Mixing
The group of units of ${a+bisqrt{2}:a,binmathbb{Z}}$ is ${1,-1}$.
The exercise is to show the group of units of the ring ${a+bisqrt{2}:a,binmathbb{Z}}$ is ${1,-1}$. I don't understand the beginning of the book's solution:
Suppose $a+bisqrt{2}$ has an inverse $x+yisqrt{2}$. Then $|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}=1$. That is, $(a^{2}+2b^{2})(x^{2}+2y^{2})=1$.
First, why the absolute value instead of just $(a+bisqrt{2})
(x+yisqrt{2})=1implies(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}=1?$
Second, how did they get $(a^{2}+2b^{2})(x^{2}+2y^{2})=1?$
When I did this:
$$begin{align}
|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}&=|(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}|\
&=|((a^{2}-2b^{2})+2abisqrt{2})((c^{2}-2d^{2})+2cdisqrt{2})|\
&=|(a^{2}-2b^{2})(x^{2}-2y^{2})-8abxy+text{other stuff with }i|.
end{align}$$
What about the $-8abxy$? Doesn't that have to be set $=1$, too?
Thanks in advance.
abstract-algebra ring-theory
edited Jan 1 at 2:18
Shaun
8,820113681
asked Jan 1 at 2:04
anonanon444anonanon444
1466
add a comment |
The exercise is to show the group of units of the ring ${a+bisqrt{2}:a,binmathbb{Z}}$ is ${1,-1}$. I don't understand the beginning of the book's solution:
Suppose $a+bisqrt{2}$ has an inverse $x+yisqrt{2}$. Then $|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}=1$. That is, $(a^{2}+2b^{2})(x^{2}+2y^{2})=1$.
First, why the absolute value instead of just $(a+bisqrt{2})
(x+yisqrt{2})=1implies(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}=1?$
Second, how did they get $(a^{2}+2b^{2})(x^{2}+2y^{2})=1?$
When I did this:
$$begin{align}
|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}&=|(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}|\
&=|((a^{2}-2b^{2})+2abisqrt{2})((c^{2}-2d^{2})+2cdisqrt{2})|\
&=|(a^{2}-2b^{2})(x^{2}-2y^{2})-8abxy+text{other stuff with }i|.
end{align}$$
What about the $-8abxy$? Doesn't that have to be set $=1$, too?
Thanks in advance.
abstract-algebra ring-theory
edited Jan 1 at 2:18
Shaun
8,820113681
asked Jan 1 at 2:04
anonanon444anonanon444
1466
Please specify which book you're using in an edit. The information is likely to be useful for others.
– Shaun
Jan 1 at 2:06
Unfortunately, I don't know the title of the book. My school has online access so I downloaded it last year and printed the book except the cover and preface, and I can't find the title of the book on the other pages. I just dug it out to read now since it's winter break, but I'm only at the first chapter, first section of it.
– anonanon444
Jan 2 at 1:44
add a comment |
The exercise is to show the group of units of the ring ${a+bisqrt{2}:a,binmathbb{Z}}$ is ${1,-1}$. I don't understand the beginning of the book's solution:
Suppose $a+bisqrt{2}$ has an inverse $x+yisqrt{2}$. Then $|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}=1$. That is, $(a^{2}+2b^{2})(x^{2}+2y^{2})=1$.
First, why the absolute value instead of just $(a+bisqrt{2})
(x+yisqrt{2})=1implies(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}=1?$
Second, how did they get $(a^{2}+2b^{2})(x^{2}+2y^{2})=1?$
When I did this:
$$begin{align}
|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}&=|(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}|\
&=|((a^{2}-2b^{2})+2abisqrt{2})((c^{2}-2d^{2})+2cdisqrt{2})|\
&=|(a^{2}-2b^{2})(x^{2}-2y^{2})-8abxy+text{other stuff with }i|.
end{align}$$
What about the $-8abxy$? Doesn't that have to be set $=1$, too?
Thanks in advance.
abstract-algebra ring-theory
edited Jan 1 at 2:18
Shaun
8,820113681
asked Jan 1 at 2:04
anonanon444anonanon444
1466
The exercise is to show the group of units of the ring ${a+bisqrt{2}:a,binmathbb{Z}}$ is ${1,-1}$. I don't understand the beginning of the book's solution:
Suppose $a+bisqrt{2}$ has an inverse $x+yisqrt{2}$. Then $|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}=1$. That is, $(a^{2}+2b^{2})(x^{2}+2y^{2})=1$.
