Invertible elements of CSA inducing Galois automorphism are linearly independent
$begingroup$
Let $A$ be a central simple algebra over a field $F$.
Let $K$ be a maximal subfield of $A$ with $[K:F]=n$ and assume $K$ is Galois extension of $F$.
Let $sigma_1,sigma_2,cdots,sigma_n$ be all the Galois automorphisms of $K$ over $F$. Then By Skolen-noether theorem, there exists invertible elements $x_i$ in $A$ such that $sigma_i$ (on $K$) are simply conjugation (on $K$) by $x_i$.
Claim: $x_1k_1+x_2k_2 + cdots + x_nk_n=0$ for $k_iin K$ implies $k_i=0$ for all $i$.
I proved this for a simple case, as follows.
If $x_1k_1+ x_2k_2=0$ for $k_1,k_2neq 0$, then we see that $x_1$ and $x_2$ differ by an element of $K^*$, i.e. $x_1=x_2(-k_2k_1^{-1})$. Then on $K$, conjugation by $x_1$ and conjugation by $x_2$ coincide, i.e. $sigma_1=sigma_2$, contradiction.
I could not proceed in the case $x_1k_1+x_2k_2+x_3k_3=0$ to get a contradiction. How to proceed for this case, and in particular for claim? Any hint is sufficient.
I was thinking to use $F$-independence of $sigma_i$'s (Dedekind's theorem), but I didn't get direction.
division-algebras
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
$endgroup$
add a comment |
$begingroup$
Let $A$ be a central simple algebra over a field $F$.
Let $K$ be a maximal subfield of $A$ with $[K:F]=n$ and assume $K$ is Galois extension of $F$.
Let $sigma_1,sigma_2,cdots,sigma_n$ be all the Galois automorphisms of $K$ over $F$. Then By Skolen-noether theorem, there exists invertible elements $x_i$ in $A$ such that $sigma_i$ (on $K$) are simply conjugation (on $K$) by $x_i$.
Claim: $x_1k_1+x_2k_2 + cdots + x_nk_n=0$ for $k_iin K$ implies $k_i=0$ for all $i$.
I proved this for a simple case, as follows.
If $x_1k_1+ x_2k_2=0$ for $k_1,k_2neq 0$, then we see that $x_1$ and $x_2$ differ by an element of $K^*$, i.e. $x_1=x_2(-k_2k_1^{-1})$. Then on $K$, conjugation by $x_1$ and conjugation by $x_2$ coincide, i.e. $sigma_1=sigma_2$, contradiction.
I could not proceed in the case $x_1k_1+x_2k_2+x_3k_3=0$ to get a contradiction. How to proceed for this case, and in particular for claim? Any hint is sufficient.
I was thinking to use $F$-independence of $sigma_i$'s (Dedekind's theorem), but I didn't get direction.
division-algebras
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
$endgroup$
1
$begingroup$
I think you can do this with ideas similar to the proof of $F$-independence of the Galois group. If you multiply the given linear dependence relation from the left by an element $kin K$, and you move the $k$s to other side of $x_i$s you get a modified relation like $$x_1sigma_1(k)k_1+x_2sigma_2(k)k_2+cdots+x_nsigma_n(k)k_n=0.$$ Then work the usual way. Assume that the linear dependency relation involves a minimal number of $x_i$, pick a useful $k$ for the above, eliminate one of the variables etc.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:09
add a comment |
$begingroup$
Let $A$ be a central simple algebra over a field $F$.
Let $K$ be a maximal subfield of $A$ with $[K:F]=n$ and assume $K$ is Galois extension of $F$.
Let $sigma_1,sigma_2,cdots,sigma_n$ be all the Galois automorphisms of $K$ over $F$. Then By Skolen-noether theorem, there exists invertible elements $x_i$ in $A$ such that $sigma_i$ (on $K$) are simply conjugation (on $K$) by $x_i$.
Claim: $x_1k_1+x_2k_2 + cdots + x_nk_n=0$ for $k_iin K$ implies $k_i=0$ for all $i$.
I proved this for a simple case, as follows.
If $x_1k_1+ x_2k_2=0$ for $k_1,k_2neq 0$, then we see that $x_1$ and $x_2$ differ by an element of $K^*$, i.e. $x_1=x_2(-k_2k_1^{-1})$. Then on $K$, conjugation by $x_1$ and conjugation by $x_2$ coincide, i.e. $sigma_1=sigma_2$, contradiction.
