Mixing
Algebraic division algebra over Euclidean field
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By Frobenius' Theorem we know that if $D$ be an algebraic non-commutative division algebra over $mathbbR$ then ,as an $mathbbR$-algebra, $D$ is isomorphic to
$mathbbH$.
We can also replace $mathbbR$ by any real closed field. if we replace $mathbbR$ by an Euclidean field can we say the same statement is true?
A Euclidean field is an ordered field $K$ for which every non-negative element is a square: that is, $0 le x$ in $K$ implies that $x = y^2$ for some y in K.
division-algebras
asked Jul 25 at 18:14
s.Bahari
384
add a comment |Â
up vote
3
down vote
favorite
By Frobenius' Theorem we know that if $D$ be an algebraic non-commutative division algebra over $mathbbR$ then ,as an $mathbbR$-algebra, $D$ is isomorphic to
$mathbbH$.
We can also replace $mathbbR$ by any real closed field. if we replace $mathbbR$ by an Euclidean field can we say the same statement is true?
A Euclidean field is an ordered field $K$ for which every non-negative element is a square: that is, $0 le x$ in $K$ implies that $x = y^2$ for some y in K.
division-algebras
asked Jul 25 at 18:14
s.Bahari
384
2
If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $Bbb Q$, and see what happened when I extended the base from $Bbb Q$ to $K$.
– Lubin
Jul 25 at 19:04
1
Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/
– Jyrki Lahtonen
Jul 26 at 8:09
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
By Frobenius' Theorem we know that if $D$ be an algebraic non-commutative division algebra over $mathbbR$ then ,as an $mathbbR$-algebra, $D$ is isomorphic to
$mathbbH$.
We can also replace $mathbbR$ by any real closed field. if we replace $mathbbR$ by an Euclidean field can we say the same statement is true?
A Euclidean field is an ordered field $K$ for which every non-negative element is a square: that is, $0 le x$ in $K$ implies that $x = y^2$ for some y in K.
division-algebras
asked Jul 25 at 18:14
s.Bahari
384
By Frobenius' Theorem we know that if $D$ be an algebraic non-commutative division algebra over $mathbbR$ then ,as an $mathbbR$-algebra, $D$ is isomorphic to
$mathbbH$.
We can also replace $mathbbR$ by any real closed field. if we replace $mathbbR$ by an Euclidean field can we say the same statement is true?
A Euclidean field is an ordered field $K$ for which every non-negative element is a square: that is, $0 le x$ in $K$ implies that $x = y^2$ for some y in K.
division-algebras
asked Jul 25 at 18:14
s.Bahari
384
asked Jul 25 at 18:14
s.Bahari
384
asked Jul 25 at 18:14
s.Bahari
384
asked Jul 25 at 18:14
s.Bahari
384
384
2
If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $Bbb Q$, and see what happened when I extended the base from $Bbb Q$ to $K$.
– Lubin
Jul 25 at 19:04
1
Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/
– Jyrki Lahtonen
Jul 26 at 8:09
add a comment |Â
2
If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $Bbb Q$, and see what happened when I extended the base from $Bbb Q$ to $K$.
– Lubin
Jul 25 at 19:04
1
Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/
– Jyrki Lahtonen
Jul 26 at 8:09
2
2
If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $Bbb Q$, and see what happened when I extended the base from $Bbb Q$ to $K$.
– Lubin
Jul 25 at 19:04
If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $Bbb Q$, and see what happened when I extended the base from $Bbb Q$ to $K$.
– Lubin
Jul 25 at 19:04
1
1
Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/
– Jyrki Lahtonen
Jul 26 at 8:09
Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/
– Jyrki Lahtonen
Jul 26 at 8:09
add a comment |Â
1 Answer
1
active
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up vote
2
down vote
accepted
I don't think the conclusion holds for all such $K$.
Consider the field $K$ of (compass and straightedge) constructible real numbers. It is Euclidean for sure. It is ordered because it is a subfield of the reals. Because the regular 9-gon is not constructible, the number $z=z_1=2cos(2pi/9)notin K$. Nor are its Galois conjugates $z_2=z^2-2=2cos(4pi/9)$
and $z_3=(z^2-2)^2-2=2cos(8pi/9)$. Their shared minimal polynomial over $BbbQ$ is the familiar
$$
m(x):=(x-z_1)(x-z_2)(x-z_3)=x^3-3x+1.
$$
This remains irreducible over $K$, so we see that $L=K(z)$ is a cubic extension. Furthermore, $L/K$ is Galois with a cyclic Galois group generated by the $K$-automorphism $sigma:z_1mapsto z_2mapsto z_3mapsto z_1$.
