Showing an isomorphism between two quadratic rings
Let $a,b$ be squarefree integers and set $R = mathbb{Z}[sqrt{a}]$ and $S = mathbb{Z}[sqrt{b}]$. Prove that
1) There is an isomorphism of abelian groups $(R,+) cong (S,+)$.
Let $varphi : R to mathbb{Z} times mathbb{Z}$ be a map of groups such that $varphi(x+ysqrt{a}) = (x,y)$. This is a homomorphism since if $x+ysqrt{a}, w+zsqrt{a} in R$, then $$varphi ((x+ysqrt{a}) + (w+zsqrt{a})) = varphi((x+w) + (y+z)sqrt{a}) = (x+w,y+z) = (x,y)+(w,z) = varphi(x+ysqrt{a}) + varphi(w+zsqrt{a}).$$
Now, let $x+ysqrt{a}in ker(varphi)$. Then $varphi(x+ysqrt{a}) = (x,y) = 0$, which is true if and only if $x=0$ and $y=0$, i.e. $x+ysqrt{a} = 0$, and so the kernel is trivial, and the map is injective.
The map is certainly surjective since $varphi(x+ysqrt{a}) = (x,y)$, and $(x,y)$ is a general element of $mathbb{Z} times mathbb{Z}$. Hence $R cong mathbb{Z} times mathbb{Z}$.
Note that this proof did not depend on the value of $a$, only that it was squarefree. Hence we also have that $S cong mathbb{Z} times mathbb{Z}$, and so $R cong S$ as groups under addition.
2) There is an isomorphism of rings $Rcong S$ if and only if $a=b$.
This is the part that I am a bit confused on. The reverse direction is clear, since if $a=b$ they are just the same group so are of course isomorphic. It's the forward direction that I am stuck on.
abstract-algebra ring-theory
asked Dec 8 at 23:48
Wesley
520313
add a comment |
Let $a,b$ be squarefree integers and set $R = mathbb{Z}[sqrt{a}]$ and $S = mathbb{Z}[sqrt{b}]$. Prove that
1) There is an isomorphism of abelian groups $(R,+) cong (S,+)$.
Let $varphi : R to mathbb{Z} times mathbb{Z}$ be a map of groups such that $varphi(x+ysqrt{a}) = (x,y)$. This is a homomorphism since if $x+ysqrt{a}, w+zsqrt{a} in R$, then $$varphi ((x+ysqrt{a}) + (w+zsqrt{a})) = varphi((x+w) + (y+z)sqrt{a}) = (x+w,y+z) = (x,y)+(w,z) = varphi(x+ysqrt{a}) + varphi(w+zsqrt{a}).$$
Now, let $x+ysqrt{a}in ker(varphi)$. Then $varphi(x+ysqrt{a}) = (x,y) = 0$, which is true if and only if $x=0$ and $y=0$, i.e. $x+ysqrt{a} = 0$, and so the kernel is trivial, and the map is injective.
The map is certainly surjective since $varphi(x+ysqrt{a}) = (x,y)$, and $(x,y)$ is a general element of $mathbb{Z} times mathbb{Z}$. Hence $R cong mathbb{Z} times mathbb{Z}$.
Note that this proof did not depend on the value of $a$, only that it was squarefree. Hence we also have that $S cong mathbb{Z} times mathbb{Z}$, and so $R cong S$ as groups under addition.
2) There is an isomorphism of rings $Rcong S$ if and only if $a=b$.
This is the part that I am a bit confused on. The reverse direction is clear, since if $a=b$ they are just the same group so are of course isomorphic. It's the forward direction that I am stuck on.
abstract-algebra ring-theory
asked Dec 8 at 23:48
Wesley
520313
That's a command not a question. What work have you done on the problem?
– Rob Arthan
Dec 9 at 1:04
Hint: Suppose $fcolon Rto S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(sqrt{a}) = r+ssqrt{b}$ for some $r,sinmathbb{Z}$, and consider $(r+ssqrt{b})^2$.
– Arturo Magidin
Dec 9 at 2:32
add a comment |
Let $a,b$ be squarefree integers and set $R = mathbb{Z}[sqrt{a}]$ and $S = mathbb{Z}[sqrt{b}]$. Prove that
1) There is an isomorphism of abelian groups $(R,+) cong (S,+)$.
Let $varphi : R to mathbb{Z} times mathbb{Z}$ be a map of groups such that $varphi(x+ysqrt{a}) = (x,y)$. This is a homomorphism since if $x+ysqrt{a}, w+zsqrt{a} in R$, then $$varphi ((x+ysqrt{a}) + (w+zsqrt{a})) = varphi((x+w) + (y+z)sqrt{a}) = (x+w,y+z) = (x,y)+(w,z) = varphi(x+ysqrt{a}) + varphi(w+zsqrt{a}).$$
Now, let $x+ysqrt{a}in ker(varphi)$. Then $varphi(x+ysqrt{a}) = (x,y) = 0$, which is true if and only if $x=0$ and $y=0$, i.e. $x+ysqrt{a} = 0$, and so the kernel is trivial, and the map is injective.
The map is certainly surjective since $varphi(x+ysqrt{a}) = (x,y)$, and $(x,y)$ is a general element of $mathbb{Z} times mathbb{Z}$. Hence $R cong mathbb{Z} times mathbb{Z}$.
Note that this proof did not depend on the value of $a$, only that it was squarefree. Hence we also have that $S cong mathbb{Z} times mathbb{Z}$, and so $R cong S$ as groups under addition.
2) There is an isomorphism of rings $Rcong S$ if and only if $a=b$.
