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Show that $mathbbQ (zeta_3 sqrt[3]2) neq mathbbQ(zeta_3^2 sqrt[3]2)$.
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This is the setting I can use:
(P1) : If $q=fracrs$ and $ q,rinmathbbQ$ then $sin mathbbQ$. This is because $fracrq = fracs qq =s in mathbbQ$ since $mathbbQ$ is closed under division.
(P2) : If $F$ is a field and $p(x) in F[x]$ is irreducible over $F$ and of degree $n$, and $alpha$ a root of $p(x)$ then $$ F(alpha)=a_0 + a_1alpha + ...+ a_n-1alpha^n-1 : a_0, a_1,...,a_n-1 in F$$
(P3): $zeta_3$ is a root of $x^2+x+1=0$ and a primitive 3rd root of unity.
Claim: $mathbbQ(zeta_3 sqrt[3]2) neq mathbbQ(zeta_3^2 sqrt[3]2)$.
Proof: Suppose $mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$.
$zeta_3 sqrt[3]2$ and $zeta_3^2 sqrt[3]2$ are roots of the polynomial $p^*(x)=x^3-2$, which is irreducible over $mathbbQ$ by the fact that the third root is $sqrt[3]2$ and none of the them lies in $mathbbQ$. Thus $textdegree(mathbbQ(zeta_3^2 sqrt[3]2))=3$ and $$ mathbbQ(zeta_3^2 sqrt[3]2) = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2: a_0,a_1,a_2 in mathbbQ$$
by (P2). It follows that
$$ zeta sqrt[3]2 = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2 quad textwith quad a_0,a_1,a_2 in mathbbQ$$
The imaginary part $mathcalIm(*)$ of this equation gives $$ mathcalIm(zeta_3)sqrt[3]2=a_1mathcalIm(zeta_3^2)sqrt[3]2 + a_2mathcalIm(zeta_3)sqrt[3]2^2$$ since $mathcalIm(sqrt[3]2)=mathcalIm(a_0)=mathcalIm(a_1)=mathcalIm(a_2)=0$.
It follows from (P3) that $mathcalIm(zeta_3) = -mathcalIm(zeta_3)$ (since $zeta_3^2+zeta_3 +1 =0$), hence we have
$$mathcalIm(zeta_3)cdot (sqrt[3]2+a_1sqrt[3]2-a_2sqrt[3]2^2)=0$$
$mathcalIm(zeta_3)neq0$ by (P3) again, and finally
$$a_2=frac1sqrt[3]2(1+a_1)notin mathbbQ$$
by (P1), which gives a contradiction since $a_2in mathbbQ$ by definition of $mathbbQ(zeta_3^2 sqrt[3]2)$.
Question: Is this proof correct? Do I use some non-trivial concept implicitly anywhere, as happens often to me?
Additionally: if anybody knows alternative proofs, please let me know.
abstract-algebra proof-verification extension-field
asked 2 days ago
C. Moos
85112
add a comment |Â
up vote
3
down vote
favorite
This is the setting I can use:
(P1) : If $q=fracrs$ and $ q,rinmathbbQ$ then $sin mathbbQ$. This is because $fracrq = fracs qq =s in mathbbQ$ since $mathbbQ$ is closed under division.
(P2) : If $F$ is a field and $p(x) in F[x]$ is irreducible over $F$ and of degree $n$, and $alpha$ a root of $p(x)$ then $$ F(alpha)=a_0 + a_1alpha + ...+ a_n-1alpha^n-1 : a_0, a_1,...,a_n-1 in F$$
(P3): $zeta_3$ is a root of $x^2+x+1=0$ and a primitive 3rd root of unity.
Claim: $mathbbQ(zeta_3 sqrt[3]2) neq mathbbQ(zeta_3^2 sqrt[3]2)$.
Proof: Suppose $mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$.
$zeta_3 sqrt[3]2$ and $zeta_3^2 sqrt[3]2$ are roots of the polynomial $p^*(x)=x^3-2$, which is irreducible over $mathbbQ$ by the fact that the third root is $sqrt[3]2$ and none of the them lies in $mathbbQ$. Thus $textdegree(mathbbQ(zeta_3^2 sqrt[3]2))=3$ and $$ mathbbQ(zeta_3^2 sqrt[3]2) = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2: a_0,a_1,a_2 in mathbbQ$$
by (P2). It follows that
$$ zeta sqrt[3]2 = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2 quad textwith quad a_0,a_1,a_2 in mathbbQ$$
The imaginary part $mathcalIm(*)$ of this equation gives $$ mathcalIm(zeta_3)sqrt[3]2=a_1mathcalIm(zeta_3^2)sqrt[3]2 + a_2mathcalIm(zeta_3)sqrt[3]2^2$$ since $mathcalIm(sqrt[3]2)=mathcalIm(a_0)=mathcalIm(a_1)=mathcalIm(a_2)=0$.