The idea is to construct a cyclic division algebra (locally see e.g. this answer) from the extension $L/K$. More specifically, I will apply a theorem of A. Albert. It calls for a so called non-norm element $ain K$ such that $aneq N_L/K(x)$ for all $xin L$. Here $N_L/K$ is the (relative) norm
$$N_L/K:Lto K, xmapsto xsigma(x)sigma^2(x).$$
I claim that $2$ is a non-norm element. Assume contrariwise that
$$2=N_L/K(x)$$
for some $x=a_0+a_1z+a_2z^2in L, a_0,a_1,a_2in K$. Let $F=BbbQ(a_0,a_1,a_2)$, and $E=F(z)$. Then $E/F$ is also a cubic cyclic Galois extension, and the assumption is also that $N_E/F(x)=2$. As the numbers $a_0,a_1,a_2$ are constructible we know that $[F:BbbQ]$ is a power of two, say $2^n$.
Furthermore, we have a tower of quadratic extensions
$$
BbbQ=F_0subset F_1subset cdotssubset F_n=F.
$$
Let $mathfrakp$ be a prime ideal of $F$ lying above the rational prime $2$. Because inertia degree is multiplicative in a tower, it follows that the inertia degree $f(mathfrakp|2)$ is also a power of two. This is because the inertia degree $f(mathfrakpcap F_i|mathfrakpcap F_i-1)$ can only be $1$ or $2$ for all $i=1,2,ldots,n$.
On the other hand, $m(x)$ is irreducible modulo two, so the prime $p=2$ is inert in the cubic extension $BbbQ(z)/BbbQ$. Therefore any prime ideal of $E$ lying above $p=2$ has inertia degree divisible by three. In particular, $mathfrakp$ is inert in $E/F$.
Consider the fractional ideal $I$ of $E$ generated by $x$. Its ideal norm
$N_E/F(I)$ was assumed to be the principal ideal $(2)$. But, the previous
result implies any prime ideal $mathfrakp|2$ appears as a factor in $N_E/F(I)$ with multiplicity divisible by three. On the other hand, trace it through the tower of intermediate fields again, in the prime ideal decomposition of $(2)$ as an ideal of $F$ all the primes appear with a multiplicity that is a power of two. This is contradiction.
Consequently Albert's theorem tells us that the cyclic algebra
$$mathcalA=Loplus Luoplus Lu^2$$
with its $K$-linear multiplication defined by the rules
- $u^3=2$, and
- $ku=usigma(k)$ for all $kin K$
is a $9$-dimensional division algebra over $K$.
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
1
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
1
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
 |Â
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I don't think the conclusion holds for all such $K$.
Consider the field $K$ of (compass and straightedge) constructible real numbers. It is Euclidean for sure. It is ordered because it is a subfield of the reals. Because the regular 9-gon is not constructible, the number $z=z_1=2cos(2pi/9)notin K$. Nor are its Galois conjugates $z_2=z^2-2=2cos(4pi/9)$
and $z_3=(z^2-2)^2-2=2cos(8pi/9)$. Their shared minimal polynomial over $BbbQ$ is the familiar
$$
m(x):=(x-z_1)(x-z_2)(x-z_3)=x^3-3x+1.
$$
This remains irreducible over $K$, so we see that $L=K(z)$ is a cubic extension. Furthermore, $L/K$ is Galois with a cyclic Galois group generated by the $K$-automorphism $sigma:z_1mapsto z_2mapsto z_3mapsto z_1$.
The idea is to construct a cyclic division algebra (locally see e.g. this answer) from the extension $L/K$. More specifically, I will apply a theorem of A. Albert. It calls for a so called non-norm element $ain K$ such that $aneq N_L/K(x)$ for all $xin L$. Here $N_L/K$ is the (relative) norm
$$N_L/K:Lto K, xmapsto xsigma(x)sigma^2(x).$$
I claim that $2$ is a non-norm element. Assume contrariwise that
$$2=N_L/K(x)$$
for some $x=a_0+a_1z+a_2z^2in L, a_0,a_1,a_2in K$. Let $F=BbbQ(a_0,a_1,a_2)$, and $E=F(z)$. Then $E/F$ is also a cubic cyclic Galois extension, and the assumption is also that $N_E/F(x)=2$. As the numbers $a_0,a_1,a_2$ are constructible we know that $[F:BbbQ]$ is a power of two, say $2^n$.
