Mixing
The units of $mathbb Z[sqrt2]$
Clash Royale CLAN TAG#URR8PPP
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How can I show that the units $u$ of $R=mathbb Z[sqrt2]$ with $u>1$ are $(1+ sqrt2)^n$ ?
I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+sqrt2$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help.
abstract-algebra number-theory algebraic-number-theory
edited Jul 8 '13 at 16:44
Potato
20.8k984183
asked Jan 17 '13 at 17:51
john
78115
add a comment |Â
up vote
15
down vote
favorite
How can I show that the units $u$ of $R=mathbb Z[sqrt2]$ with $u>1$ are $(1+ sqrt2)^n$ ?
I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+sqrt2$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help.
abstract-algebra number-theory algebraic-number-theory
edited Jul 8 '13 at 16:44
Potato
20.8k984183
asked Jan 17 '13 at 17:51
john
78115
5
Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag...
– awllower
Jan 17 '13 at 18:08
en.wikipedia.org/wiki/Continued_fraction See its theorem 3
– awllower
Jan 17 '13 at 18:17
2
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=pm 1$. Indeed, $1+sqrt2$ corresponds to such a solution for $x^2-2y^2=-1$
– Thomas Andrews
Jan 17 '13 at 18:21
@awllower, I would like to read about the relations you mentioned above. I realized that it has been 4 years but still I appreciate if you could answer.
– Ninja
Nov 19 '17 at 22:22
add a comment |Â
up vote
15
down vote
favorite
up vote
15
down vote
favorite
How can I show that the units $u$ of $R=mathbb Z[sqrt2]$ with $u>1$ are $(1+ sqrt2)^n$ ?
I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+sqrt2$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help.
abstract-algebra number-theory algebraic-number-theory
edited Jul 8 '13 at 16:44
Potato
20.8k984183
asked Jan 17 '13 at 17:51
john
78115
How can I show that the units $u$ of $R=mathbb Z[sqrt2]$ with $u>1$ are $(1+ sqrt2)^n$ ?
I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+sqrt2$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help.
abstract-algebra number-theory algebraic-number-theory
edited Jul 8 '13 at 16:44
Potato
20.8k984183
asked Jan 17 '13 at 17:51
john
78115
edited Jul 8 '13 at 16:44
Potato
20.8k984183
edited Jul 8 '13 at 16:44
Potato
20.8k984183
edited Jul 8 '13 at 16:44
Potato
20.8k984183
20.8k984183
asked Jan 17 '13 at 17:51
john
78115
asked Jan 17 '13 at 17:51
john
78115
asked Jan 17 '13 at 17:51
john
78115
78115
5
Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag...
– awllower
Jan 17 '13 at 18:08
en.wikipedia.org/wiki/Continued_fraction See its theorem 3
– awllower
Jan 17 '13 at 18:17
2
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=pm 1$. Indeed, $1+sqrt2$ corresponds to such a solution for $x^2-2y^2=-1$
– Thomas Andrews
Jan 17 '13 at 18:21
@awllower, I would like to read about the relations you mentioned above. I realized that it has been 4 years but still I appreciate if you could answer.
– Ninja
Nov 19 '17 at 22:22
add a comment |Â
5
Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag...
– awllower
Jan 17 '13 at 18:08
en.wikipedia.org/wiki/Continued_fraction See its theorem 3
– awllower
Jan 17 '13 at 18:17
2
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=pm 1$. Indeed, $1+sqrt2$ corresponds to such a solution for $x^2-2y^2=-1$
– Thomas Andrews
Jan 17 '13 at 18:21
@awllower, I would like to read about the relations you mentioned above. I realized that it has been 4 years but still I appreciate if you could answer.
– Ninja
Nov 19 '17 at 22:22
5
5
Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag...
– awllower
Jan 17 '13 at 18:08
Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag...
