Mixing
Maximal subfield of a central simple algebra which is not Galois
$begingroup$
In the book Algebra IX: Finite Groups of Lie type and Finite Dimensional Algebra, the authors Kostrikin-Shafarevich mention (p. 159) that
If $A$ is a central simple algebra over $F$ of finite dimension and if $K$ is a maximal subfield of $A$, then $K$ may not be Galois extension of $F$.
On the other hand, in a paper on Central Simple Algebras, Rowen mentiones
Every division algebra of degree $2, 3, 4, 6$, or $12$ is a crossed product (i.e. maximal subfields are Galois extensions; am I right? Here degree of $D$ over $F$ is $n$ means $dim_F(D)=n^2$.)
Q. What is example of a non-commutative simple (or division) algebra of finite dimension over $F$ in which a maximal subfield is not Galois extension of $F$?
(I have no idea of how difficult is this problem in the literature, and I have never seen simple algebras other than quaternions and matrix algebras over quaternions or fields.)
division-algebras
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
$endgroup$
|
show 1 more comment
$begingroup$
In the book Algebra IX: Finite Groups of Lie type and Finite Dimensional Algebra, the authors Kostrikin-Shafarevich mention (p. 159) that
If $A$ is a central simple algebra over $F$ of finite dimension and if $K$ is a maximal subfield of $A$, then $K$ may not be Galois extension of $F$.
On the other hand, in a paper on Central Simple Algebras, Rowen mentiones
Every division algebra of degree $2, 3, 4, 6$, or $12$ is a crossed product (i.e. maximal subfields are Galois extensions; am I right? Here degree of $D$ over $F$ is $n$ means $dim_F(D)=n^2$.)
Q. What is example of a non-commutative simple (or division) algebra of finite dimension over $F$ in which a maximal subfield is not Galois extension of $F$?
(I have no idea of how difficult is this problem in the literature, and I have never seen simple algebras other than quaternions and matrix algebras over quaternions or fields.)
division-algebras
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
$endgroup$
2
$begingroup$
In anothe question you asked for a 9-dimensional division algebra, central over $Bbb{Q}$ containing a copy of the non-Galois extension field $Bbb{Q}(root3of2)$. Is that not an answer to this question as well :-)
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:11
2
$begingroup$
Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 18:33
2
$begingroup$
Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $Lvert K$ with degree $[L:K] le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 19:00
$begingroup$
@Torsten The double centralizer theorem guarantees that any $L/K$ such that $[L:K]=n$ is maximal in $M_n(K)$. Granted, calling upon that result may be overkill :-).
$endgroup$
– Jyrki Lahtonen
Jan 30 at 16:19
2
$begingroup$
@Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and...
$endgroup$
– Jyrki Lahtonen
Jan 30 at 19:43
|
show 1 more comment
$begingroup$
In the book Algebra IX: Finite Groups of Lie type and Finite Dimensional Algebra, the authors Kostrikin-Shafarevich mention (p. 159) that
If $A$ is a central simple algebra over $F$ of finite dimension and if $K$ is a maximal subfield of $A$, then $K$ may not be Galois extension of $F$.
On the other hand, in a paper on Central Simple Algebras, Rowen mentiones
Every division algebra of degree $2, 3, 4, 6$, or $12$ is a crossed product (i.e. maximal subfields are Galois extensions; am I right? Here degree of $D$ over $F$ is $n$ means $dim_F(D)=n^2$.)
Q. What is example of a non-commutative simple (or division) algebra of finite dimension over $F$ in which a maximal subfield is not Galois extension of $F$?
(I have no idea of how difficult is this problem in the literature, and I have never seen simple algebras other than quaternions and matrix algebras over quaternions or fields.)
division-algebras
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
$endgroup$
In the book Algebra IX: Finite Groups of Lie type and Finite Dimensional Algebra, the authors Kostrikin-Shafarevich mention (p. 159) that
If $A$ is a central simple algebra over $F$ of finite dimension and if $K$ is a maximal subfield of $A$, then $K$ may not be Galois extension of $F$.
On the other hand, in a paper on Central Simple Algebras, Rowen mentiones
Every division algebra of degree $2, 3, 4, 6$, or $12$ is a crossed product (i.e. maximal subfields are Galois extensions; am I right? Here degree of $D$ over $F$ is $n$ means $dim_F(D)=n^2$.)