Furthermore, we have a tower of quadratic extensions
$$
BbbQ=F_0subset F_1subset cdotssubset F_n=F.
$$
Let $mathfrakp$ be a prime ideal of $F$ lying above the rational prime $2$. Because inertia degree is multiplicative in a tower, it follows that the inertia degree $f(mathfrakp|2)$ is also a power of two. This is because the inertia degree $f(mathfrakpcap F_i|mathfrakpcap F_i-1)$ can only be $1$ or $2$ for all $i=1,2,ldots,n$.
On the other hand, $m(x)$ is irreducible modulo two, so the prime $p=2$ is inert in the cubic extension $BbbQ(z)/BbbQ$. Therefore any prime ideal of $E$ lying above $p=2$ has inertia degree divisible by three. In particular, $mathfrakp$ is inert in $E/F$.
Consider the fractional ideal $I$ of $E$ generated by $x$. Its ideal norm
$N_E/F(I)$ was assumed to be the principal ideal $(2)$. But, the previous
result implies any prime ideal $mathfrakp|2$ appears as a factor in $N_E/F(I)$ with multiplicity divisible by three. On the other hand, trace it through the tower of intermediate fields again, in the prime ideal decomposition of $(2)$ as an ideal of $F$ all the primes appear with a multiplicity that is a power of two. This is contradiction.
Consequently Albert's theorem tells us that the cyclic algebra
$$mathcalA=Loplus Luoplus Lu^2$$
with its $K$-linear multiplication defined by the rules
- $u^3=2$, and
- $ku=usigma(k)$ for all $kin K$
is a $9$-dimensional division algebra over $K$.
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
1
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
1
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
 |Â
show 1 more comment
up vote
2
down vote
accepted
I don't think the conclusion holds for all such $K$.
Consider the field $K$ of (compass and straightedge) constructible real numbers. It is Euclidean for sure. It is ordered because it is a subfield of the reals. Because the regular 9-gon is not constructible, the number $z=z_1=2cos(2pi/9)notin K$. Nor are its Galois conjugates $z_2=z^2-2=2cos(4pi/9)$
and $z_3=(z^2-2)^2-2=2cos(8pi/9)$. Their shared minimal polynomial over $BbbQ$ is the familiar
$$
m(x):=(x-z_1)(x-z_2)(x-z_3)=x^3-3x+1.
$$
This remains irreducible over $K$, so we see that $L=K(z)$ is a cubic extension. Furthermore, $L/K$ is Galois with a cyclic Galois group generated by the $K$-automorphism $sigma:z_1mapsto z_2mapsto z_3mapsto z_1$.
The idea is to construct a cyclic division algebra (locally see e.g. this answer) from the extension $L/K$. More specifically, I will apply a theorem of A. Albert. It calls for a so called non-norm element $ain K$ such that $aneq N_L/K(x)$ for all $xin L$. Here $N_L/K$ is the (relative) norm
$$N_L/K:Lto K, xmapsto xsigma(x)sigma^2(x).$$
I claim that $2$ is a non-norm element. Assume contrariwise that
$$2=N_L/K(x)$$
for some $x=a_0+a_1z+a_2z^2in L, a_0,a_1,a_2in K$. Let $F=BbbQ(a_0,a_1,a_2)$, and $E=F(z)$. Then $E/F$ is also a cubic cyclic Galois extension, and the assumption is also that $N_E/F(x)=2$. As the numbers $a_0,a_1,a_2$ are constructible we know that $[F:BbbQ]$ is a power of two, say $2^n$.
Furthermore, we have a tower of quadratic extensions
$$
BbbQ=F_0subset F_1subset cdotssubset F_n=F.
$$
Let $mathfrakp$ be a prime ideal of $F$ lying above the rational prime $2$. Because inertia degree is multiplicative in a tower, it follows that the inertia degree $f(mathfrakp|2)$ is also a power of two. This is because the inertia degree $f(mathfrakpcap F_i|mathfrakpcap F_i-1)$ can only be $1$ or $2$ for all $i=1,2,ldots,n$.
On the other hand, $m(x)$ is irreducible modulo two, so the prime $p=2$ is inert in the cubic extension $BbbQ(z)/BbbQ$. Therefore any prime ideal of $E$ lying above $p=2$ has inertia degree divisible by three. In particular, $mathfrakp$ is inert in $E/F$.