– awllower
Jan 17 '13 at 18:08
en.wikipedia.org/wiki/Continued_fraction See its theorem 3
– awllower
Jan 17 '13 at 18:17
en.wikipedia.org/wiki/Continued_fraction See its theorem 3
– awllower
Jan 17 '13 at 18:17
2
2
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=pm 1$. Indeed, $1+sqrt2$ corresponds to such a solution for $x^2-2y^2=-1$
– Thomas Andrews
Jan 17 '13 at 18:21
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=pm 1$. Indeed, $1+sqrt2$ corresponds to such a solution for $x^2-2y^2=-1$
– Thomas Andrews
Jan 17 '13 at 18:21
@awllower, I would like to read about the relations you mentioned above. I realized that it has been 4 years but still I appreciate if you could answer.
– Ninja
Nov 19 '17 at 22:22
@awllower, I would like to read about the relations you mentioned above. I realized that it has been 4 years but still I appreciate if you could answer.
– Ninja
Nov 19 '17 at 22:22
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
18
down vote
accepted
First, note that $a+bsqrt2$ is a unit if and only if $a^2-2b^2=pm1$. Use that to show that if $bneq 0$ then $|b|leq |a|< 2|b|$.
Now we first restrict ourselves to $a,bgeq 0$, and prove by induction on $b$.
If $b=0$ then $a=pm 1$, and $u>0$ implies $a=1$, so $u=(1+sqrt 2)^0$.
If $a,b>0$ and $a+bsqrt2$ is a unit, then $$(a+bsqrt2)(sqrt2-1) = (2b-a)+(a-b)sqrt2$$
is also a unit.
Since we know that $bleq a< 2b$ and $b<a$, we have that $2b-a>0$ and $0<a-b<b$, so by induction:
$$(a+bsqrt2)(sqrt2-1) =(1+sqrt2)^n$$
But multiplying both sides by $1+sqrt2$ you get:
$$a+bsqrt2=(1+sqrt2)^n+1$$
Then you have to deal with the case where one of $a,b$ is negative...
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
1
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
 |Â
show 2 more comments
up vote
5
down vote
Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+sqrt2$. Clearly the nonnegative powers of $1+sqrt2$ monotonically tend to $infty$ starting from $1$, so we can write $$(1+sqrt2)^kle u <(1+sqrt2)^k+1$$ for some $kinmathbfZ^+cup0$. Dividing by $(1+sqrt2)^k$ yields $$1le u(1+sqrt2)^-k<1+sqrt2.$$ Note that $u(1+sqrt2)^-kinmathbfZ[sqrt2]^times$, and since $1+sqrt2$ is the smallest unit greater than $1$, we must have $u(1+sqrt2)^-k=1implies u=(1+sqrt2)^k$. Due to norm being multiplicative, all powers of $1+sqrt2$ are units, so we are done.
answered Nov 27 '13 at 2:47
tc1729
2,66611332
4
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
First, note that $a+bsqrt2$ is a unit if and only if $a^2-2b^2=pm1$. Use that to show that if $bneq 0$ then $|b|leq |a|< 2|b|$.
Now we first restrict ourselves to $a,bgeq 0$, and prove by induction on $b$.
If $b=0$ then $a=pm 1$, and $u>0$ implies $a=1$, so $u=(1+sqrt 2)^0$.
If $a,b>0$ and $a+bsqrt2$ is a unit, then $$(a+bsqrt2)(sqrt2-1) = (2b-a)+(a-b)sqrt2$$
is also a unit.
Since we know that $bleq a< 2b$ and $b<a$, we have that $2b-a>0$ and $0<a-b<b$, so by induction:
$$(a+bsqrt2)(sqrt2-1) =(1+sqrt2)^n$$
But multiplying both sides by $1+sqrt2$ you get:
$$a+bsqrt2=(1+sqrt2)^n+1$$
Then you have to deal with the case where one of $a,b$ is negative...
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
1
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
 |Â
show 2 more comments
up vote
18
down vote
accepted
First, note that $a+bsqrt2$ is a unit if and only if $a^2-2b^2=pm1$. Use that to show that if $bneq 0$ then $|b|leq |a|< 2|b|$.
Now we first restrict ourselves to $a,bgeq 0$, and prove by induction on $b$.
If $b=0$ then $a=pm 1$, and $u>0$ implies $a=1$, so $u=(1+sqrt 2)^0$.