Q. What is example of a non-commutative simple (or division) algebra of finite dimension over $F$ in which a maximal subfield is not Galois extension of $F$?
(I have no idea of how difficult is this problem in the literature, and I have never seen simple algebras other than quaternions and matrix algebras over quaternions or fields.)
division-algebras
division-algebras
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
asked Jan 21 at 12:06
BeginnerBeginner
3,94611225
3,94611225
2
$begingroup$
In anothe question you asked for a 9-dimensional division algebra, central over $Bbb{Q}$ containing a copy of the non-Galois extension field $Bbb{Q}(root3of2)$. Is that not an answer to this question as well :-)
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:11
2
$begingroup$
Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 18:33
2
$begingroup$
Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $Lvert K$ with degree $[L:K] le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 19:00
$begingroup$
@Torsten The double centralizer theorem guarantees that any $L/K$ such that $[L:K]=n$ is maximal in $M_n(K)$. Granted, calling upon that result may be overkill :-).
$endgroup$
– Jyrki Lahtonen
Jan 30 at 16:19
2
$begingroup$
@Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and...
$endgroup$
– Jyrki Lahtonen
Jan 30 at 19:43
|
show 1 more comment
2
$begingroup$
In anothe question you asked for a 9-dimensional division algebra, central over $Bbb{Q}$ containing a copy of the non-Galois extension field $Bbb{Q}(root3of2)$. Is that not an answer to this question as well :-)
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:11
2
$begingroup$
Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 18:33
2
$begingroup$
Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $Lvert K$ with degree $[L:K] le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 19:00
$begingroup$
@Torsten The double centralizer theorem guarantees that any $L/K$ such that $[L:K]=n$ is maximal in $M_n(K)$. Granted, calling upon that result may be overkill :-).
$endgroup$
– Jyrki Lahtonen
Jan 30 at 16:19
2
$begingroup$
@Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and...
$endgroup$
– Jyrki Lahtonen
Jan 30 at 19:43
2
2
$begingroup$
In anothe question you asked for a 9-dimensional division algebra, central over $Bbb{Q}$ containing a copy of the non-Galois extension field $Bbb{Q}(root3of2)$. Is that not an answer to this question as well :-)
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:11
$begingroup$
In anothe question you asked for a 9-dimensional division algebra, central over $Bbb{Q}$ containing a copy of the non-Galois extension field $Bbb{Q}(root3of2)$. Is that not an answer to this question as well :-)
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:11
2
2
$begingroup$
Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 18:33
$begingroup$
Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 18:33
2
2
$begingroup$
Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $Lvert K$ with degree $[L:K] le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 19:00
$begingroup$
Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $Lvert K$ with degree $[L:K] le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 19:00
$begingroup$
@Torsten The double centralizer theorem guarantees that any $L/K$ such that $[L:K]=n$ is maximal in $M_n(K)$. Granted, calling upon that result may be overkill :-).
$endgroup$
– Jyrki Lahtonen
Jan 30 at 16:19
$begingroup$
@Torsten The double centralizer theorem guarantees that any $L/K$ such that $[L:K]=n$ is maximal in $M_n(K)$. Granted, calling upon that result may be overkill :-).
$endgroup$
– Jyrki Lahtonen
Jan 30 at 16:19
2
2
$begingroup$
@Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and...
$endgroup$
– Jyrki Lahtonen
Jan 30 at 19:43
$begingroup$
@Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and...
$endgroup$
– Jyrki Lahtonen
Jan 30 at 19:43
|
show 1 more comment
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$begingroup$
In anothe question you asked for a 9-dimensional division algebra, central over $Bbb{Q}$ containing a copy of the non-Galois extension field $Bbb{Q}(root3of2)$. Is that not an answer to this question as well :-)
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:11
2
$begingroup$
Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 18:33
2
$begingroup$
Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $Lvert K$ with degree $[L:K] le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now.
$endgroup$
– Torsten Schoeneberg
Jan 23 at 19:00
$begingroup$
@Torsten The double centralizer theorem guarantees that any $L/K$ such that $[L:K]=n$ is maximal in $M_n(K)$. Granted, calling upon that result may be overkill :-).
$endgroup$
– Jyrki Lahtonen
Jan 30 at 16:19
2
$begingroup$
@Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and...
$endgroup$
– Jyrki Lahtonen
Jan 30 at 19:43