First, why the absolute value instead of just $(a+bisqrt{2})
(x+yisqrt{2})=1implies(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}=1?$
Second, how did they get $(a^{2}+2b^{2})(x^{2}+2y^{2})=1?$
When I did this:
$$begin{align}
|a+bisqrt{2}|^{2}|x+yisqrt{2}|^{2}&=|(a+bisqrt{2})^{2}(x+yisqrt{2})^{2}|\
&=|((a^{2}-2b^{2})+2abisqrt{2})((c^{2}-2d^{2})+2cdisqrt{2})|\
&=|(a^{2}-2b^{2})(x^{2}-2y^{2})-8abxy+text{other stuff with }i|.
end{align}$$
What about the $-8abxy$? Doesn't that have to be set $=1$, too?
Thanks in advance.
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Jan 1 at 2:18
Shaun
8,820113681
asked Jan 1 at 2:04
anonanon444anonanon444
1466
edited Jan 1 at 2:18
Shaun
8,820113681
asked Jan 1 at 2:04
anonanon444anonanon444
1466
edited Jan 1 at 2:18
Shaun
8,820113681
edited Jan 1 at 2:18
Shaun
8,820113681
edited Jan 1 at 2:18
Shaun
8,820113681
8,820113681
asked Jan 1 at 2:04
anonanon444anonanon444
1466
asked Jan 1 at 2:04
anonanon444anonanon444
1466
asked Jan 1 at 2:04
anonanon444anonanon444
1466
1466
Please specify which book you're using in an edit. The information is likely to be useful for others.
– Shaun
Jan 1 at 2:06
Unfortunately, I don't know the title of the book. My school has online access so I downloaded it last year and printed the book except the cover and preface, and I can't find the title of the book on the other pages. I just dug it out to read now since it's winter break, but I'm only at the first chapter, first section of it.
– anonanon444
Jan 2 at 1:44
add a comment |
Please specify which book you're using in an edit. The information is likely to be useful for others.
– Shaun
Jan 1 at 2:06
Unfortunately, I don't know the title of the book. My school has online access so I downloaded it last year and printed the book except the cover and preface, and I can't find the title of the book on the other pages. I just dug it out to read now since it's winter break, but I'm only at the first chapter, first section of it.
– anonanon444
Jan 2 at 1:44
Please specify which book you're using in an edit. The information is likely to be useful for others.
– Shaun
Jan 1 at 2:06
Please specify which book you're using in an edit. The information is likely to be useful for others.
– Shaun
Jan 1 at 2:06
Unfortunately, I don't know the title of the book. My school has online access so I downloaded it last year and printed the book except the cover and preface, and I can't find the title of the book on the other pages. I just dug it out to read now since it's winter break, but I'm only at the first chapter, first section of it.
– anonanon444
Jan 2 at 1:44
Unfortunately, I don't know the title of the book. My school has online access so I downloaded it last year and printed the book except the cover and preface, and I can't find the title of the book on the other pages. I just dug it out to read now since it's winter break, but I'm only at the first chapter, first section of it.
– anonanon444
Jan 2 at 1:44
add a comment |
2 Answers
2
active
oldest
votes
There is a norm with values in $mathbb{N}$ on that ring. It is defined by $N(a+isqrt{2}b)=a^2+2b^2$. This norm is multiplicative, you can check that. So if $x$ and $y$ are inverses of each other $i.e.$ $xy=1$ then $1=N(1)=N(xy)=N(x)N(y) $ which implies $N(x)=N(y)=1$. From this you get that the only invertible elements are $pm 1$.
answered Jan 1 at 2:12
moutheticsmouthetics
50127
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
1
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
add a comment |
You are working with complex numbers here, and the absolute value of a complex number is defined by $|a+bi| = sqrt{a^2+b^2}$ ($a$ and $b$ being the real and imaginary parts), the length of the vector from (0,0) to $(a,b)$.
You seem to be saying that $|z|^2 = z^2$, which is incorrect; that only works with real numbers.
answered Jan 1 at 2:24
TedTed
21.4k13259
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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votes
There is a norm with values in $mathbb{N}$ on that ring. It is defined by $N(a+isqrt{2}b)=a^2+2b^2$. This norm is multiplicative, you can check that. So if $x$ and $y$ are inverses of each other $i.e.$ $xy=1$ then $1=N(1)=N(xy)=N(x)N(y) $ which implies $N(x)=N(y)=1$. From this you get that the only invertible elements are $pm 1$.