If $a,b>0$ and $a+bsqrt2$ is a unit, then $$(a+bsqrt2)(sqrt2-1) = (2b-a)+(a-b)sqrt2$$
is also a unit.
Since we know that $bleq a< 2b$ and $b<a$, we have that $2b-a>0$ and $0<a-b<b$, so by induction:
$$(a+bsqrt2)(sqrt2-1) =(1+sqrt2)^n$$
But multiplying both sides by $1+sqrt2$ you get:
$$a+bsqrt2=(1+sqrt2)^n+1$$
Then you have to deal with the case where one of $a,b$ is negative...
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
1
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
 |Â
show 2 more comments
up vote
18
down vote
accepted
up vote
18
down vote
accepted
First, note that $a+bsqrt2$ is a unit if and only if $a^2-2b^2=pm1$. Use that to show that if $bneq 0$ then $|b|leq |a|< 2|b|$.
Now we first restrict ourselves to $a,bgeq 0$, and prove by induction on $b$.
If $b=0$ then $a=pm 1$, and $u>0$ implies $a=1$, so $u=(1+sqrt 2)^0$.
If $a,b>0$ and $a+bsqrt2$ is a unit, then $$(a+bsqrt2)(sqrt2-1) = (2b-a)+(a-b)sqrt2$$
is also a unit.
Since we know that $bleq a< 2b$ and $b<a$, we have that $2b-a>0$ and $0<a-b<b$, so by induction:
$$(a+bsqrt2)(sqrt2-1) =(1+sqrt2)^n$$
But multiplying both sides by $1+sqrt2$ you get:
$$a+bsqrt2=(1+sqrt2)^n+1$$
Then you have to deal with the case where one of $a,b$ is negative...
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
First, note that $a+bsqrt2$ is a unit if and only if $a^2-2b^2=pm1$. Use that to show that if $bneq 0$ then $|b|leq |a|< 2|b|$.
Now we first restrict ourselves to $a,bgeq 0$, and prove by induction on $b$.
If $b=0$ then $a=pm 1$, and $u>0$ implies $a=1$, so $u=(1+sqrt 2)^0$.
If $a,b>0$ and $a+bsqrt2$ is a unit, then $$(a+bsqrt2)(sqrt2-1) = (2b-a)+(a-b)sqrt2$$
is also a unit.
Since we know that $bleq a< 2b$ and $b<a$, we have that $2b-a>0$ and $0<a-b<b$, so by induction:
$$(a+bsqrt2)(sqrt2-1) =(1+sqrt2)^n$$
But multiplying both sides by $1+sqrt2$ you get:
$$a+bsqrt2=(1+sqrt2)^n+1$$
Then you have to deal with the case where one of $a,b$ is negative...
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
edited Aug 13 '15 at 12:52
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
answered Jan 17 '13 at 18:11
Thomas Andrews
128k10143284
128k10143284
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
1
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
 |Â
show 2 more comments
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
1
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
$a+bsqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$?
– Temitope.A
Jan 18 '13 at 23:17
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
thomas Whhy is it so?
– Temitope.A
Jan 19 '13 at 10:34
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A If $a+bsqrt2$ is a unit, show that $a-bsqrt2$ is also a unit, so $a^2-2b^2=(a+bsqrt2)(a-bsqrt2)$ is a unit. But the only elements of $mathbb Z$ that are units in $mathbb Z[sqrt2]$ are $pm 1$
– Thomas Andrews
Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+bsqrt2$ cannot be a unit.
– Thomas Andrews
Jan 19 '13 at 12:52
1
1
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
Let $a+b sqrt2$ be a unit. To show $a^2-2b^2=pm 1$, define $N(a+b sqrt2):=a^2-2b^2$. It can be verified that $N(alpha beta)=N(alpha)N(beta), forall alpha, beta in mathbbZ[sqrt2]$. If $alpha=a+b sqrt2$ is a unit, then $alpha beta=1$ for some $beta$, whence $N(alpha)N(beta)=1$ in $mathbbZ$. Thus, $N(alpha)=pm 1$.
– AG.
