Show that $mathscr{O}_{mathbb{Q}(sqrt{-7})}$ is a UFD
$begingroup$
It is known that the ring of integer is a Dedekind domain which means that it is a UFD iff it is a PID. Since $-7equiv1$ mod $4$, we have that $mathscr{O}_{mathbb{Q}(sqrt{-7})}=mathbb{Z}left[frac{1+sqrt{-7}}{2}right]$. Now I read something in the sense of: if $alpha:=frac{1+sqrt{-7}}{2}$ has an irreducible minimal polynomial mod $2$ and mod $3$, then we have a PID; I don't know anything about that. I think I have stated that wrong since the minimal polynomial is $f_{alpha}=x^2-x+2$ which is reducible mod $2$.
Dr. Math:
We pick an arbitrary complex number $x + iyinmathbb{Z}[alpha]$, and we must find a
suitable lattice point:
$$z = r + salpha = (r+s/2) + i(ssqrt{7})/2.$$
It is natural to try to have the real and imaginary parts of
$(x + yi - z)$ as small as possible.
Let's start with the imaginary part $ y - ssqrt{7}/2$. We take $s$ as
the closest integer to $2y/sqrt{7}$. This will give us the following:
begin{align*}
| 2y/sqrt{7} - s | &leqslant 1/2\
| y - ssqrt{7}/2 | &leqslant sqrt{7}/4.
end{align*}
Now, we turn to the real part $x - r - s/2$. If we select $r$ as the
integer closest to $(x - s/2)$, we will have:
begin{align*}
| x - r - s/2 | leqslant 1/2.
end{align*}
Putting both relations together, we get:
$$
N(x + yi - z) = (x - r - s/2)^2 + (y - ssqrt{7}/2)^2
leqslant 1/4 + 7/16
< 1$$
as desired. Hence, Euclidean domain, so PID, so UFD.
Is this proof correct and can it be applied in all cases of showing that $mathbb{Z}left[frac{1+sqrt{d}}{2}right]$, $squareneq dinmathbb{N}$ is a Euclidean domain?
number-theory ring-theory algebraic-number-theory unique-factorization-domains euclidean-domain
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
$endgroup$
add a comment |
$begingroup$
It is known that the ring of integer is a Dedekind domain which means that it is a UFD iff it is a PID. Since $-7equiv1$ mod $4$, we have that $mathscr{O}_{mathbb{Q}(sqrt{-7})}=mathbb{Z}left[frac{1+sqrt{-7}}{2}right]$. Now I read something in the sense of: if $alpha:=frac{1+sqrt{-7}}{2}$ has an irreducible minimal polynomial mod $2$ and mod $3$, then we have a PID; I don't know anything about that. I think I have stated that wrong since the minimal polynomial is $f_{alpha}=x^2-x+2$ which is reducible mod $2$.
Dr. Math:
We pick an arbitrary complex number $x + iyinmathbb{Z}[alpha]$, and we must find a
suitable lattice point:
$$z = r + salpha = (r+s/2) + i(ssqrt{7})/2.$$
It is natural to try to have the real and imaginary parts of
$(x + yi - z)$ as small as possible.
Let's start with the imaginary part $ y - ssqrt{7}/2$. We take $s$ as
the closest integer to $2y/sqrt{7}$. This will give us the following:
begin{align*}
| 2y/sqrt{7} - s | &leqslant 1/2\
| y - ssqrt{7}/2 | &leqslant sqrt{7}/4.
end{align*}
Now, we turn to the real part $x - r - s/2$. If we select $r$ as the
integer closest to $(x - s/2)$, we will have:
begin{align*}
| x - r - s/2 | leqslant 1/2.
end{align*}
Putting both relations together, we get:
$$
N(x + yi - z) = (x - r - s/2)^2 + (y - ssqrt{7}/2)^2
leqslant 1/4 + 7/16
< 1$$
as desired. Hence, Euclidean domain, so PID, so UFD.
Is this proof correct and can it be applied in all cases of showing that $mathbb{Z}left[frac{1+sqrt{d}}{2}right]$, $squareneq dinmathbb{N}$ is a Euclidean domain?
number-theory ring-theory algebraic-number-theory unique-factorization-domains euclidean-domain
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
$endgroup$
1
$begingroup$
It's a Euclidean domain.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 20:22
1
$begingroup$
@LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such?
$endgroup$
– Algebear
Dec 16 '18 at 20:26
$begingroup$
I didn't find the exact place yet but the conclusion that all $mathbb Z[ 1+sqrt d / 2]$ for $d equiv 1 pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large.
