How to prove that $mathbb{Z}[(1+sqrt{2})/2]$ is a Euclidean domain?
$begingroup$
I know how to prove Euclidean domain for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{-2}]$, etc, but I am getting stuck on this notation. From what I've read,
We let $R_d$ be the ring $mathbb{Z}[(1+sqrt{2})/2]$ defined as $R_d={x + yw : x ,y ∈ mathbb{Z}}$, where $w = sqrt{d}$ if $d$ is not equivalent to $1$ mod $4$, and $(1+sqrt{d})/2$ if $d$ is equivalent to $1$ mod $4$. So clearly, in this problem, we are assuming that $d= 2$ is equiv. to $1$ mod $4$. But, I really don't know what to do with this information. Does the norm change? What is the norm? Is it the same as it would be for $sqrt{2}$? I just need a little more information to get going, and I would really appreciate any help you're willing to provide. Thank you.
abstract-algebra ring-theory euclidean-domain
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
$endgroup$
add a comment |
$begingroup$
I know how to prove Euclidean domain for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{-2}]$, etc, but I am getting stuck on this notation. From what I've read,
We let $R_d$ be the ring $mathbb{Z}[(1+sqrt{2})/2]$ defined as $R_d={x + yw : x ,y ∈ mathbb{Z}}$, where $w = sqrt{d}$ if $d$ is not equivalent to $1$ mod $4$, and $(1+sqrt{d})/2$ if $d$ is equivalent to $1$ mod $4$. So clearly, in this problem, we are assuming that $d= 2$ is equiv. to $1$ mod $4$. But, I really don't know what to do with this information. Does the norm change? What is the norm? Is it the same as it would be for $sqrt{2}$? I just need a little more information to get going, and I would really appreciate any help you're willing to provide. Thank you.
abstract-algebra ring-theory euclidean-domain
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
$endgroup$
1
$begingroup$
"So clearly, in this problem, we are assuming that $d=2$ is equiv. to $1$ mod $4$." But $2$ is not equivalent to $1$ mod $4$, so you are in the first case and should be considering $mathbb{Z}[sqrt{2}]$.
$endgroup$
– André 3000
Nov 30 '18 at 6:27
add a comment |
$begingroup$
I know how to prove Euclidean domain for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{-2}]$, etc, but I am getting stuck on this notation. From what I've read,
We let $R_d$ be the ring $mathbb{Z}[(1+sqrt{2})/2]$ defined as $R_d={x + yw : x ,y ∈ mathbb{Z}}$, where $w = sqrt{d}$ if $d$ is not equivalent to $1$ mod $4$, and $(1+sqrt{d})/2$ if $d$ is equivalent to $1$ mod $4$. So clearly, in this problem, we are assuming that $d= 2$ is equiv. to $1$ mod $4$. But, I really don't know what to do with this information. Does the norm change? What is the norm? Is it the same as it would be for $sqrt{2}$? I just need a little more information to get going, and I would really appreciate any help you're willing to provide. Thank you.
abstract-algebra ring-theory euclidean-domain
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
$endgroup$
I know how to prove Euclidean domain for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{-2}]$, etc, but I am getting stuck on this notation. From what I've read,
We let $R_d$ be the ring $mathbb{Z}[(1+sqrt{2})/2]$ defined as $R_d={x + yw : x ,y ∈ mathbb{Z}}$, where $w = sqrt{d}$ if $d$ is not equivalent to $1$ mod $4$, and $(1+sqrt{d})/2$ if $d$ is equivalent to $1$ mod $4$. So clearly, in this problem, we are assuming that $d= 2$ is equiv. to $1$ mod $4$. But, I really don't know what to do with this information. Does the norm change? What is the norm? Is it the same as it would be for $sqrt{2}$? I just need a little more information to get going, and I would really appreciate any help you're willing to provide. Thank you.
abstract-algebra ring-theory euclidean-domain
abstract-algebra ring-theory euclidean-domain
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
edited Nov 30 '18 at 6:27
André 3000
12.6k22243
12.6k22243
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
asked Nov 30 '18 at 4:42
Zoë SZoë S
61
61
1
$begingroup$
"So clearly, in this problem, we are assuming that $d=2$ is equiv. to $1$ mod $4$." But $2$ is not equivalent to $1$ mod $4$, so you are in the first case and should be considering $mathbb{Z}[sqrt{2}]$.
$endgroup$
– André 3000
Nov 30 '18 at 6:27
add a comment |
1
$begingroup$
"So clearly, in this problem, we are assuming that $d=2$ is equiv. to $1$ mod $4$." But $2$ is not equivalent to $1$ mod $4$, so you are in the first case and should be considering $mathbb{Z}[sqrt{2}]$.
$endgroup$
– André 3000
Nov 30 '18 at 6:27
1
1
$begingroup$
"So clearly, in this problem, we are assuming that $d=2$ is equiv. to $1$ mod $4$." But $2$ is not equivalent to $1$ mod $4$, so you are in the first case and should be considering $mathbb{Z}[sqrt{2}]$.
$endgroup$
– André 3000
Nov 30 '18 at 6:27
$begingroup$
"So clearly, in this problem, we are assuming that $d=2$ is equiv. to $1$ mod $4$." But $2$ is not equivalent to $1$ mod $4$, so you are in the first case and should be considering $mathbb{Z}[sqrt{2}]$.
$endgroup$
– André 3000
Nov 30 '18 at 6:27
add a comment |
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$begingroup$
"So clearly, in this problem, we are assuming that $d=2$ is equiv. to $1$ mod $4$." But $2$ is not equivalent to $1$ mod $4$, so you are in the first case and should be considering $mathbb{Z}[sqrt{2}]$.
$endgroup$
– André 3000
Nov 30 '18 at 6:27