Consider the fractional ideal $I$ of $E$ generated by $x$. Its ideal norm
$N_E/F(I)$ was assumed to be the principal ideal $(2)$. But, the previous
result implies any prime ideal $mathfrakp|2$ appears as a factor in $N_E/F(I)$ with multiplicity divisible by three. On the other hand, trace it through the tower of intermediate fields again, in the prime ideal decomposition of $(2)$ as an ideal of $F$ all the primes appear with a multiplicity that is a power of two. This is contradiction.
Consequently Albert's theorem tells us that the cyclic algebra
$$mathcalA=Loplus Luoplus Lu^2$$
with its $K$-linear multiplication defined by the rules
- $u^3=2$, and
- $ku=usigma(k)$ for all $kin K$
is a $9$-dimensional division algebra over $K$.
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
1
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
1
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I don't think the conclusion holds for all such $K$.
Consider the field $K$ of (compass and straightedge) constructible real numbers. It is Euclidean for sure. It is ordered because it is a subfield of the reals. Because the regular 9-gon is not constructible, the number $z=z_1=2cos(2pi/9)notin K$. Nor are its Galois conjugates $z_2=z^2-2=2cos(4pi/9)$
and $z_3=(z^2-2)^2-2=2cos(8pi/9)$. Their shared minimal polynomial over $BbbQ$ is the familiar
$$
m(x):=(x-z_1)(x-z_2)(x-z_3)=x^3-3x+1.
$$
This remains irreducible over $K$, so we see that $L=K(z)$ is a cubic extension. Furthermore, $L/K$ is Galois with a cyclic Galois group generated by the $K$-automorphism $sigma:z_1mapsto z_2mapsto z_3mapsto z_1$.
The idea is to construct a cyclic division algebra (locally see e.g. this answer) from the extension $L/K$. More specifically, I will apply a theorem of A. Albert. It calls for a so called non-norm element $ain K$ such that $aneq N_L/K(x)$ for all $xin L$. Here $N_L/K$ is the (relative) norm
$$N_L/K:Lto K, xmapsto xsigma(x)sigma^2(x).$$
I claim that $2$ is a non-norm element. Assume contrariwise that
$$2=N_L/K(x)$$
for some $x=a_0+a_1z+a_2z^2in L, a_0,a_1,a_2in K$. Let $F=BbbQ(a_0,a_1,a_2)$, and $E=F(z)$. Then $E/F$ is also a cubic cyclic Galois extension, and the assumption is also that $N_E/F(x)=2$. As the numbers $a_0,a_1,a_2$ are constructible we know that $[F:BbbQ]$ is a power of two, say $2^n$.
Furthermore, we have a tower of quadratic extensions
$$
BbbQ=F_0subset F_1subset cdotssubset F_n=F.
$$
Let $mathfrakp$ be a prime ideal of $F$ lying above the rational prime $2$. Because inertia degree is multiplicative in a tower, it follows that the inertia degree $f(mathfrakp|2)$ is also a power of two. This is because the inertia degree $f(mathfrakpcap F_i|mathfrakpcap F_i-1)$ can only be $1$ or $2$ for all $i=1,2,ldots,n$.
On the other hand, $m(x)$ is irreducible modulo two, so the prime $p=2$ is inert in the cubic extension $BbbQ(z)/BbbQ$. Therefore any prime ideal of $E$ lying above $p=2$ has inertia degree divisible by three. In particular, $mathfrakp$ is inert in $E/F$.
Consider the fractional ideal $I$ of $E$ generated by $x$. Its ideal norm
$N_E/F(I)$ was assumed to be the principal ideal $(2)$. But, the previous
result implies any prime ideal $mathfrakp|2$ appears as a factor in $N_E/F(I)$ with multiplicity divisible by three. On the other hand, trace it through the tower of intermediate fields again, in the prime ideal decomposition of $(2)$ as an ideal of $F$ all the primes appear with a multiplicity that is a power of two. This is contradiction.
Consequently Albert's theorem tells us that the cyclic algebra
$$mathcalA=Loplus Luoplus Lu^2$$
with its $K$-linear multiplication defined by the rules
- $u^3=2$, and
- $ku=usigma(k)$ for all $kin K$
is a $9$-dimensional division algebra over $K$.
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
I don't think the conclusion holds for all such $K$.