$endgroup$
– Alex J Best
Dec 16 '18 at 21:10
$begingroup$
@AlexJBest: You're right and an example for what you say is $ d = -19 $.
$endgroup$
– hellHound
Dec 17 '18 at 17:47
$begingroup$
The Minkowski bound for $Bbb Q(sqrt{-7})$ is ~1.68..., if you have this machinery available to you.
$endgroup$
– ÍgjøgnumMeg
Dec 22 '18 at 11:30
add a comment |
$begingroup$
It is known that the ring of integer is a Dedekind domain which means that it is a UFD iff it is a PID. Since $-7equiv1$ mod $4$, we have that $mathscr{O}_{mathbb{Q}(sqrt{-7})}=mathbb{Z}left[frac{1+sqrt{-7}}{2}right]$. Now I read something in the sense of: if $alpha:=frac{1+sqrt{-7}}{2}$ has an irreducible minimal polynomial mod $2$ and mod $3$, then we have a PID; I don't know anything about that. I think I have stated that wrong since the minimal polynomial is $f_{alpha}=x^2-x+2$ which is reducible mod $2$.
Dr. Math:
We pick an arbitrary complex number $x + iyinmathbb{Z}[alpha]$, and we must find a
suitable lattice point:
$$z = r + salpha = (r+s/2) + i(ssqrt{7})/2.$$
It is natural to try to have the real and imaginary parts of
$(x + yi - z)$ as small as possible.
Let's start with the imaginary part $ y - ssqrt{7}/2$. We take $s$ as
the closest integer to $2y/sqrt{7}$. This will give us the following:
begin{align*}
| 2y/sqrt{7} - s | &leqslant 1/2\
| y - ssqrt{7}/2 | &leqslant sqrt{7}/4.
end{align*}
Now, we turn to the real part $x - r - s/2$. If we select $r$ as the
integer closest to $(x - s/2)$, we will have:
begin{align*}
| x - r - s/2 | leqslant 1/2.
end{align*}
Putting both relations together, we get:
$$
N(x + yi - z) = (x - r - s/2)^2 + (y - ssqrt{7}/2)^2
leqslant 1/4 + 7/16
< 1$$
as desired. Hence, Euclidean domain, so PID, so UFD.
Is this proof correct and can it be applied in all cases of showing that $mathbb{Z}left[frac{1+sqrt{d}}{2}right]$, $squareneq dinmathbb{N}$ is a Euclidean domain?
number-theory ring-theory algebraic-number-theory unique-factorization-domains euclidean-domain
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
$endgroup$
It is known that the ring of integer is a Dedekind domain which means that it is a UFD iff it is a PID. Since $-7equiv1$ mod $4$, we have that $mathscr{O}_{mathbb{Q}(sqrt{-7})}=mathbb{Z}left[frac{1+sqrt{-7}}{2}right]$. Now I read something in the sense of: if $alpha:=frac{1+sqrt{-7}}{2}$ has an irreducible minimal polynomial mod $2$ and mod $3$, then we have a PID; I don't know anything about that. I think I have stated that wrong since the minimal polynomial is $f_{alpha}=x^2-x+2$ which is reducible mod $2$.
Dr. Math:
We pick an arbitrary complex number $x + iyinmathbb{Z}[alpha]$, and we must find a
suitable lattice point:
$$z = r + salpha = (r+s/2) + i(ssqrt{7})/2.$$
It is natural to try to have the real and imaginary parts of
$(x + yi - z)$ as small as possible.
Let's start with the imaginary part $ y - ssqrt{7}/2$. We take $s$ as
the closest integer to $2y/sqrt{7}$. This will give us the following:
begin{align*}
| 2y/sqrt{7} - s | &leqslant 1/2\
| y - ssqrt{7}/2 | &leqslant sqrt{7}/4.
end{align*}
Now, we turn to the real part $x - r - s/2$. If we select $r$ as the
integer closest to $(x - s/2)$, we will have:
begin{align*}
| x - r - s/2 | leqslant 1/2.
end{align*}
Putting both relations together, we get:
$$
N(x + yi - z) = (x - r - s/2)^2 + (y - ssqrt{7}/2)^2
leqslant 1/4 + 7/16
< 1$$
as desired. Hence, Euclidean domain, so PID, so UFD.
