Mixing
Units of Gaussian integers
$begingroup$
How can we show that $pm 1, pm i$ are the only units in the ring of Gaussian integers, $mathbb Z[i]$?
Thank you.
abstract-algebra ring-theory gaussian-integers
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
$endgroup$
add a comment |
$begingroup$
How can we show that $pm 1, pm i$ are the only units in the ring of Gaussian integers, $mathbb Z[i]$?
Thank you.
abstract-algebra ring-theory gaussian-integers
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
$endgroup$
7
$begingroup$
Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer.
$endgroup$
– André Nicolas
Feb 11 '12 at 7:53
add a comment |
$begingroup$
How can we show that $pm 1, pm i$ are the only units in the ring of Gaussian integers, $mathbb Z[i]$?
Thank you.
abstract-algebra ring-theory gaussian-integers
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
$endgroup$
How can we show that $pm 1, pm i$ are the only units in the ring of Gaussian integers, $mathbb Z[i]$?
Thank you.
abstract-algebra ring-theory gaussian-integers
abstract-algebra ring-theory gaussian-integers
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
edited Jan 21 at 16:03
J. W. Tanner
2,9171217
2,9171217
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
asked Feb 11 '12 at 7:50
gaussiangaussian
51113
51113
7
$begingroup$
Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer.
$endgroup$
– André Nicolas
Feb 11 '12 at 7:53
add a comment |
7
$begingroup$
Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer.
$endgroup$
– André Nicolas
Feb 11 '12 at 7:53
7
7
$begingroup$
Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer.
$endgroup$
– André Nicolas
Feb 11 '12 at 7:53
$begingroup$
Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer.
$endgroup$
– André Nicolas
Feb 11 '12 at 7:53
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Units are those elements in a ring that are invertible. Assume $a+bi$ is a unit.
Then $exists c+di in mathbb Z[i]$ such that $(a+bi)(c+di)=1$. This implies $$begin{cases}ac-bd=1\bc+ad=0 end{cases}$$
Now solve this system and remember that $a,b,c,d in mathbb Z$.
We get from the second equation that $bc=-ad$. This means that, from the first equation, $$text{If $b neq0$, then, } d(a^2+b^2)=-b.$$
This leaves you with two possibilities: $d=-1$ 0r $d=-b$.
But, if $d=-b$ then, we have that $c=a$ and $a^2+b^2=1$. Now since we have been given that $b neq 0$ the only solution for $a^2+b^2=1$ with $a,b in mathbb Z$ will be $a=0$ and $b=pm 1$.
Now, if $d=-1$, then note that if $c=0$, we immediately have that $b=1$.
More generally, the last equation we derived from the system gives you that, $b(1-b)=a^2$ which has no integral solutions if $b neq 0$.
For the last case, if $b=0$, then, the Gaussian integer is just $a$ and $a$ has an inverse in $mathbb Z[i]$ iff $a=pm 1$. This is because, the multiplicative inverse is $frac{1}{a}$, which will be in $mathbb Z$ iff $a=pm 1$.
edited Jun 15 '16 at 9:08
user26857
39.3k124183
$endgroup$
add a comment |
$begingroup$
If $z,winmathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or
$$(a^2+b^2)(c^2+d^2)=1, quad z=a+bi,; w=c+di.$$
Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2le1$. Check by hand the only solutions here correspond to $(a,b)=(pm1,0)$ or $(0,pm1)$.
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
$endgroup$
7
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
add a comment |
$begingroup$
Alternatively, $mathbb{Z}[i] = { a + bi : a,b in mathbb{Z}}$. So the norm of $mathbb{Z}[i]$ is $a^{2} + b^{2}$. Now, $a+bi$ is a unit if and only if $a+bi$ divides 1. Hence $N(a+bi)=pm 1. $
So, $a^{2} + b^{2} = pm 1$
Since $a,b in mathbb{Z}$, the only solutions are $a=pm 1$, $b=0$ and $a=0$, $b=pm1$. So the units are $1, -1, i, -i$
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Units are those elements in a ring that are invertible. Assume $a+bi$ is a unit.
Then $exists c+di in mathbb Z[i]$ such that $(a+bi)(c+di)=1$. This implies $$begin{cases}ac-bd=1\bc+ad=0 end{cases}$$
Now solve this system and remember that $a,b,c,d in mathbb Z$.
We get from the second equation that $bc=-ad$. This means that, from the first equation, $$text{If $b neq0$, then, } d(a^2+b^2)=-b.$$
This leaves you with two possibilities: $d=-1$ 0r $d=-b$.
But, if $d=-b$ then, we have that $c=a$ and $a^2+b^2=1$. Now since we have been given that $b neq 0$ the only solution for $a^2+b^2=1$ with $a,b in mathbb Z$ will be $a=0$ and $b=pm 1$.