Consider the field $K$ of (compass and straightedge) constructible real numbers. It is Euclidean for sure. It is ordered because it is a subfield of the reals. Because the regular 9-gon is not constructible, the number $z=z_1=2cos(2pi/9)notin K$. Nor are its Galois conjugates $z_2=z^2-2=2cos(4pi/9)$
and $z_3=(z^2-2)^2-2=2cos(8pi/9)$. Their shared minimal polynomial over $BbbQ$ is the familiar
$$
m(x):=(x-z_1)(x-z_2)(x-z_3)=x^3-3x+1.
$$
This remains irreducible over $K$, so we see that $L=K(z)$ is a cubic extension. Furthermore, $L/K$ is Galois with a cyclic Galois group generated by the $K$-automorphism $sigma:z_1mapsto z_2mapsto z_3mapsto z_1$.
The idea is to construct a cyclic division algebra (locally see e.g. this answer) from the extension $L/K$. More specifically, I will apply a theorem of A. Albert. It calls for a so called non-norm element $ain K$ such that $aneq N_L/K(x)$ for all $xin L$. Here $N_L/K$ is the (relative) norm
$$N_L/K:Lto K, xmapsto xsigma(x)sigma^2(x).$$
I claim that $2$ is a non-norm element. Assume contrariwise that
$$2=N_L/K(x)$$
for some $x=a_0+a_1z+a_2z^2in L, a_0,a_1,a_2in K$. Let $F=BbbQ(a_0,a_1,a_2)$, and $E=F(z)$. Then $E/F$ is also a cubic cyclic Galois extension, and the assumption is also that $N_E/F(x)=2$. As the numbers $a_0,a_1,a_2$ are constructible we know that $[F:BbbQ]$ is a power of two, say $2^n$.
Furthermore, we have a tower of quadratic extensions
$$
BbbQ=F_0subset F_1subset cdotssubset F_n=F.
$$
Let $mathfrakp$ be a prime ideal of $F$ lying above the rational prime $2$. Because inertia degree is multiplicative in a tower, it follows that the inertia degree $f(mathfrakp|2)$ is also a power of two. This is because the inertia degree $f(mathfrakpcap F_i|mathfrakpcap F_i-1)$ can only be $1$ or $2$ for all $i=1,2,ldots,n$.
On the other hand, $m(x)$ is irreducible modulo two, so the prime $p=2$ is inert in the cubic extension $BbbQ(z)/BbbQ$. Therefore any prime ideal of $E$ lying above $p=2$ has inertia degree divisible by three. In particular, $mathfrakp$ is inert in $E/F$.
Consider the fractional ideal $I$ of $E$ generated by $x$. Its ideal norm
$N_E/F(I)$ was assumed to be the principal ideal $(2)$. But, the previous
result implies any prime ideal $mathfrakp|2$ appears as a factor in $N_E/F(I)$ with multiplicity divisible by three. On the other hand, trace it through the tower of intermediate fields again, in the prime ideal decomposition of $(2)$ as an ideal of $F$ all the primes appear with a multiplicity that is a power of two. This is contradiction.
Consequently Albert's theorem tells us that the cyclic algebra
$$mathcalA=Loplus Luoplus Lu^2$$
with its $K$-linear multiplication defined by the rules
- $u^3=2$, and
- $ku=usigma(k)$ for all $kin K$
is a $9$-dimensional division algebra over $K$.
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
edited Jul 26 at 8:17
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
answered Jul 26 at 8:03
Jyrki Lahtonen
105k12161355
105k12161355
1
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
1
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
 |Â
show 1 more comment
1
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
1
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
1
1
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
There may be 1) easier ways of showing that $N_L/K$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods.
– Jyrki Lahtonen
Jul 26 at 8:06
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra?
– s.Bahari
Jul 26 at 8:16
1
1
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
@s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(sqrt-d)$ for some positive element $din E$. But $sqrtdin E$, so $L=E(sqrt-1)$. This should imply that $D$ is the usual quaternion algebra over $E$.
– Jyrki Lahtonen
Jul 26 at 8:24
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
Thank again. This help me a lot.
– s.Bahari
Jul 26 at 8:28
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
@s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $jin D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$?
– Jyrki Lahtonen
Jul 27 at 5:14
 |Â
show 1 more comment
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2
If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $Bbb Q$, and see what happened when I extended the base from $Bbb Q$ to $K$.
– Lubin
Jul 25 at 19:04
1
Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/
– Jyrki Lahtonen
Jul 26 at 8:09