Mixing
Prove $mathbb{Z} [sqrt{2}]$ is a Euclidean ring.
$begingroup$
The definition of Euclidean ring:
An integral domain R is called Euclidean ring if $exists delta$ : $R${$0$} -> $mathbb{N} cup{0}$ satisfying:
(1) $delta (a) leqslant delta (ab)$ if a, b $in R${$0$};
(2) $forall a,b in R${$0$} => $exists q,r in R$ s.t. $a=bq+r$, where either $r=0$ or $delta (r) < delta (b)$.
How do we usually set the delta function? The norm?
Let $x=a_1+b_1sqrt{2}, y=a_2+b_2sqrt{2}$, and $delta(x)=|a_1^2-2b_1^2|, delta(y)=|a_2^2-2b_2^2|$, then, $delta(xy)=|(a_1+b_1sqrt{2})(a_1-b_1sqrt{2})(a_2+b_2sqrt{2})(a_2-b_2sqrt{2})|$, since $delta(a)geq 1$, for any $aneq 0$
Consider $yoverline{x}=xoverline{x}q_1+r_1$,
$delta(r_1)=0$ or $delta(r_1)=delta((y-xq_1)(overline{x}))=delta(y-xq_1)delta(overline{x})<delta(xoverline{x})=delta(x)delta(overline{x})$,
$delta(r_0)=delta(y-xq_1)<delta(x)$
I'm not sure what am I doing... Orz
abstract-algebra ring-theory
edited Apr 19 '15 at 15:45
user26857
39.3k124183
asked Apr 19 '15 at 12:29
RichardRichard
316211
$endgroup$
|
show 3 more comments
$begingroup$
The definition of Euclidean ring:
An integral domain R is called Euclidean ring if $exists delta$ : $R${$0$} -> $mathbb{N} cup{0}$ satisfying:
(1) $delta (a) leqslant delta (ab)$ if a, b $in R${$0$};
(2) $forall a,b in R${$0$} => $exists q,r in R$ s.t. $a=bq+r$, where either $r=0$ or $delta (r) < delta (b)$.
How do we usually set the delta function? The norm?
Let $x=a_1+b_1sqrt{2}, y=a_2+b_2sqrt{2}$, and $delta(x)=|a_1^2-2b_1^2|, delta(y)=|a_2^2-2b_2^2|$, then, $delta(xy)=|(a_1+b_1sqrt{2})(a_1-b_1sqrt{2})(a_2+b_2sqrt{2})(a_2-b_2sqrt{2})|$, since $delta(a)geq 1$, for any $aneq 0$
Consider $yoverline{x}=xoverline{x}q_1+r_1$,
$delta(r_1)=0$ or $delta(r_1)=delta((y-xq_1)(overline{x}))=delta(y-xq_1)delta(overline{x})<delta(xoverline{x})=delta(x)delta(overline{x})$,
$delta(r_0)=delta(y-xq_1)<delta(x)$
I'm not sure what am I doing... Orz
abstract-algebra ring-theory
edited Apr 19 '15 at 15:45
user26857
39.3k124183
asked Apr 19 '15 at 12:29
RichardRichard
316211
$endgroup$
4
$begingroup$
The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet?
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 12:31
$begingroup$
Do you know a proof that any other quadratic ring $mathbf Z[sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)?
$endgroup$
– KCd
Apr 19 '15 at 12:39
$begingroup$
@DanielFischer, I tried and it can work. But, more than that, how about $mathbb{Z} [i]$ or something more typically? I want to know some $delta$ function instead of the norm.
$endgroup$
– Richard
Apr 19 '15 at 13:06
$begingroup$
@KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :)
$endgroup$
– Richard
Apr 19 '15 at 13:08
2
$begingroup$
For $mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $mathbb{Z}[sqrt{d}]$ are Euclidean (they aren't, e.g. $mathbb{Z}[sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example.
