. $L/mathbb{Q}$ is Galois and we have the following isomorphism of Galois groups
$$operatorname{Gal}(L/mathbb{Q})stackrel{simeq}{longrightarrow}operatorname{Gal}(K_{1}/mathbb{Q})timesoperatorname{Gal}(K_{2}/mathbb{Q}),$$
which you can use to deduce the answer.
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
$endgroup$
add a comment |
$begingroup$
Both things you've noticed is of help. The second note just shows that the roots of $(x+1)^{2}-2$ are $x=pmsqrt{2}-1$ because $x'=x+1$ and $x'=pmsqrt{2}$ are the roots of $(x')^{2}-2$ and you can show that $K_{1}=mathbb{Q}(sqrt{2})=mathbb{Q}(sqrt{2}-1)$. Likewise for the second factor (where the splitting field is $K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3})$ where $zeta_{3}$ is a primitive third root of unity). Now just show that $$mathbb{Q}(sqrt[3]{7},zeta_{3})capmathbb{Q}(sqrt{2})=mathbb{Q},$$ and consider the compositum $L=K_{1}K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3},sqrt{2})$ as an extension over $mathbb{Q}$. $L/mathbb{Q}$ is Galois and we have the following isomorphism of Galois groups
$$operatorname{Gal}(L/mathbb{Q})stackrel{simeq}{longrightarrow}operatorname{Gal}(K_{1}/mathbb{Q})timesoperatorname{Gal}(K_{2}/mathbb{Q}),$$
which you can use to deduce the answer.
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
$endgroup$
Both things you've noticed is of help. The second note just shows that the roots of $(x+1)^{2}-2$ are $x=pmsqrt{2}-1$ because $x'=x+1$ and $x'=pmsqrt{2}$ are the roots of $(x')^{2}-2$ and you can show that $K_{1}=mathbb{Q}(sqrt{2})=mathbb{Q}(sqrt{2}-1)$. Likewise for the second factor (where the splitting field is $K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3})$ where $zeta_{3}$ is a primitive third root of unity). Now just show that $$mathbb{Q}(sqrt[3]{7},zeta_{3})capmathbb{Q}(sqrt{2})=mathbb{Q},$$ and consider the compositum $L=K_{1}K_{2}=mathbb{Q}(sqrt[3]{7},zeta_{3},sqrt{2})$ as an extension over $mathbb{Q}$. $L/mathbb{Q}$ is Galois and we have the following isomorphism of Galois groups
$$operatorname{Gal}(L/mathbb{Q})stackrel{simeq}{longrightarrow}operatorname{Gal}(K_{1}/mathbb{Q})timesoperatorname{Gal}(K_{2}/mathbb{Q}),$$
which you can use to deduce the answer.
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
answered Mar 23 at 19:12
YumekuiMathYumekuiMath
366114
366114
add a comment |
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