Is this proof correct and can it be applied in all cases of showing that $mathbb{Z}left[frac{1+sqrt{d}}{2}right]$, $squareneq dinmathbb{N}$ is a Euclidean domain?
number-theory ring-theory algebraic-number-theory unique-factorization-domains euclidean-domain
number-theory ring-theory algebraic-number-theory unique-factorization-domains euclidean-domain
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
edited Dec 17 '18 at 1:42
Batominovski
33.1k33293
33.1k33293
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
asked Dec 16 '18 at 20:21
AlgebearAlgebear
704419
704419
1
$begingroup$
It's a Euclidean domain.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 20:22
1
$begingroup$
@LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such?
$endgroup$
– Algebear
Dec 16 '18 at 20:26
$begingroup$
I didn't find the exact place yet but the conclusion that all $mathbb Z[ 1+sqrt d / 2]$ for $d equiv 1 pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large.
$endgroup$
– Alex J Best
Dec 16 '18 at 21:10
$begingroup$
@AlexJBest: You're right and an example for what you say is $ d = -19 $.
$endgroup$
– hellHound
Dec 17 '18 at 17:47
$begingroup$
The Minkowski bound for $Bbb Q(sqrt{-7})$ is ~1.68..., if you have this machinery available to you.
$endgroup$
– ÍgjøgnumMeg
Dec 22 '18 at 11:30
add a comment |
1
$begingroup$
It's a Euclidean domain.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 20:22
1
$begingroup$
@LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such?
$endgroup$
– Algebear
Dec 16 '18 at 20:26
$begingroup$
I didn't find the exact place yet but the conclusion that all $mathbb Z[ 1+sqrt d / 2]$ for $d equiv 1 pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large.
$endgroup$
– Alex J Best
Dec 16 '18 at 21:10
$begingroup$
@AlexJBest: You're right and an example for what you say is $ d = -19 $.
$endgroup$
– hellHound
Dec 17 '18 at 17:47
$begingroup$
The Minkowski bound for $Bbb Q(sqrt{-7})$ is ~1.68..., if you have this machinery available to you.
$endgroup$
– ÍgjøgnumMeg
Dec 22 '18 at 11:30
1
1
$begingroup$
It's a Euclidean domain.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 20:22
$begingroup$
It's a Euclidean domain.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 20:22
1
1
$begingroup$
@LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such?
$endgroup$
– Algebear
Dec 16 '18 at 20:26
$begingroup$
@LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such?
$endgroup$
– Algebear
Dec 16 '18 at 20:26
$begingroup$
I didn't find the exact place yet but the conclusion that all $mathbb Z[ 1+sqrt d / 2]$ for $d equiv 1 pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large.
$endgroup$
– Alex J Best
Dec 16 '18 at 21:10
$begingroup$
I didn't find the exact place yet but the conclusion that all $mathbb Z[ 1+sqrt d / 2]$ for $d equiv 1 pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large.
$endgroup$
– Alex J Best
Dec 16 '18 at 21:10
$begingroup$
@AlexJBest: You're right and an example for what you say is $ d = -19 $.
$endgroup$
– hellHound
Dec 17 '18 at 17:47
$begingroup$
@AlexJBest: You're right and an example for what you say is $ d = -19 $.
$endgroup$
– hellHound
Dec 17 '18 at 17:47
$begingroup$
The Minkowski bound for $Bbb Q(sqrt{-7})$ is ~1.68..., if you have this machinery available to you.
$endgroup$
– ÍgjøgnumMeg
Dec 22 '18 at 11:30
$begingroup$
The Minkowski bound for $Bbb Q(sqrt{-7})$ is ~1.68..., if you have this machinery available to you.
$endgroup$
– ÍgjøgnumMeg
Dec 22 '18 at 11:30
add a comment |
0
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1
$begingroup$
It's a Euclidean domain.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 20:22
1
$begingroup$
@LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such?
$endgroup$
– Algebear
Dec 16 '18 at 20:26
$begingroup$
I didn't find the exact place yet but the conclusion that all $mathbb Z[ 1+sqrt d / 2]$ for $d equiv 1 pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large.
$endgroup$
– Alex J Best
Dec 16 '18 at 21:10
$begingroup$
@AlexJBest: You're right and an example for what you say is $ d = -19 $.
$endgroup$
– hellHound
Dec 17 '18 at 17:47
$begingroup$
The Minkowski bound for $Bbb Q(sqrt{-7})$ is ~1.68..., if you have this machinery available to you.
$endgroup$
– ÍgjøgnumMeg
Dec 22 '18 at 11:30