Now, if $d=-1$, then note that if $c=0$, we immediately have that $b=1$.
More generally, the last equation we derived from the system gives you that, $b(1-b)=a^2$ which has no integral solutions if $b neq 0$.
For the last case, if $b=0$, then, the Gaussian integer is just $a$ and $a$ has an inverse in $mathbb Z[i]$ iff $a=pm 1$. This is because, the multiplicative inverse is $frac{1}{a}$, which will be in $mathbb Z$ iff $a=pm 1$.
edited Jun 15 '16 at 9:08
user26857
39.3k124183
$endgroup$
add a comment |
$begingroup$
Units are those elements in a ring that are invertible. Assume $a+bi$ is a unit.
Then $exists c+di in mathbb Z[i]$ such that $(a+bi)(c+di)=1$. This implies $$begin{cases}ac-bd=1\bc+ad=0 end{cases}$$
Now solve this system and remember that $a,b,c,d in mathbb Z$.
We get from the second equation that $bc=-ad$. This means that, from the first equation, $$text{If $b neq0$, then, } d(a^2+b^2)=-b.$$
This leaves you with two possibilities: $d=-1$ 0r $d=-b$.
But, if $d=-b$ then, we have that $c=a$ and $a^2+b^2=1$. Now since we have been given that $b neq 0$ the only solution for $a^2+b^2=1$ with $a,b in mathbb Z$ will be $a=0$ and $b=pm 1$.
Now, if $d=-1$, then note that if $c=0$, we immediately have that $b=1$.
More generally, the last equation we derived from the system gives you that, $b(1-b)=a^2$ which has no integral solutions if $b neq 0$.
For the last case, if $b=0$, then, the Gaussian integer is just $a$ and $a$ has an inverse in $mathbb Z[i]$ iff $a=pm 1$. This is because, the multiplicative inverse is $frac{1}{a}$, which will be in $mathbb Z$ iff $a=pm 1$.
edited Jun 15 '16 at 9:08
user26857
39.3k124183
$endgroup$
add a comment |
$begingroup$
Units are those elements in a ring that are invertible. Assume $a+bi$ is a unit.
Then $exists c+di in mathbb Z[i]$ such that $(a+bi)(c+di)=1$. This implies $$begin{cases}ac-bd=1\bc+ad=0 end{cases}$$
Now solve this system and remember that $a,b,c,d in mathbb Z$.
We get from the second equation that $bc=-ad$. This means that, from the first equation, $$text{If $b neq0$, then, } d(a^2+b^2)=-b.$$
This leaves you with two possibilities: $d=-1$ 0r $d=-b$.
But, if $d=-b$ then, we have that $c=a$ and $a^2+b^2=1$. Now since we have been given that $b neq 0$ the only solution for $a^2+b^2=1$ with $a,b in mathbb Z$ will be $a=0$ and $b=pm 1$.
Now, if $d=-1$, then note that if $c=0$, we immediately have that $b=1$.
More generally, the last equation we derived from the system gives you that, $b(1-b)=a^2$ which has no integral solutions if $b neq 0$.
For the last case, if $b=0$, then, the Gaussian integer is just $a$ and $a$ has an inverse in $mathbb Z[i]$ iff $a=pm 1$. This is because, the multiplicative inverse is $frac{1}{a}$, which will be in $mathbb Z$ iff $a=pm 1$.
edited Jun 15 '16 at 9:08
user26857
39.3k124183
$endgroup$
Units are those elements in a ring that are invertible. Assume $a+bi$ is a unit.
Then $exists c+di in mathbb Z[i]$ such that $(a+bi)(c+di)=1$. This implies $$begin{cases}ac-bd=1\bc+ad=0 end{cases}$$
Now solve this system and remember that $a,b,c,d in mathbb Z$.
We get from the second equation that $bc=-ad$. This means that, from the first equation, $$text{If $b neq0$, then, } d(a^2+b^2)=-b.$$
This leaves you with two possibilities: $d=-1$ 0r $d=-b$.
But, if $d=-b$ then, we have that $c=a$ and $a^2+b^2=1$. Now since we have been given that $b neq 0$ the only solution for $a^2+b^2=1$ with $a,b in mathbb Z$ will be $a=0$ and $b=pm 1$.
Now, if $d=-1$, then note that if $c=0$, we immediately have that $b=1$.
More generally, the last equation we derived from the system gives you that, $b(1-b)=a^2$ which has no integral solutions if $b neq 0$.