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 13:19
|
show 3 more comments
$begingroup$
The definition of Euclidean ring:
An integral domain R is called Euclidean ring if $exists delta$ : $R${$0$} -> $mathbb{N} cup{0}$ satisfying:
(1) $delta (a) leqslant delta (ab)$ if a, b $in R${$0$};
(2) $forall a,b in R${$0$} => $exists q,r in R$ s.t. $a=bq+r$, where either $r=0$ or $delta (r) < delta (b)$.
How do we usually set the delta function? The norm?
Let $x=a_1+b_1sqrt{2}, y=a_2+b_2sqrt{2}$, and $delta(x)=|a_1^2-2b_1^2|, delta(y)=|a_2^2-2b_2^2|$, then, $delta(xy)=|(a_1+b_1sqrt{2})(a_1-b_1sqrt{2})(a_2+b_2sqrt{2})(a_2-b_2sqrt{2})|$, since $delta(a)geq 1$, for any $aneq 0$
Consider $yoverline{x}=xoverline{x}q_1+r_1$,
$delta(r_1)=0$ or $delta(r_1)=delta((y-xq_1)(overline{x}))=delta(y-xq_1)delta(overline{x})<delta(xoverline{x})=delta(x)delta(overline{x})$,
$delta(r_0)=delta(y-xq_1)<delta(x)$
I'm not sure what am I doing... Orz
abstract-algebra ring-theory
edited Apr 19 '15 at 15:45
user26857
39.3k124183
asked Apr 19 '15 at 12:29
RichardRichard
316211
$endgroup$
The definition of Euclidean ring:
An integral domain R is called Euclidean ring if $exists delta$ : $R${$0$} -> $mathbb{N} cup{0}$ satisfying:
(1) $delta (a) leqslant delta (ab)$ if a, b $in R${$0$};
(2) $forall a,b in R${$0$} => $exists q,r in R$ s.t. $a=bq+r$, where either $r=0$ or $delta (r) < delta (b)$.
How do we usually set the delta function? The norm?
Let $x=a_1+b_1sqrt{2}, y=a_2+b_2sqrt{2}$, and $delta(x)=|a_1^2-2b_1^2|, delta(y)=|a_2^2-2b_2^2|$, then, $delta(xy)=|(a_1+b_1sqrt{2})(a_1-b_1sqrt{2})(a_2+b_2sqrt{2})(a_2-b_2sqrt{2})|$, since $delta(a)geq 1$, for any $aneq 0$
Consider $yoverline{x}=xoverline{x}q_1+r_1$,
$delta(r_1)=0$ or $delta(r_1)=delta((y-xq_1)(overline{x}))=delta(y-xq_1)delta(overline{x})<delta(xoverline{x})=delta(x)delta(overline{x})$,
$delta(r_0)=delta(y-xq_1)<delta(x)$
I'm not sure what am I doing... Orz
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Apr 19 '15 at 15:45
user26857
39.3k124183
asked Apr 19 '15 at 12:29
RichardRichard
316211
edited Apr 19 '15 at 15:45
user26857
39.3k124183
asked Apr 19 '15 at 12:29
RichardRichard
316211
edited Apr 19 '15 at 15:45
user26857
39.3k124183
edited Apr 19 '15 at 15:45
user26857
39.3k124183
edited Apr 19 '15 at 15:45
user26857
39.3k124183
39.3k124183
asked Apr 19 '15 at 12:29
RichardRichard
316211
asked Apr 19 '15 at 12:29
RichardRichard
316211
asked Apr 19 '15 at 12:29
RichardRichard
316211
316211
4
$begingroup$
The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet?
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 12:31
$begingroup$
Do you know a proof that any other quadratic ring $mathbf Z[sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)?
$endgroup$
– KCd
Apr 19 '15 at 12:39
$begingroup$
@DanielFischer, I tried and it can work. But, more than that, how about $mathbb{Z} [i]$ or something more typically? I want to know some $delta$ function instead of the norm.
