Find all units in the ring $R$ as defined.
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2
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$R = mathbb{Z}left[sqrt{-7}right] le mathbb{C}$
In other words, $R$ is the set of all integers and integers multiplied by the square root of $-7$. I believe this might be called $mathbb{Z}$ adjoin root $-7$ but I am not completely sure.
I am asked to find all units in $R$. In $mathbb{Z}$, the only units are $1$ and $-1$ which are obviously units of $R$ as well. I know $mathbb{Z}[i]$ has units $1, -1, i$, and $-i$ but the latter two are not in $R$.
Are there any non-trivial units in this ring? All I can find is $1$ and $-1$, but that seems too simple.
ring-theory
edited Nov 14 at 10:57
mechanodroid
24.9k62245
asked Apr 2 '15 at 1:20
ocyeung
825
add a comment |
up vote
2
down vote
favorite
$R = mathbb{Z}left[sqrt{-7}right] le mathbb{C}$
In other words, $R$ is the set of all integers and integers multiplied by the square root of $-7$. I believe this might be called $mathbb{Z}$ adjoin root $-7$ but I am not completely sure.
I am asked to find all units in $R$. In $mathbb{Z}$, the only units are $1$ and $-1$ which are obviously units of $R$ as well. I know $mathbb{Z}[i]$ has units $1, -1, i$, and $-i$ but the latter two are not in $R$.
Are there any non-trivial units in this ring? All I can find is $1$ and $-1$, but that seems too simple.
ring-theory
edited Nov 14 at 10:57
mechanodroid
24.9k62245
asked Apr 2 '15 at 1:20
ocyeung
825
1
The set $ mathbb Z[sqrt{-7}]$ is the set of all complex numbers of the form $a+bsqrt{-7}$, with $a,bin mathbb Z$.
– lhf
Apr 2 '15 at 1:24
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$R = mathbb{Z}left[sqrt{-7}right] le mathbb{C}$
In other words, $R$ is the set of all integers and integers multiplied by the square root of $-7$. I believe this might be called $mathbb{Z}$ adjoin root $-7$ but I am not completely sure.
I am asked to find all units in $R$. In $mathbb{Z}$, the only units are $1$ and $-1$ which are obviously units of $R$ as well. I know $mathbb{Z}[i]$ has units $1, -1, i$, and $-i$ but the latter two are not in $R$.
Are there any non-trivial units in this ring? All I can find is $1$ and $-1$, but that seems too simple.
ring-theory
edited Nov 14 at 10:57
mechanodroid
24.9k62245
asked Apr 2 '15 at 1:20
ocyeung
825
$R = mathbb{Z}left[sqrt{-7}right] le mathbb{C}$
In other words, $R$ is the set of all integers and integers multiplied by the square root of $-7$. I believe this might be called $mathbb{Z}$ adjoin root $-7$ but I am not completely sure.
I am asked to find all units in $R$. In $mathbb{Z}$, the only units are $1$ and $-1$ which are obviously units of $R$ as well. I know $mathbb{Z}[i]$ has units $1, -1, i$, and $-i$ but the latter two are not in $R$.
Are there any non-trivial units in this ring? All I can find is $1$ and $-1$, but that seems too simple.
ring-theory
ring-theory
edited Nov 14 at 10:57
mechanodroid
24.9k62245
asked Apr 2 '15 at 1:20
ocyeung
825
edited Nov 14 at 10:57
mechanodroid
24.9k62245
asked Apr 2 '15 at 1:20
ocyeung
825
edited Nov 14 at 10:57
mechanodroid
24.9k62245
edited Nov 14 at 10:57
mechanodroid
24.9k62245
edited Nov 14 at 10:57
mechanodroid
24.9k62245
24.9k62245
asked Apr 2 '15 at 1:20
ocyeung
825
asked Apr 2 '15 at 1:20
ocyeung
825
asked Apr 2 '15 at 1:20
ocyeung
825
825
1
The set $ mathbb Z[sqrt{-7}]$ is the set of all complex numbers of the form $a+bsqrt{-7}$, with $a,bin mathbb Z$.
– lhf
Apr 2 '15 at 1:24
add a comment |
1
The set $ mathbb Z[sqrt{-7}]$ is the set of all complex numbers of the form $a+bsqrt{-7}$, with $a,bin mathbb Z$.
– lhf
Apr 2 '15 at 1:24
1
1
The set $ mathbb Z[sqrt{-7}]$ is the set of all complex numbers of the form $a+bsqrt{-7}$, with $a,bin mathbb Z$.
– lhf
Apr 2 '15 at 1:24
The set $ mathbb Z[sqrt{-7}]$ is the set of all complex numbers of the form $a+bsqrt{-7}$, with $a,bin mathbb Z$.
– lhf
Apr 2 '15 at 1:24
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Hint:
Consider $N(a+bsqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+bsqrt{-7}$.
Conclude that $N$ is multiplicative.
Prove that $N(alpha)=pm1$ iff $alpha$ is a unit in $mathbb Z[sqrt{-7}]$.
Conclude that you need to solve $a^2+7b^2=pm 1$ with $a,bin mathbb Z$.
Solve it.
answered Apr 2 '15 at 1:31
lhf
161k9164383
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint:
Consider $N(a+bsqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+bsqrt{-7}$.
Conclude that $N$ is multiplicative.
Prove that $N(alpha)=pm1$ iff $alpha$ is a unit in $mathbb Z[sqrt{-7}]$.
Conclude that you need to solve $a^2+7b^2=pm 1$ with $a,bin mathbb Z$.
Solve it.
answered Apr 2 '15 at 1:31
lhf
161k9164383
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
add a comment |
up vote
3
down vote
accepted
Hint:
Consider $N(a+bsqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+bsqrt{-7}$.
Conclude that $N$ is multiplicative.
Prove that $N(alpha)=pm1$ iff $alpha$ is a unit in $mathbb Z[sqrt{-7}]$.
Conclude that you need to solve $a^2+7b^2=pm 1$ with $a,bin mathbb Z$.
Solve it.
answered Apr 2 '15 at 1:31
lhf
161k9164383
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint:
Consider $N(a+bsqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+bsqrt{-7}$.
Conclude that $N$ is multiplicative.
Prove that $N(alpha)=pm1$ iff $alpha$ is a unit in $mathbb Z[sqrt{-7}]$.
Conclude that you need to solve $a^2+7b^2=pm 1$ with $a,bin mathbb Z$.
Solve it.
answered Apr 2 '15 at 1:31
lhf
161k9164383
Hint:
Consider $N(a+bsqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+bsqrt{-7}$.
Conclude that $N$ is multiplicative.
Prove that $N(alpha)=pm1$ iff $alpha$ is a unit in $mathbb Z[sqrt{-7}]$.
Conclude that you need to solve $a^2+7b^2=pm 1$ with $a,bin mathbb Z$.
Solve it.
answered Apr 2 '15 at 1:31
lhf
161k9164383
answered Apr 2 '15 at 1:31
lhf
161k9164383
answered Apr 2 '15 at 1:31
lhf
161k9164383
answered Apr 2 '15 at 1:31
lhf
161k9164383
161k9164383
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
add a comment |
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units.
– ocyeung
Apr 2 '15 at 1:44
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$).
– lhf
Apr 2 '15 at 2:36
add a comment |
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The set $ mathbb Z[sqrt{-7}]$ is the set of all complex numbers of the form $a+bsqrt{-7}$, with $a,bin mathbb Z$.
– lhf
Apr 2 '15 at 1:24