I could not proceed in the case $x_1k_1+x_2k_2+x_3k_3=0$ to get a contradiction. How to proceed for this case, and in particular for claim? Any hint is sufficient.
I was thinking to use $F$-independence of $sigma_i$'s (Dedekind's theorem), but I didn't get direction.
division-algebras
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
$endgroup$
Let $A$ be a central simple algebra over a field $F$.
Let $K$ be a maximal subfield of $A$ with $[K:F]=n$ and assume $K$ is Galois extension of $F$.
Let $sigma_1,sigma_2,cdots,sigma_n$ be all the Galois automorphisms of $K$ over $F$. Then By Skolen-noether theorem, there exists invertible elements $x_i$ in $A$ such that $sigma_i$ (on $K$) are simply conjugation (on $K$) by $x_i$.
Claim: $x_1k_1+x_2k_2 + cdots + x_nk_n=0$ for $k_iin K$ implies $k_i=0$ for all $i$.
I proved this for a simple case, as follows.
If $x_1k_1+ x_2k_2=0$ for $k_1,k_2neq 0$, then we see that $x_1$ and $x_2$ differ by an element of $K^*$, i.e. $x_1=x_2(-k_2k_1^{-1})$. Then on $K$, conjugation by $x_1$ and conjugation by $x_2$ coincide, i.e. $sigma_1=sigma_2$, contradiction.
I could not proceed in the case $x_1k_1+x_2k_2+x_3k_3=0$ to get a contradiction. How to proceed for this case, and in particular for claim? Any hint is sufficient.
I was thinking to use $F$-independence of $sigma_i$'s (Dedekind's theorem), but I didn't get direction.
division-algebras
division-algebras
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
asked Jan 19 at 7:24
BeginnerBeginner
3,93611225
3,93611225
1
$begingroup$
I think you can do this with ideas similar to the proof of $F$-independence of the Galois group. If you multiply the given linear dependence relation from the left by an element $kin K$, and you move the $k$s to other side of $x_i$s you get a modified relation like $$x_1sigma_1(k)k_1+x_2sigma_2(k)k_2+cdots+x_nsigma_n(k)k_n=0.$$ Then work the usual way. Assume that the linear dependency relation involves a minimal number of $x_i$, pick a useful $k$ for the above, eliminate one of the variables etc.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:09
add a comment |
1
$begingroup$
I think you can do this with ideas similar to the proof of $F$-independence of the Galois group. If you multiply the given linear dependence relation from the left by an element $kin K$, and you move the $k$s to other side of $x_i$s you get a modified relation like $$x_1sigma_1(k)k_1+x_2sigma_2(k)k_2+cdots+x_nsigma_n(k)k_n=0.$$ Then work the usual way. Assume that the linear dependency relation involves a minimal number of $x_i$, pick a useful $k$ for the above, eliminate one of the variables etc.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:09
1
1
$begingroup$
I think you can do this with ideas similar to the proof of $F$-independence of the Galois group. If you multiply the given linear dependence relation from the left by an element $kin K$, and you move the $k$s to other side of $x_i$s you get a modified relation like $$x_1sigma_1(k)k_1+x_2sigma_2(k)k_2+cdots+x_nsigma_n(k)k_n=0.$$ Then work the usual way. Assume that the linear dependency relation involves a minimal number of $x_i$, pick a useful $k$ for the above, eliminate one of the variables etc.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:09
$begingroup$
I think you can do this with ideas similar to the proof of $F$-independence of the Galois group. If you multiply the given linear dependence relation from the left by an element $kin K$, and you move the $k$s to other side of $x_i$s you get a modified relation like $$x_1sigma_1(k)k_1+x_2sigma_2(k)k_2+cdots+x_nsigma_n(k)k_n=0.$$ Then work the usual way. Assume that the linear dependency relation involves a minimal number of $x_i$, pick a useful $k$ for the above, eliminate one of the variables etc.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:09
add a comment |
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$begingroup$
I think you can do this with ideas similar to the proof of $F$-independence of the Galois group. If you multiply the given linear dependence relation from the left by an element $kin K$, and you move the $k$s to other side of $x_i$s you get a modified relation like $$x_1sigma_1(k)k_1+x_2sigma_2(k)k_2+cdots+x_nsigma_n(k)k_n=0.$$ Then work the usual way. Assume that the linear dependency relation involves a minimal number of $x_i$, pick a useful $k$ for the above, eliminate one of the variables etc.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:09