Is $mathbb Z_3+mathbb Z_3mathbf i+mathbb Z_3mathbf j+mathbb Z_3mathbf k$ a local ring?
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
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
asked Dec 25 '18 at 0:50
joaopa
33618
|
show 1 more comment
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
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
asked Dec 25 '18 at 0:50
joaopa
33618
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 $3$.
– rschwieb
Dec 25 '18 at 1:31
$mathbb Z_3=3$-adic numbers.
– joaopa
Dec 25 '18 at 1:41
1
By local non commutative ring you mean ...
– Charlie Frohman
Dec 25 '18 at 2:15
That every element is a unit or nilpotent?
– Charlie Frohman
Dec 25 '18 at 2:17
1
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
– rschwieb
Dec 25 '18 at 3:17
|
show 1 more comment
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
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
asked Dec 25 '18 at 0:50
joaopa
33618
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
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
asked Dec 25 '18 at 0:50
joaopa
33618
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
asked Dec 25 '18 at 0:50
joaopa
33618
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
edited Dec 26 '18 at 17:24
André 3000
12.4k22042
12.4k22042
asked Dec 25 '18 at 0:50
joaopa
33618
asked Dec 25 '18 at 0:50
joaopa
33618
asked Dec 25 '18 at 0:50
joaopa
33618
33618
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 $3$.
– rschwieb
Dec 25 '18 at 1:31
$mathbb Z_3=3$-adic numbers.
– joaopa
Dec 25 '18 at 1:41
1
By local non commutative ring you mean ...
– Charlie Frohman
Dec 25 '18 at 2:15
That every element is a unit or nilpotent?
– Charlie Frohman
Dec 25 '18 at 2:17
1
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
– rschwieb
Dec 25 '18 at 3:17
|
show 1 more comment
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 $3$.
– rschwieb
Dec 25 '18 at 1:31
$mathbb Z_3=3$-adic numbers.
– joaopa
Dec 25 '18 at 1:41
1
By local non commutative ring you mean ...
– Charlie Frohman
Dec 25 '18 at 2:15
That every element is a unit or nilpotent?
– Charlie Frohman
Dec 25 '18 at 2:17
1
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
– rschwieb
Dec 25 '18 at 3:17
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 $3$.
– rschwieb
Dec 25 '18 at 1:31
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 $3$.
– rschwieb
Dec 25 '18 at 1:31
$mathbb Z_3=3$-adic numbers.
– joaopa
Dec 25 '18 at 1:41
$mathbb Z_3=3$-adic numbers.
– joaopa
Dec 25 '18 at 1:41
1
1
By local non commutative ring you mean ...
– Charlie Frohman
Dec 25 '18 at 2:15
By local non commutative ring you mean ...
– Charlie Frohman
Dec 25 '18 at 2:15
That every element is a unit or nilpotent?
– Charlie Frohman
Dec 25 '18 at 2:17
That every element is a unit or nilpotent?
– Charlie Frohman
Dec 25 '18 at 2:17
1
1
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
– rschwieb
Dec 25 '18 at 3:17
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
– rschwieb
Dec 25 '18 at 3:17
|
show 1 more comment
1 Answer
1
active
oldest
votes
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.
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
1
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
2
@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.
– André 3000
Dec 25 '18 at 8:54
1
(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.)
– André 3000
Dec 25 '18 at 8:56
1
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
2
+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.
– rschwieb
Dec 25 '18 at 14:42
|
show 1 more comment
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1 Answer
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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.
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
1
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
2
@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.
– André 3000
Dec 25 '18 at 8:54
1
(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.)
– André 3000
Dec 25 '18 at 8:56
1
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
2
+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.
– rschwieb
Dec 25 '18 at 14:42
|
show 1 more comment
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.
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
1
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
2
@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.
– André 3000
Dec 25 '18 at 8:54
1
(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.)
– André 3000
Dec 25 '18 at 8:56
1
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
2
+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.
– rschwieb
Dec 25 '18 at 14:42
|
show 1 more comment
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.
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
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.
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
edited Dec 26 '18 at 17:25
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
answered Dec 25 '18 at 8:14
André 3000
12.4k22042
12.4k22042
1
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
2
@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.
– André 3000
Dec 25 '18 at 8:54
1
(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.)
– André 3000
Dec 25 '18 at 8:56
1
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
2
+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.
– rschwieb
Dec 25 '18 at 14:42
|
show 1 more comment
1
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
2
@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.
– André 3000
Dec 25 '18 at 8:54
1
(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.)
– André 3000
Dec 25 '18 at 8:56
1
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
2
+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.
– rschwieb
Dec 25 '18 at 14:42
1
1
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
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.
– Jyrki Lahtonen
Dec 25 '18 at 8:42
2
2
@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.
– André 3000
Dec 25 '18 at 8:54
@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.
– André 3000
Dec 25 '18 at 8:54
1
1
(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.)
– André 3000
Dec 25 '18 at 8:56
(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.)
– André 3000
Dec 25 '18 at 8:56
1
1
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
Yeah. Thanks for the clarification.
– Jyrki Lahtonen
Dec 25 '18 at 13:23
2
2
+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.
– rschwieb
Dec 25 '18 at 14:42
+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.
– rschwieb
Dec 25 '18 at 14:42
|
show 1 more comment
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 $3$.
– rschwieb
Dec 25 '18 at 1:31
$mathbb Z_3=3$-adic numbers.
– joaopa
Dec 25 '18 at 1:41
1
By local non commutative ring you mean ...
– Charlie Frohman
Dec 25 '18 at 2:15
That every element is a unit or nilpotent?
– Charlie Frohman
Dec 25 '18 at 2:17
1
@joaopa "every element a unit or nilpotent" is not equivalent to what you said (and what you said is the standard definition, imo)
– rschwieb
Dec 25 '18 at 3:17