For the last case, if $b=0$, then, the Gaussian integer is just $a$ and $a$ has an inverse in $mathbb Z[i]$ iff $a=pm 1$. This is because, the multiplicative inverse is $frac{1}{a}$, which will be in $mathbb Z$ iff $a=pm 1$.
edited Jun 15 '16 at 9:08
user26857
39.3k124183
edited Jun 15 '16 at 9:08
user26857
39.3k124183
edited Jun 15 '16 at 9:08
user26857
39.3k124183
edited Jun 15 '16 at 9:08
user26857
39.3k124183
39.3k124183
answered Feb 11 '12 at 7:59
user21436
add a comment |
add a comment |
$begingroup$
If $z,winmathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or
$$(a^2+b^2)(c^2+d^2)=1, quad z=a+bi,; w=c+di.$$
Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2le1$. Check by hand the only solutions here correspond to $(a,b)=(pm1,0)$ or $(0,pm1)$.
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
$endgroup$
7
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
add a comment |
$begingroup$
If $z,winmathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or
$$(a^2+b^2)(c^2+d^2)=1, quad z=a+bi,; w=c+di.$$
Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2le1$. Check by hand the only solutions here correspond to $(a,b)=(pm1,0)$ or $(0,pm1)$.
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
$endgroup$
7
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
add a comment |
$begingroup$
If $z,winmathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or
$$(a^2+b^2)(c^2+d^2)=1, quad z=a+bi,; w=c+di.$$
Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2le1$. Check by hand the only solutions here correspond to $(a,b)=(pm1,0)$ or $(0,pm1)$.
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
$endgroup$
If $z,winmathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or
$$(a^2+b^2)(c^2+d^2)=1, quad z=a+bi,; w=c+di.$$
Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2le1$. Check by hand the only solutions here correspond to $(a,b)=(pm1,0)$ or $(0,pm1)$.
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
answered Feb 11 '12 at 7:58
anonanon
72.4k5111214
72.4k5111214
7
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
add a comment |
7
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
7
7
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
$begingroup$
Equivalently, by taking conjugates we conclude that $bar{z}bar{w}=1$ and therefore $(zbar z)(wbar{w})=1$ and therefore $zbar{z}=1$.
$endgroup$
– André Nicolas
Feb 11 '12 at 8:14
add a comment |
$begingroup$
Alternatively, $mathbb{Z}[i] = { a + bi : a,b in mathbb{Z}}$. So the norm of $mathbb{Z}[i]$ is $a^{2} + b^{2}$. Now, $a+bi$ is a unit if and only if $a+bi$ divides 1. Hence $N(a+bi)=pm 1. $
So, $a^{2} + b^{2} = pm 1$
Since $a,b in mathbb{Z}$, the only solutions are $a=pm 1$, $b=0$ and $a=0$, $b=pm1$. So the units are $1, -1, i, -i$
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
$endgroup$
add a comment |
$begingroup$
Alternatively, $mathbb{Z}[i] = { a + bi : a,b in mathbb{Z}}$. So the norm of $mathbb{Z}[i]$ is $a^{2} + b^{2}$. Now, $a+bi$ is a unit if and only if $a+bi$ divides 1. Hence $N(a+bi)=pm 1. $
So, $a^{2} + b^{2} = pm 1$
Since $a,b in mathbb{Z}$, the only solutions are $a=pm 1$, $b=0$ and $a=0$, $b=pm1$. So the units are $1, -1, i, -i$
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
$endgroup$
add a comment |
$begingroup$
Alternatively, $mathbb{Z}[i] = { a + bi : a,b in mathbb{Z}}$. So the norm of $mathbb{Z}[i]$ is $a^{2} + b^{2}$. Now, $a+bi$ is a unit if and only if $a+bi$ divides 1. Hence $N(a+bi)=pm 1. $
So, $a^{2} + b^{2} = pm 1$
Since $a,b in mathbb{Z}$, the only solutions are $a=pm 1$, $b=0$ and $a=0$, $b=pm1$. So the units are $1, -1, i, -i$
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
$endgroup$
Alternatively, $mathbb{Z}[i] = { a + bi : a,b in mathbb{Z}}$. So the norm of $mathbb{Z}[i]$ is $a^{2} + b^{2}$. Now, $a+bi$ is a unit if and only if $a+bi$ divides 1. Hence $N(a+bi)=pm 1. $
So, $a^{2} + b^{2} = pm 1$
Since $a,b in mathbb{Z}$, the only solutions are $a=pm 1$, $b=0$ and $a=0$, $b=pm1$. So the units are $1, -1, i, -i$
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
edited Apr 19 '16 at 18:29
user217174
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
answered Apr 19 '16 at 18:11
DMSTADMSTA
7112
7112
add a comment |
add a comment |
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7
$begingroup$
Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer.
$endgroup$
– André Nicolas
Feb 11 '12 at 7:53