answered Jan 1 at 2:12
moutheticsmouthetics
50127
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
1
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
add a comment |
There is a norm with values in $mathbb{N}$ on that ring. It is defined by $N(a+isqrt{2}b)=a^2+2b^2$. This norm is multiplicative, you can check that. So if $x$ and $y$ are inverses of each other $i.e.$ $xy=1$ then $1=N(1)=N(xy)=N(x)N(y) $ which implies $N(x)=N(y)=1$. From this you get that the only invertible elements are $pm 1$.
answered Jan 1 at 2:12
moutheticsmouthetics
50127
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
1
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
add a comment |
There is a norm with values in $mathbb{N}$ on that ring. It is defined by $N(a+isqrt{2}b)=a^2+2b^2$. This norm is multiplicative, you can check that. So if $x$ and $y$ are inverses of each other $i.e.$ $xy=1$ then $1=N(1)=N(xy)=N(x)N(y) $ which implies $N(x)=N(y)=1$. From this you get that the only invertible elements are $pm 1$.
answered Jan 1 at 2:12
moutheticsmouthetics
50127
There is a norm with values in $mathbb{N}$ on that ring. It is defined by $N(a+isqrt{2}b)=a^2+2b^2$. This norm is multiplicative, you can check that. So if $x$ and $y$ are inverses of each other $i.e.$ $xy=1$ then $1=N(1)=N(xy)=N(x)N(y) $ which implies $N(x)=N(y)=1$. From this you get that the only invertible elements are $pm 1$.
answered Jan 1 at 2:12
moutheticsmouthetics
50127
edited Jan 1 at 2:20
answered Jan 1 at 2:12
moutheticsmouthetics
50127
answered Jan 1 at 2:12
moutheticsmouthetics
50127
answered Jan 1 at 2:12
moutheticsmouthetics
50127
50127
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
1
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
add a comment |
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
1
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
But $a,b$ should be integers, so only $pm 1$ are units.
– coffeemath
Jan 1 at 2:19
1
1
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Edited! I was thinking of the Gauss integers. Sorry.
– mouthetics
Jan 1 at 2:21
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
Oh, thanks! I thought $a+ibsqrt{2}mapsto a^{2}+2b^{2}$ might be a norm but I wasn't sure.
– anonanon444
Jan 2 at 1:38
add a comment |
You are working with complex numbers here, and the absolute value of a complex number is defined by $|a+bi| = sqrt{a^2+b^2}$ ($a$ and $b$ being the real and imaginary parts), the length of the vector from (0,0) to $(a,b)$.
You seem to be saying that $|z|^2 = z^2$, which is incorrect; that only works with real numbers.
answered Jan 1 at 2:24
TedTed
21.4k13259
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
add a comment |
You are working with complex numbers here, and the absolute value of a complex number is defined by $|a+bi| = sqrt{a^2+b^2}$ ($a$ and $b$ being the real and imaginary parts), the length of the vector from (0,0) to $(a,b)$.
You seem to be saying that $|z|^2 = z^2$, which is incorrect; that only works with real numbers.
answered Jan 1 at 2:24
TedTed
21.4k13259
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
add a comment |
You are working with complex numbers here, and the absolute value of a complex number is defined by $|a+bi| = sqrt{a^2+b^2}$ ($a$ and $b$ being the real and imaginary parts), the length of the vector from (0,0) to $(a,b)$.
You seem to be saying that $|z|^2 = z^2$, which is incorrect; that only works with real numbers.
answered Jan 1 at 2:24
TedTed
21.4k13259
You are working with complex numbers here, and the absolute value of a complex number is defined by $|a+bi| = sqrt{a^2+b^2}$ ($a$ and $b$ being the real and imaginary parts), the length of the vector from (0,0) to $(a,b)$.
You seem to be saying that $|z|^2 = z^2$, which is incorrect; that only works with real numbers.
answered Jan 1 at 2:24
TedTed
21.4k13259
answered Jan 1 at 2:24
TedTed
21.4k13259
answered Jan 1 at 2:24
TedTed
21.4k13259
answered Jan 1 at 2:24
TedTed
21.4k13259
21.4k13259
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
add a comment |
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
Thanks, I totally forgot about that.
– anonanon444
Jan 2 at 1:36
add a comment |
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Please specify which book you're using in an edit. The information is likely to be useful for others.
– Shaun
Jan 1 at 2:06
Unfortunately, I don't know the title of the book. My school has online access so I downloaded it last year and printed the book except the cover and preface, and I can't find the title of the book on the other pages. I just dug it out to read now since it's winter break, but I'm only at the first chapter, first section of it.
– anonanon444
Jan 2 at 1:44