Is $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k$ a local ring?












1












$begingroup$


I consider the quaternion division ring on $mathbb Q_3$: that is
$$mathbb H_{mathbb Q_3}={a+bmathbf i+cmathbf j+dmathbf k mid a,b,c,dinmathbb Q_3}$$
with $mathbf i^2=mathbf j^2=mathbf k^2=mathbf imathbf jmathbf k=-1$. Is the subring
$$mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k$$
a noncommutative local ring, i.e., does it have a unique maximal left ideal (or equivalently a unique maximal right ideal)?



Thanks in advance for any help.






abstract-algebra p-adic-number-theory noncommutative-algebra quaternions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Wait... what are $mathbb Q_3$ and $mathbb Z_3$ supposed to be? At first I thought $mathbb Z_3$ was supposed to be the integers mod $3$.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 1:31












  • $begingroup$
    $mathbb Z_3=3$-adic numbers.
    $endgroup$
    – joaopa
    Dec 25 '18 at 1:41








  • 1




    $begingroup$
    By local non commutative ring you mean ...
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:15










  • $begingroup$
    That every element is a unit or nilpotent?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:17






  • 1




    $begingroup$
    @joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
    $endgroup$
    – rschwieb
    Dec 25 '18 at 3:17
















1












$begingroup$


I consider the quaternion division ring on $mathbb Q_3$: that is
$$mathbb H_{mathbb Q_3}={a+bmathbf i+cmathbf j+dmathbf k mid a,b,c,dinmathbb Q_3}$$
with $mathbf i^2=mathbf j^2=mathbf k^2=mathbf imathbf jmathbf k=-1$. Is the subring
$$mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k$$
a noncommutative local ring, i.e., does it have a unique maximal left ideal (or equivalently a unique maximal right ideal)?



Thanks in advance for any help.






abstract-algebra p-adic-number-theory noncommutative-algebra quaternions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Wait... what are $mathbb Q_3$ and $mathbb Z_3$ supposed to be? At first I thought $mathbb Z_3$ was supposed to be the integers mod $3$.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 1:31












  • $begingroup$
    $mathbb Z_3=3$-adic numbers.
    $endgroup$
    – joaopa
    Dec 25 '18 at 1:41








  • 1




    $begingroup$
    By local non commutative ring you mean ...
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:15










  • $begingroup$
    That every element is a unit or nilpotent?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:17






  • 1




    $begingroup$
    @joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
    $endgroup$
    – rschwieb
    Dec 25 '18 at 3:17














1












1








1





$begingroup$


I consider the quaternion division ring on $mathbb Q_3$: that is
$$mathbb H_{mathbb Q_3}={a+bmathbf i+cmathbf j+dmathbf k mid a,b,c,dinmathbb Q_3}$$
with $mathbf i^2=mathbf j^2=mathbf k^2=mathbf imathbf jmathbf k=-1$. Is the subring
$$mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k$$
a noncommutative local ring, i.e., does it have a unique maximal left ideal (or equivalently a unique maximal right ideal)?



Thanks in advance for any help.






abstract-algebra p-adic-number-theory noncommutative-algebra quaternions







share|cite|improve this question











$endgroup$




I consider the quaternion division ring on $mathbb Q_3$: that is
$$mathbb H_{mathbb Q_3}={a+bmathbf i+cmathbf j+dmathbf k mid a,b,c,dinmathbb Q_3}$$
with $mathbf i^2=mathbf j^2=mathbf k^2=mathbf imathbf jmathbf k=-1$. Is the subring
$$mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k$$
a noncommutative local ring, i.e., does it have a unique maximal left ideal (or equivalently a unique maximal right ideal)?



Thanks in advance for any help.






abstract-algebra p-adic-number-theory noncommutative-algebra quaternions




abstract-algebra p-adic-number-theory noncommutative-algebra quaternions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 17:24









André 3000

12.6k22243




12.6k22243










asked Dec 25 '18 at 0:50









joaopajoaopa

35118




35118












  • $begingroup$
    Wait... what are $mathbb Q_3$ and $mathbb Z_3$ supposed to be? At first I thought $mathbb Z_3$ was supposed to be the integers mod $3$.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 1:31












  • $begingroup$
    $mathbb Z_3=3$-adic numbers.
    $endgroup$
    – joaopa
    Dec 25 '18 at 1:41








  • 1




    $begingroup$
    By local non commutative ring you mean ...
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:15










  • $begingroup$
    That every element is a unit or nilpotent?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:17






  • 1




    $begingroup$
    @joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
    $endgroup$
    – rschwieb
    Dec 25 '18 at 3:17


















  • $begingroup$
    Wait... what are $mathbb Q_3$ and $mathbb Z_3$ supposed to be? At first I thought $mathbb Z_3$ was supposed to be the integers mod $3$.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 1:31