It follows from (P3) that $mathcalIm(zeta_3) = -mathcalIm(zeta_3)$ (since $zeta_3^2+zeta_3 +1 =0$), hence we have
$$mathcalIm(zeta_3)cdot (sqrt[3]2+a_1sqrt[3]2-a_2sqrt[3]2^2)=0$$
$mathcalIm(zeta_3)neq0$ by (P3) again, and finally
$$a_2=frac1sqrt[3]2(1+a_1)notin mathbbQ$$
by (P1), which gives a contradiction since $a_2in mathbbQ$ by definition of $mathbbQ(zeta_3^2 sqrt[3]2)$.
Question: Is this proof correct? Do I use some non-trivial concept implicitly anywhere, as happens often to me?
Additionally: if anybody knows alternative proofs, please let me know.
abstract-algebra proof-verification extension-field
asked 2 days ago
C. Moos
85112
Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3).
– C. Moos
2 days ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is the setting I can use:
(P1) : If $q=fracrs$ and $ q,rinmathbbQ$ then $sin mathbbQ$. This is because $fracrq = fracs qq =s in mathbbQ$ since $mathbbQ$ is closed under division.
(P2) : If $F$ is a field and $p(x) in F[x]$ is irreducible over $F$ and of degree $n$, and $alpha$ a root of $p(x)$ then $$ F(alpha)=a_0 + a_1alpha + ...+ a_n-1alpha^n-1 : a_0, a_1,...,a_n-1 in F$$
(P3): $zeta_3$ is a root of $x^2+x+1=0$ and a primitive 3rd root of unity.
Claim: $mathbbQ(zeta_3 sqrt[3]2) neq mathbbQ(zeta_3^2 sqrt[3]2)$.
Proof: Suppose $mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$.
$zeta_3 sqrt[3]2$ and $zeta_3^2 sqrt[3]2$ are roots of the polynomial $p^*(x)=x^3-2$, which is irreducible over $mathbbQ$ by the fact that the third root is $sqrt[3]2$ and none of the them lies in $mathbbQ$. Thus $textdegree(mathbbQ(zeta_3^2 sqrt[3]2))=3$ and $$ mathbbQ(zeta_3^2 sqrt[3]2) = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2: a_0,a_1,a_2 in mathbbQ$$
by (P2). It follows that
$$ zeta sqrt[3]2 = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2 quad textwith quad a_0,a_1,a_2 in mathbbQ$$
The imaginary part $mathcalIm(*)$ of this equation gives $$ mathcalIm(zeta_3)sqrt[3]2=a_1mathcalIm(zeta_3^2)sqrt[3]2 + a_2mathcalIm(zeta_3)sqrt[3]2^2$$ since $mathcalIm(sqrt[3]2)=mathcalIm(a_0)=mathcalIm(a_1)=mathcalIm(a_2)=0$.
It follows from (P3) that $mathcalIm(zeta_3) = -mathcalIm(zeta_3)$ (since $zeta_3^2+zeta_3 +1 =0$), hence we have
$$mathcalIm(zeta_3)cdot (sqrt[3]2+a_1sqrt[3]2-a_2sqrt[3]2^2)=0$$
$mathcalIm(zeta_3)neq0$ by (P3) again, and finally
$$a_2=frac1sqrt[3]2(1+a_1)notin mathbbQ$$
by (P1), which gives a contradiction since $a_2in mathbbQ$ by definition of $mathbbQ(zeta_3^2 sqrt[3]2)$.
Question: Is this proof correct? Do I use some non-trivial concept implicitly anywhere, as happens often to me?
Additionally: if anybody knows alternative proofs, please let me know.
abstract-algebra proof-verification extension-field
asked 2 days ago
C. Moos
85112
This is the setting I can use:
(P1) : If $q=fracrs$ and $ q,rinmathbbQ$ then $sin mathbbQ$. This is because $fracrq = fracs qq =s in mathbbQ$ since $mathbbQ$ is closed under division.
(P2) : If $F$ is a field and $p(x) in F[x]$ is irreducible over $F$ and of degree $n$, and $alpha$ a root of $p(x)$ then $$ F(alpha)=a_0 + a_1alpha + ...+ a_n-1alpha^n-1 : a_0, a_1,...,a_n-1 in F$$
(P3): $zeta_3$ is a root of $x^2+x+1=0$ and a primitive 3rd root of unity.