$endgroup$
– Richard
Apr 19 '15 at 13:06
$begingroup$
@KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :)
$endgroup$
– Richard
Apr 19 '15 at 13:08
2
$begingroup$
For $mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $mathbb{Z}[sqrt{d}]$ are Euclidean (they aren't, e.g. $mathbb{Z}[sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example.
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 13:19
|
show 3 more comments
4
$begingroup$
The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet?
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 12:31
$begingroup$
Do you know a proof that any other quadratic ring $mathbf Z[sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)?
$endgroup$
– KCd
Apr 19 '15 at 12:39
$begingroup$
@DanielFischer, I tried and it can work. But, more than that, how about $mathbb{Z} [i]$ or something more typically? I want to know some $delta$ function instead of the norm.
$endgroup$
– Richard
Apr 19 '15 at 13:06
$begingroup$
@KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :)
$endgroup$
– Richard
Apr 19 '15 at 13:08
2
$begingroup$
For $mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $mathbb{Z}[sqrt{d}]$ are Euclidean (they aren't, e.g. $mathbb{Z}[sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example.
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 13:19
4
4
$begingroup$
The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet?
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 12:31
$begingroup$
The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet?
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 12:31
$begingroup$
Do you know a proof that any other quadratic ring $mathbf Z[sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)?
$endgroup$
– KCd
Apr 19 '15 at 12:39
$begingroup$
Do you know a proof that any other quadratic ring $mathbf Z[sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)?
$endgroup$
– KCd
Apr 19 '15 at 12:39
$begingroup$
@DanielFischer, I tried and it can work. But, more than that, how about $mathbb{Z} [i]$ or something more typically? I want to know some $delta$ function instead of the norm.
$endgroup$
– Richard
Apr 19 '15 at 13:06
$begingroup$
@DanielFischer, I tried and it can work. But, more than that, how about $mathbb{Z} [i]$ or something more typically? I want to know some $delta$ function instead of the norm.
$endgroup$
– Richard
Apr 19 '15 at 13:06
$begingroup$
@KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :)
$endgroup$
– Richard
Apr 19 '15 at 13:08
$begingroup$
@KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :)
$endgroup$
– Richard
Apr 19 '15 at 13:08
2
2
$begingroup$
For $mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $mathbb{Z}[sqrt{d}]$ are Euclidean (they aren't, e.g. $mathbb{Z}[sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example.
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 13:19
$begingroup$
For $mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $mathbb{Z}[sqrt{d}]$ are Euclidean (they aren't, e.g. $mathbb{Z}[sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example.
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 13:19
|
show 3 more comments
1 Answer
1
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oldest
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$begingroup$
In order to mark the question answered I note that your proof is finished by Kowser in this answer.
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
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$begingroup$
In order to mark the question answered I note that your proof is finished by Kowser in this answer.
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
$begingroup$
In order to mark the question answered I note that your proof is finished by Kowser in this answer.
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
$begingroup$
In order to mark the question answered I note that your proof is finished by Kowser in this answer.
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
In order to mark the question answered I note that your proof is finished by Kowser in this answer.
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
answered Jan 5 at 9:02
Alex RavskyAlex Ravsky
39.9k32282
39.9k32282
add a comment |
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$begingroup$
The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet?
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 12:31
$begingroup$
Do you know a proof that any other quadratic ring $mathbf Z[sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)?
$endgroup$
– KCd
Apr 19 '15 at 12:39
$begingroup$
@DanielFischer, I tried and it can work. But, more than that, how about $mathbb{Z} [i]$ or something more typically? I want to know some $delta$ function instead of the norm.
$endgroup$
– Richard
Apr 19 '15 at 13:06
$begingroup$
@KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :)
$endgroup$
– Richard
Apr 19 '15 at 13:08
2
$begingroup$
For $mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $mathbb{Z}[sqrt{d}]$ are Euclidean (they aren't, e.g. $mathbb{Z}[sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example.
$endgroup$
– Daniel Fischer♦
Apr 19 '15 at 13:19