  • $begingroup$
    $mathbb Z_3=3$-adic numbers.
    $endgroup$
    – joaopa
    Dec 25 '18 at 1:41








  • 1




    $begingroup$
    By local non commutative ring you mean ...
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:15










  • $begingroup$
    That every element is a unit or nilpotent?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:17






  • 1




    $begingroup$
    @joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
    $endgroup$
    – rschwieb
    Dec 25 '18 at 3:17
















$begingroup$
Wait... what are $mathbb Q_3$ and $mathbb Z_3$ supposed to be? At first I thought $mathbb Z_3$ was supposed to be the integers mod $3$.
$endgroup$
– rschwieb
Dec 25 '18 at 1:31






$begingroup$
Wait... what are $mathbb Q_3$ and $mathbb Z_3$ supposed to be? At first I thought $mathbb Z_3$ was supposed to be the integers mod $3$.
$endgroup$
– rschwieb
Dec 25 '18 at 1:31














$begingroup$
$mathbb Z_3=3$-adic numbers.
$endgroup$
– joaopa
Dec 25 '18 at 1:41






$begingroup$
$mathbb Z_3=3$-adic numbers.
$endgroup$
– joaopa
Dec 25 '18 at 1:41






1




1




$begingroup$
By local non commutative ring you mean ...
$endgroup$
– Charlie Frohman
Dec 25 '18 at 2:15




$begingroup$
By local non commutative ring you mean ...
$endgroup$
– Charlie Frohman
Dec 25 '18 at 2:15












$begingroup$
That every element is a unit or nilpotent?
$endgroup$
– Charlie Frohman
Dec 25 '18 at 2:17




$begingroup$
That every element is a unit or nilpotent?
$endgroup$
– Charlie Frohman
Dec 25 '18 at 2:17




1




1




$begingroup$
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
$endgroup$
– rschwieb
Dec 25 '18 at 3:17




$begingroup$
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
$endgroup$
– rschwieb
Dec 25 '18 at 3:17










1 Answer
1






active

oldest

votes


















4












$begingroup$

Since $-1$ is a unit in $mathbb{Q}_3$, then the Hilbert symbol is trivial: $(-1, -1)_{mathbb{Q}_3} = 1$. This means that $mathbb{H}_{mathbb{Q}_3}$ is split, i.e., $mathbb{H}_{mathbb{Q}_3} cong M_2(mathbb{Q}_3)$, the ring of $2 times 2$ matrices. (So in particular it is not a division ring.) Then $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k cong M_2(mathbb{Z}_3)$, which indeed has the unique maximal ideal $M_2(3 mathbb{Z}_3)$.



For reference, see $S12.4$ of this book.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 8:42






  • 2




    $begingroup$
    @JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:54








  • 1




    $begingroup$
    (I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:56






  • 1




    $begingroup$
    Yeah. Thanks for the clarification.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 13:23






  • 2




    $begingroup$
    +1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 14:42













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Since $-1$ is a unit in $mathbb{Q}_3$, then the Hilbert symbol is trivial: $(-1, -1)_{mathbb{Q}_3} = 1$. This means that $mathbb{H}_{mathbb{Q}_3}$ is split, i.e., $mathbb{H}_{mathbb{Q}_3} cong M_2(mathbb{Q}_3)$, the ring of $2 times 2$ matrices. (So in particular it is not a division ring.) Then $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k cong M_2(mathbb{Z}_3)$, which indeed has the unique maximal ideal $M_2(3 mathbb{Z}_3)$.



For reference, see $S12.4$ of this book.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 8:42






  • 2




    $begingroup$
    @JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:54








  • 1




    $begingroup$
    (I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:56






  • 1




    $begingroup$
    Yeah. Thanks for the clarification.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 13:23






  • 2




    $begingroup$
    +1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 14:42


















4












$begingroup$

Since $-1$ is a unit in $mathbb{Q}_3$, then the Hilbert symbol is trivial: $(-1, -1)_{mathbb{Q}_3} = 1$. This means that $mathbb{H}_{mathbb{Q}_3}$ is split, i.e., $mathbb{H}_{mathbb{Q}_3} cong M_2(mathbb{Q}_3)$, the ring of $2 times 2$ matrices. (So in particular it is not a division ring.) Then $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k cong M_2(mathbb{Z}_3)$, which indeed has the unique maximal ideal $M_2(3 mathbb{Z}_3)$.