Claim: $mathbbQ(zeta_3 sqrt[3]2) neq mathbbQ(zeta_3^2 sqrt[3]2)$.
Proof: Suppose $mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$.
$zeta_3 sqrt[3]2$ and $zeta_3^2 sqrt[3]2$ are roots of the polynomial $p^*(x)=x^3-2$, which is irreducible over $mathbbQ$ by the fact that the third root is $sqrt[3]2$ and none of the them lies in $mathbbQ$. Thus $textdegree(mathbbQ(zeta_3^2 sqrt[3]2))=3$ and $$ mathbbQ(zeta_3^2 sqrt[3]2) = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2: a_0,a_1,a_2 in mathbbQ$$
by (P2). It follows that
$$ zeta sqrt[3]2 = a_0 + a_1 (zeta_3^2 sqrt[3]2)+a_2(zeta_3^2 sqrt[3]2)^2 quad textwith quad a_0,a_1,a_2 in mathbbQ$$
The imaginary part $mathcalIm(*)$ of this equation gives $$ mathcalIm(zeta_3)sqrt[3]2=a_1mathcalIm(zeta_3^2)sqrt[3]2 + a_2mathcalIm(zeta_3)sqrt[3]2^2$$ since $mathcalIm(sqrt[3]2)=mathcalIm(a_0)=mathcalIm(a_1)=mathcalIm(a_2)=0$.
It follows from (P3) that $mathcalIm(zeta_3) = -mathcalIm(zeta_3)$ (since $zeta_3^2+zeta_3 +1 =0$), hence we have
$$mathcalIm(zeta_3)cdot (sqrt[3]2+a_1sqrt[3]2-a_2sqrt[3]2^2)=0$$
$mathcalIm(zeta_3)neq0$ by (P3) again, and finally
$$a_2=frac1sqrt[3]2(1+a_1)notin mathbbQ$$
by (P1), which gives a contradiction since $a_2in mathbbQ$ by definition of $mathbbQ(zeta_3^2 sqrt[3]2)$.
Question: Is this proof correct? Do I use some non-trivial concept implicitly anywhere, as happens often to me?
Additionally: if anybody knows alternative proofs, please let me know.
abstract-algebra proof-verification extension-field
asked 2 days ago
C. Moos
85112
edited 2 days ago
asked 2 days ago
C. Moos
85112
asked 2 days ago
C. Moos
85112
asked 2 days ago
C. Moos
85112
85112
Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3).
– C. Moos
2 days ago
add a comment |Â
Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3).
– C. Moos
2 days ago
Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3).
– C. Moos
2 days ago
Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3).
– C. Moos
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Alternate proof, using Galois theory:
$Bbb Q(zeta_3,sqrt[3]2)$ is Galois over $Bbb Q$. Its Galois group is generated by the two automorphisms
$$
tau:zeta_3mapstozeta_3^2,sqrt[3]2mapsto sqrt[3]2\
sigma:zeta_3mapstozeta_3, sqrt[3]2mapsto zeta_3sqrt[3]2
$$
The automorphism $tausigma$ keeps one of your two fields fixed and not the other, so they can't be the same subfield of $Bbb Q(zeta_3,sqrt[3]2)$.
answered 2 days ago
Arthur
97.9k792173
1
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
add a comment |Â
up vote
4
down vote
Another proof: if $K:=mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$, then $zeta_3in K$, so $$mathbb Q subset mathbbQ(zeta_3) subseteq mathbbQ(zeta_3sqrt[3]2).$$
But $[mathbbQ(zeta_3 sqrt[3]2) : mathbbQ]=3$, since $ zeta_3 sqrt[3]2$ is a root of $X^3-2$, and $[ mathbbQ(zeta_3): mathbbQ]=2$ analogously, so the hypothesis would imply that 2 divides 3, which is absurd.
answered 2 days ago
xarles
83548
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Alternate proof, using Galois theory:
$Bbb Q(zeta_3,sqrt[3]2)$ is Galois over $Bbb Q$. Its Galois group is generated by the two automorphisms
$$
tau:zeta_3mapstozeta_3^2,sqrt[3]2mapsto sqrt[3]2\
sigma:zeta_3mapstozeta_3, sqrt[3]2mapsto zeta_3sqrt[3]2
$$
The automorphism $tausigma$ keeps one of your two fields fixed and not the other, so they can't be the same subfield of $Bbb Q(zeta_3,sqrt[3]2)$.