Jul 13 '13 at 0:07
 |Â
show 2 more comments
up vote
5
down vote
Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+sqrt2$. Clearly the nonnegative powers of $1+sqrt2$ monotonically tend to $infty$ starting from $1$, so we can write $$(1+sqrt2)^kle u <(1+sqrt2)^k+1$$ for some $kinmathbfZ^+cup0$. Dividing by $(1+sqrt2)^k$ yields $$1le u(1+sqrt2)^-k<1+sqrt2.$$ Note that $u(1+sqrt2)^-kinmathbfZ[sqrt2]^times$, and since $1+sqrt2$ is the smallest unit greater than $1$, we must have $u(1+sqrt2)^-k=1implies u=(1+sqrt2)^k$. Due to norm being multiplicative, all powers of $1+sqrt2$ are units, so we are done.
answered Nov 27 '13 at 2:47
tc1729
2,66611332
4
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
add a comment |Â
up vote
5
down vote
Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+sqrt2$. Clearly the nonnegative powers of $1+sqrt2$ monotonically tend to $infty$ starting from $1$, so we can write $$(1+sqrt2)^kle u <(1+sqrt2)^k+1$$ for some $kinmathbfZ^+cup0$. Dividing by $(1+sqrt2)^k$ yields $$1le u(1+sqrt2)^-k<1+sqrt2.$$ Note that $u(1+sqrt2)^-kinmathbfZ[sqrt2]^times$, and since $1+sqrt2$ is the smallest unit greater than $1$, we must have $u(1+sqrt2)^-k=1implies u=(1+sqrt2)^k$. Due to norm being multiplicative, all powers of $1+sqrt2$ are units, so we are done.
answered Nov 27 '13 at 2:47
tc1729
2,66611332
4
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+sqrt2$. Clearly the nonnegative powers of $1+sqrt2$ monotonically tend to $infty$ starting from $1$, so we can write $$(1+sqrt2)^kle u <(1+sqrt2)^k+1$$ for some $kinmathbfZ^+cup0$. Dividing by $(1+sqrt2)^k$ yields $$1le u(1+sqrt2)^-k<1+sqrt2.$$ Note that $u(1+sqrt2)^-kinmathbfZ[sqrt2]^times$, and since $1+sqrt2$ is the smallest unit greater than $1$, we must have $u(1+sqrt2)^-k=1implies u=(1+sqrt2)^k$. Due to norm being multiplicative, all powers of $1+sqrt2$ are units, so we are done.
answered Nov 27 '13 at 2:47
tc1729
2,66611332
Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+sqrt2$. Clearly the nonnegative powers of $1+sqrt2$ monotonically tend to $infty$ starting from $1$, so we can write $$(1+sqrt2)^kle u <(1+sqrt2)^k+1$$ for some $kinmathbfZ^+cup0$. Dividing by $(1+sqrt2)^k$ yields $$1le u(1+sqrt2)^-k<1+sqrt2.$$ Note that $u(1+sqrt2)^-kinmathbfZ[sqrt2]^times$, and since $1+sqrt2$ is the smallest unit greater than $1$, we must have $u(1+sqrt2)^-k=1implies u=(1+sqrt2)^k$. Due to norm being multiplicative, all powers of $1+sqrt2$ are units, so we are done.
answered Nov 27 '13 at 2:47
tc1729
2,66611332
answered Nov 27 '13 at 2:47
tc1729
2,66611332
answered Nov 27 '13 at 2:47
tc1729
2,66611332
answered Nov 27 '13 at 2:47
tc1729
2,66611332
2,66611332
4
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
add a comment |Â
4
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
4
4
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
How do you get that $1+sqrt2$ is the smallest unit greater than 1?
– Lao-tzu
Jan 8 '15 at 3:07
add a comment |Â
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5
Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag...
– awllower
Jan 17 '13 at 18:08
en.wikipedia.org/wiki/Continued_fraction See its theorem 3
– awllower
Jan 17 '13 at 18:17
2
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=pm 1$. Indeed, $1+sqrt2$ corresponds to such a solution for $x^2-2y^2=-1$
– Thomas Andrews
Jan 17 '13 at 18:21
@awllower, I would like to read about the relations you mentioned above. I realized that it has been 4 years but still I appreciate if you could answer.
– Ninja
Nov 19 '17 at 22:22