In GCD domain every invertible ideal is principal
$begingroup$
This is the exercise in the book Commutative Rings by Kaplansky.
Prove that in a GCD domain every invertible ideal is principal.
I'm looking for some hints.
Edit
After understanding the hint, here is my approach:
Let $I$ be an invertible ideal of a GCD domain $R$. Because $I$ is finitely generated as $R$-module we can write $I=(a_1/b_1,...,a_n/b_n)R$, where $a_i, b_i$ are elements in $R$.
Since $R$ is a GCD domain we can choose $a_i, b_i$ such that $(a_i,b_i)=1$.
By hypothesis $R$ is also an LCM domain. Let $c$ be the least common multiple of $b_i$'s, $d$ be the greatest common divisor of $a_i$'s. It is easy to see that $I^{-1}=(c/d)R$.
Because of invertibility there exist $m_i$'s of $I$ such that $m_1(c/d)+cdots+m_n(c/d)=1$.
We conclude that $I=uR$, where $u=m_1+cdots+m_n$, for if $xin I$, $x=xcdot1=xm_1(c/d)+cdots+xm_n(c/d)=x(c/d)u$.
commutative-algebra
edited Dec 15 '18 at 8:00
user26857
39.5k124284
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
$endgroup$
add a comment |
$begingroup$
This is the exercise in the book Commutative Rings by Kaplansky.
Prove that in a GCD domain every invertible ideal is principal.
I'm looking for some hints.
Edit
After understanding the hint, here is my approach:
Let $I$ be an invertible ideal of a GCD domain $R$. Because $I$ is finitely generated as $R$-module we can write $I=(a_1/b_1,...,a_n/b_n)R$, where $a_i, b_i$ are elements in $R$.
Since $R$ is a GCD domain we can choose $a_i, b_i$ such that $(a_i,b_i)=1$.
By hypothesis $R$ is also an LCM domain. Let $c$ be the least common multiple of $b_i$'s, $d$ be the greatest common divisor of $a_i$'s. It is easy to see that $I^{-1}=(c/d)R$.
Because of invertibility there exist $m_i$'s of $I$ such that $m_1(c/d)+cdots+m_n(c/d)=1$.
We conclude that $I=uR$, where $u=m_1+cdots+m_n$, for if $xin I$, $x=xcdot1=xm_1(c/d)+cdots+xm_n(c/d)=x(c/d)u$.
commutative-algebra
edited Dec 15 '18 at 8:00
user26857
39.5k124284
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
$endgroup$
1
$begingroup$
An integral domain is called GCD domain if any two its elements have a greatest common divisor.
$endgroup$
– Anh_Rose 1210
Apr 4 '17 at 19:14
1
$begingroup$
@Axlp1210 Hint: $I^{-1}=d^{-1}R$ where $d$ is..., you know.
$endgroup$
– user26857
Apr 4 '17 at 21:00
$begingroup$
Actually you can start with $Isubseteq R$.
$endgroup$
– user26857
Apr 7 '17 at 13:29
$begingroup$
@user26857 Thank you. I love that kind of hint. It got me think seriously to understand how things work.
$endgroup$
– Anh_Rose 1210
Apr 7 '17 at 13:45
add a comment |
$begingroup$
This is the exercise in the book Commutative Rings by Kaplansky.
Prove that in a GCD domain every invertible ideal is principal.
I'm looking for some hints.
Edit
After understanding the hint, here is my approach:
Let $I$ be an invertible ideal of a GCD domain $R$. Because $I$ is finitely generated as $R$-module we can write $I=(a_1/b_1,...,a_n/b_n)R$, where $a_i, b_i$ are elements in $R$.
Since $R$ is a GCD domain we can choose $a_i, b_i$ such that $(a_i,b_i)=1$.
By hypothesis $R$ is also an LCM domain. Let $c$ be the least common multiple of $b_i$'s, $d$ be the greatest common divisor of $a_i$'s. It is easy to see that $I^{-1}=(c/d)R$.
Because of invertibility there exist $m_i$'s of $I$ such that $m_1(c/d)+cdots+m_n(c/d)=1$.
We conclude that $I=uR$, where $u=m_1+cdots+m_n$, for if $xin I$, $x=xcdot1=xm_1(c/d)+cdots+xm_n(c/d)=x(c/d)u$.
commutative-algebra
edited Dec 15 '18 at 8:00
user26857
39.5k124284
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
$endgroup$
This is the exercise in the book Commutative Rings by Kaplansky.
Prove that in a GCD domain every invertible ideal is principal.
I'm looking for some hints.
Edit
After understanding the hint, here is my approach:
Let $I$ be an invertible ideal of a GCD domain $R$. Because $I$ is finitely generated as $R$-module we can write $I=(a_1/b_1,...,a_n/b_n)R$, where $a_i, b_i$ are elements in $R$.