For reference, see $S12.4$ of this book.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 8:42






  • 2




    $begingroup$
    @JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:54








  • 1




    $begingroup$
    (I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:56






  • 1




    $begingroup$
    Yeah. Thanks for the clarification.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 13:23






  • 2




    $begingroup$
    +1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 14:42
















4












4








4





$begingroup$

Since $-1$ is a unit in $mathbb{Q}_3$, then the Hilbert symbol is trivial: $(-1, -1)_{mathbb{Q}_3} = 1$. This means that $mathbb{H}_{mathbb{Q}_3}$ is split, i.e., $mathbb{H}_{mathbb{Q}_3} cong M_2(mathbb{Q}_3)$, the ring of $2 times 2$ matrices. (So in particular it is not a division ring.) Then $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k cong M_2(mathbb{Z}_3)$, which indeed has the unique maximal ideal $M_2(3 mathbb{Z}_3)$.



For reference, see $S12.4$ of this book.






share|cite|improve this answer











$endgroup$



Since $-1$ is a unit in $mathbb{Q}_3$, then the Hilbert symbol is trivial: $(-1, -1)_{mathbb{Q}_3} = 1$. This means that $mathbb{H}_{mathbb{Q}_3}$ is split, i.e., $mathbb{H}_{mathbb{Q}_3} cong M_2(mathbb{Q}_3)$, the ring of $2 times 2$ matrices. (So in particular it is not a division ring.) Then $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k cong M_2(mathbb{Z}_3)$, which indeed has the unique maximal ideal $M_2(3 mathbb{Z}_3)$.



For reference, see $S12.4$ of this book.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 17:25

























answered Dec 25 '18 at 8:14









André 3000André 3000

12.6k22243




12.6k22243








  • 1




    $begingroup$
    I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 8:42






  • 2




    $begingroup$
    @JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:54








  • 1




    $begingroup$
    (I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:56






  • 1




    $begingroup$
    Yeah. Thanks for the clarification.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 13:23






  • 2




    $begingroup$
    +1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 14:42
















  • 1




    $begingroup$
    I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 8:42






  • 2




    $begingroup$
    @JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:54








  • 1




    $begingroup$
    (I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
    $endgroup$
    – André 3000
    Dec 25 '18 at 8:56






  • 1




    $begingroup$
    Yeah. Thanks for the clarification.
    $endgroup$
    – Jyrki Lahtonen
    Dec 25 '18 at 13:23






  • 2




    $begingroup$
    +1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
    $endgroup$
    – rschwieb
    Dec 25 '18 at 14:42










1




1




$begingroup$
I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
$endgroup$
– Jyrki Lahtonen
Dec 25 '18 at 8:42




$begingroup$
I agree with the conclusion that $Bbb{H}_{Bbb{Q}}$ splits when extended to $Bbb{Q}_3$. But what is the reasoning again? $-1$ is a square in $Bbb{Q}_5$, but that extension also splits the rational quaternions. Also, $-1$ is not a square in $Bbb{Q}_2$, but that extension does not split $Bbb{H}$. The discriminant of $Bbb{H}_{Bbb{Q}}$ is a power of two (Hurwitz order is maximal) meaning that $Bbb{Q}_2$ and, of course, $Bbb{R}$ are the only completions where the division algebra is not split.
$endgroup$
– Jyrki Lahtonen
Dec 25 '18 at 8:42




2




2




$begingroup$
@JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
$endgroup$
– André 3000
Dec 25 '18 at 8:54






$begingroup$
@JyrkiLahtonen I guess you're right and the fact that $-1$ isn't a square doesn't really matter: $(a,b)_{mathbb{Q}_3} = 1$ for all $a, b in mathbb{Q}_3^times$. The various possibilities for the value of the Hilbert symbol are given on p. 170 of my reference, and I just focused on the entry relevant to this question. But the underlying reasoning is the same as in Keith Conrad's comment: the quadratic form $langle -1, -1, -1 rangle$ is isotropic, which allows us to find a zero divisor.
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– André 3000
Dec 25 '18 at 8:54






1




1




$begingroup$
(I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
$endgroup$
– André 3000
Dec 25 '18 at 8:56




$begingroup$
(I'm a little wary of $mathbb{Q}_2$ since things are strange when the residue field has characteristic $2$, but the table for the Hilbert symbol is also given in that case on the following page.)
$endgroup$
– André 3000
Dec 25 '18 at 8:56




1




1




$begingroup$
Yeah. Thanks for the clarification.
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– Jyrki Lahtonen
Dec 25 '18 at 13:23




$begingroup$
Yeah. Thanks for the clarification.
$endgroup$
– Jyrki Lahtonen
Dec 25 '18 at 13:23




2




2




$begingroup$
+1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
$endgroup$
– rschwieb
Dec 25 '18 at 14:42






$begingroup$
+1 very helpful answer, although in the end the ring is not local in the standard sense (by having a unique maximal right ideal). It’s easy to display numerous nontrivial idempotents in this matrix ring.
$endgroup$
– rschwieb
Dec 25 '18 at 14:42




















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