This is the part that I am a bit confused on. The reverse direction is clear, since if $a=b$ they are just the same group so are of course isomorphic. It's the forward direction that I am stuck on.
abstract-algebra ring-theory
asked Dec 8 at 23:48
Wesley
520313
Let $a,b$ be squarefree integers and set $R = mathbb{Z}[sqrt{a}]$ and $S = mathbb{Z}[sqrt{b}]$. Prove that
1) There is an isomorphism of abelian groups $(R,+) cong (S,+)$.
Let $varphi : R to mathbb{Z} times mathbb{Z}$ be a map of groups such that $varphi(x+ysqrt{a}) = (x,y)$. This is a homomorphism since if $x+ysqrt{a}, w+zsqrt{a} in R$, then $$varphi ((x+ysqrt{a}) + (w+zsqrt{a})) = varphi((x+w) + (y+z)sqrt{a}) = (x+w,y+z) = (x,y)+(w,z) = varphi(x+ysqrt{a}) + varphi(w+zsqrt{a}).$$
Now, let $x+ysqrt{a}in ker(varphi)$. Then $varphi(x+ysqrt{a}) = (x,y) = 0$, which is true if and only if $x=0$ and $y=0$, i.e. $x+ysqrt{a} = 0$, and so the kernel is trivial, and the map is injective.
The map is certainly surjective since $varphi(x+ysqrt{a}) = (x,y)$, and $(x,y)$ is a general element of $mathbb{Z} times mathbb{Z}$. Hence $R cong mathbb{Z} times mathbb{Z}$.
Note that this proof did not depend on the value of $a$, only that it was squarefree. Hence we also have that $S cong mathbb{Z} times mathbb{Z}$, and so $R cong S$ as groups under addition.
2) There is an isomorphism of rings $Rcong S$ if and only if $a=b$.
This is the part that I am a bit confused on. The reverse direction is clear, since if $a=b$ they are just the same group so are of course isomorphic. It's the forward direction that I am stuck on.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Dec 8 at 23:48
Wesley
520313
asked Dec 8 at 23:48
Wesley
520313
edited Dec 9 at 2:18
asked Dec 8 at 23:48
Wesley
520313
asked Dec 8 at 23:48
Wesley
520313
asked Dec 8 at 23:48
Wesley
520313
520313
That's a command not a question. What work have you done on the problem?
– Rob Arthan
Dec 9 at 1:04
Hint: Suppose $fcolon Rto S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(sqrt{a}) = r+ssqrt{b}$ for some $r,sinmathbb{Z}$, and consider $(r+ssqrt{b})^2$.
– Arturo Magidin
Dec 9 at 2:32
add a comment |
That's a command not a question. What work have you done on the problem?
– Rob Arthan
Dec 9 at 1:04
Hint: Suppose $fcolon Rto S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(sqrt{a}) = r+ssqrt{b}$ for some $r,sinmathbb{Z}$, and consider $(r+ssqrt{b})^2$.
– Arturo Magidin
Dec 9 at 2:32
That's a command not a question. What work have you done on the problem?
– Rob Arthan
Dec 9 at 1:04
That's a command not a question. What work have you done on the problem?
– Rob Arthan
Dec 9 at 1:04
Hint: Suppose $fcolon Rto S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(sqrt{a}) = r+ssqrt{b}$ for some $r,sinmathbb{Z}$, and consider $(r+ssqrt{b})^2$.
– Arturo Magidin
Dec 9 at 2:32
Hint: Suppose $fcolon Rto S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(sqrt{a}) = r+ssqrt{b}$ for some $r,sinmathbb{Z}$, and consider $(r+ssqrt{b})^2$.
– Arturo Magidin
Dec 9 at 2:32
add a comment |
1 Answer
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Hint 1: as abelian groups, $mathbb{Z}[sqrt{a}]congmathbb{Z}oplusmathbb{Z}$.
Hint 2: what integers have a square root in $mathbb{Z}[sqrt{a}]$?
answered Dec 8 at 23:53
egreg
178k1484200
add a comment |
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Hint 1: as abelian groups, $mathbb{Z}[sqrt{a}]congmathbb{Z}oplusmathbb{Z}$.
Hint 2: what integers have a square root in $mathbb{Z}[sqrt{a}]$?
answered Dec 8 at 23:53
egreg
178k1484200
add a comment |
Hint 1: as abelian groups, $mathbb{Z}[sqrt{a}]congmathbb{Z}oplusmathbb{Z}$.
Hint 2: what integers have a square root in $mathbb{Z}[sqrt{a}]$?
answered Dec 8 at 23:53
egreg
178k1484200
add a comment |
Hint 1: as abelian groups, $mathbb{Z}[sqrt{a}]congmathbb{Z}oplusmathbb{Z}$.
Hint 2: what integers have a square root in $mathbb{Z}[sqrt{a}]$?
answered Dec 8 at 23:53
egreg
178k1484200
Hint 1: as abelian groups, $mathbb{Z}[sqrt{a}]congmathbb{Z}oplusmathbb{Z}$.
Hint 2: what integers have a square root in $mathbb{Z}[sqrt{a}]$?
answered Dec 8 at 23:53
egreg
178k1484200
answered Dec 8 at 23:53
egreg
178k1484200
answered Dec 8 at 23:53
egreg
178k1484200
answered Dec 8 at 23:53
egreg
178k1484200
178k1484200
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That's a command not a question. What work have you done on the problem?
– Rob Arthan
Dec 9 at 1:04
Hint: Suppose $fcolon Rto S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(sqrt{a}) = r+ssqrt{b}$ for some $r,sinmathbb{Z}$, and consider $(r+ssqrt{b})^2$.
– Arturo Magidin
Dec 9 at 2:32