Since $R$ is a GCD domain we can choose $a_i, b_i$ such that $(a_i,b_i)=1$.
By hypothesis $R$ is also an LCM domain. Let $c$ be the least common multiple of $b_i$'s, $d$ be the greatest common divisor of $a_i$'s. It is easy to see that $I^{-1}=(c/d)R$.
Because of invertibility there exist $m_i$'s of $I$ such that $m_1(c/d)+cdots+m_n(c/d)=1$.
We conclude that $I=uR$, where $u=m_1+cdots+m_n$, for if $xin I$, $x=xcdot1=xm_1(c/d)+cdots+xm_n(c/d)=x(c/d)u$.
commutative-algebra
commutative-algebra
edited Dec 15 '18 at 8:00
user26857
39.5k124284
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
edited Dec 15 '18 at 8:00
user26857
39.5k124284
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
edited Dec 15 '18 at 8:00
user26857
39.5k124284
edited Dec 15 '18 at 8:00
user26857
39.5k124284
edited Dec 15 '18 at 8:00
user26857
39.5k124284
39.5k124284
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
asked Apr 4 '17 at 18:53
Anh_Rose 1210Anh_Rose 1210
374210
374210
1
$begingroup$
An integral domain is called GCD domain if any two its elements have a greatest common divisor.
$endgroup$
– Anh_Rose 1210
Apr 4 '17 at 19:14
1
$begingroup$
@Axlp1210 Hint: $I^{-1}=d^{-1}R$ where $d$ is..., you know.
$endgroup$
– user26857
Apr 4 '17 at 21:00
$begingroup$
Actually you can start with $Isubseteq R$.
$endgroup$
– user26857
Apr 7 '17 at 13:29
$begingroup$
@user26857 Thank you. I love that kind of hint. It got me think seriously to understand how things work.
$endgroup$
– Anh_Rose 1210
Apr 7 '17 at 13:45
add a comment |
1
$begingroup$
An integral domain is called GCD domain if any two its elements have a greatest common divisor.
$endgroup$
– Anh_Rose 1210
Apr 4 '17 at 19:14
1
$begingroup$
@Axlp1210 Hint: $I^{-1}=d^{-1}R$ where $d$ is..., you know.
$endgroup$
– user26857
Apr 4 '17 at 21:00
$begingroup$
Actually you can start with $Isubseteq R$.
$endgroup$
– user26857
Apr 7 '17 at 13:29
$begingroup$
@user26857 Thank you. I love that kind of hint. It got me think seriously to understand how things work.
$endgroup$
– Anh_Rose 1210
Apr 7 '17 at 13:45
1
1
$begingroup$
An integral domain is called GCD domain if any two its elements have a greatest common divisor.
$endgroup$
– Anh_Rose 1210
Apr 4 '17 at 19:14
$begingroup$
An integral domain is called GCD domain if any two its elements have a greatest common divisor.
$endgroup$
– Anh_Rose 1210
Apr 4 '17 at 19:14
1
1
$begingroup$
@Axlp1210 Hint: $I^{-1}=d^{-1}R$ where $d$ is..., you know.
$endgroup$
– user26857
Apr 4 '17 at 21:00
$begingroup$
@Axlp1210 Hint: $I^{-1}=d^{-1}R$ where $d$ is..., you know.
$endgroup$
– user26857
Apr 4 '17 at 21:00
$begingroup$
Actually you can start with $Isubseteq R$.
$endgroup$
– user26857
Apr 7 '17 at 13:29
$begingroup$
Actually you can start with $Isubseteq R$.
$endgroup$
– user26857
Apr 7 '17 at 13:29
$begingroup$
@user26857 Thank you. I love that kind of hint. It got me think seriously to understand how things work.
$endgroup$
– Anh_Rose 1210
Apr 7 '17 at 13:45
$begingroup$
@user26857 Thank you. I love that kind of hint. It got me think seriously to understand how things work.
$endgroup$
– Anh_Rose 1210
Apr 7 '17 at 13:45
add a comment |
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1
$begingroup$
An integral domain is called GCD domain if any two its elements have a greatest common divisor.
$endgroup$
– Anh_Rose 1210
Apr 4 '17 at 19:14
1
$begingroup$
@Axlp1210 Hint: $I^{-1}=d^{-1}R$ where $d$ is..., you know.
$endgroup$
– user26857
Apr 4 '17 at 21:00
$begingroup$
Actually you can start with $Isubseteq R$.
$endgroup$
– user26857
Apr 7 '17 at 13:29
$begingroup$
@user26857 Thank you. I love that kind of hint. It got me think seriously to understand how things work.
$endgroup$
– Anh_Rose 1210
Apr 7 '17 at 13:45