answered 2 days ago
Arthur
97.9k792173
1
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
add a comment |Â
up vote
4
down vote
accepted
Alternate proof, using Galois theory:
$Bbb Q(zeta_3,sqrt[3]2)$ is Galois over $Bbb Q$. Its Galois group is generated by the two automorphisms
$$
tau:zeta_3mapstozeta_3^2,sqrt[3]2mapsto sqrt[3]2\
sigma:zeta_3mapstozeta_3, sqrt[3]2mapsto zeta_3sqrt[3]2
$$
The automorphism $tausigma$ keeps one of your two fields fixed and not the other, so they can't be the same subfield of $Bbb Q(zeta_3,sqrt[3]2)$.
answered 2 days ago
Arthur
97.9k792173
1
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Alternate proof, using Galois theory:
$Bbb Q(zeta_3,sqrt[3]2)$ is Galois over $Bbb Q$. Its Galois group is generated by the two automorphisms
$$
tau:zeta_3mapstozeta_3^2,sqrt[3]2mapsto sqrt[3]2\
sigma:zeta_3mapstozeta_3, sqrt[3]2mapsto zeta_3sqrt[3]2
$$
The automorphism $tausigma$ keeps one of your two fields fixed and not the other, so they can't be the same subfield of $Bbb Q(zeta_3,sqrt[3]2)$.
answered 2 days ago
Arthur
97.9k792173
Alternate proof, using Galois theory:
$Bbb Q(zeta_3,sqrt[3]2)$ is Galois over $Bbb Q$. Its Galois group is generated by the two automorphisms
$$
tau:zeta_3mapstozeta_3^2,sqrt[3]2mapsto sqrt[3]2\
sigma:zeta_3mapstozeta_3, sqrt[3]2mapsto zeta_3sqrt[3]2
$$
The automorphism $tausigma$ keeps one of your two fields fixed and not the other, so they can't be the same subfield of $Bbb Q(zeta_3,sqrt[3]2)$.
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
97.9k792173
1
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
add a comment |Â
1
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
1
1
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks
– C. Moos
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
I agree (+1)....
– Dietrich Burde
2 days ago
add a comment |Â
up vote
4
down vote
Another proof: if $K:=mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$, then $zeta_3in K$, so $$mathbb Q subset mathbbQ(zeta_3) subseteq mathbbQ(zeta_3sqrt[3]2).$$
But $[mathbbQ(zeta_3 sqrt[3]2) : mathbbQ]=3$, since $ zeta_3 sqrt[3]2$ is a root of $X^3-2$, and $[ mathbbQ(zeta_3): mathbbQ]=2$ analogously, so the hypothesis would imply that 2 divides 3, which is absurd.
answered 2 days ago
xarles
83548
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
add a comment |Â
up vote
4
down vote
Another proof: if $K:=mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$, then $zeta_3in K$, so $$mathbb Q subset mathbbQ(zeta_3) subseteq mathbbQ(zeta_3sqrt[3]2).$$
But $[mathbbQ(zeta_3 sqrt[3]2) : mathbbQ]=3$, since $ zeta_3 sqrt[3]2$ is a root of $X^3-2$, and $[ mathbbQ(zeta_3): mathbbQ]=2$ analogously, so the hypothesis would imply that 2 divides 3, which is absurd.
answered 2 days ago
xarles
83548
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Another proof: if $K:=mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$, then $zeta_3in K$, so $$mathbb Q subset mathbbQ(zeta_3) subseteq mathbbQ(zeta_3sqrt[3]2).$$
But $[mathbbQ(zeta_3 sqrt[3]2) : mathbbQ]=3$, since $ zeta_3 sqrt[3]2$ is a root of $X^3-2$, and $[ mathbbQ(zeta_3): mathbbQ]=2$ analogously, so the hypothesis would imply that 2 divides 3, which is absurd.
answered 2 days ago
xarles
83548
Another proof: if $K:=mathbbQ(zeta_3 sqrt[3]2) = mathbbQ(zeta_3^2 sqrt[3]2)$, then $zeta_3in K$, so $$mathbb Q subset mathbbQ(zeta_3) subseteq mathbbQ(zeta_3sqrt[3]2).$$
But $[mathbbQ(zeta_3 sqrt[3]2) : mathbbQ]=3$, since $ zeta_3 sqrt[3]2$ is a root of $X^3-2$, and $[ mathbbQ(zeta_3): mathbbQ]=2$ analogously, so the hypothesis would imply that 2 divides 3, which is absurd.
answered 2 days ago
xarles
83548
edited 2 days ago
answered 2 days ago
xarles
83548
answered 2 days ago
xarles
83548
answered 2 days ago
xarles
83548
83548
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
add a comment |Â
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you.
– C. Moos
2 days ago
add a comment |Â
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Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3).
– C. Moos
2 days ago