Mixing
Properties of the first syzygy over Dedekind domains
Clash Royale CLAN TAG#URR8PPP
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Let $L$ be a Dedekind domain and $I = (a) + (b)$ a non-principal proper ideal in $L$. Consider equation of the form $xa + yb = 0$. Is it true that then there exists a proper ideal $J$ such that any $x, y$ that satisfy the equation belong in $J$?
commutative-algebra ideals dedekind-domain
asked Jul 20 at 18:06
UnrealVillager
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Let $L$ be a Dedekind domain and $I = (a) + (b)$ a non-principal proper ideal in $L$. Consider equation of the form $xa + yb = 0$. Is it true that then there exists a proper ideal $J$ such that any $x, y$ that satisfy the equation belong in $J$?
commutative-algebra ideals dedekind-domain
asked Jul 20 at 18:06
UnrealVillager
256
I have edited the question to make it clearer.
– UnrealVillager
Jul 20 at 18:29
Is it clear to you that such a proper ideal does not exist if $I$ is principal? If yes, can you reduce to that case by localizing?
– Mohan
Jul 20 at 19:21
Well, looks like I just found a counterexample to that logic, Mohan. For $I = (5, 2sqrt-5)$ in $mathbbZ[sqrt-5]$ the equation seems to imply that $x, y in (sqrt-5)$. Unless I made a mistake somewhere in calculations. Could you please expand a little on your logic? Maybe I'm missing something... Thank you for your answer.
– UnrealVillager
Jul 20 at 19:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $L$ be a Dedekind domain and $I = (a) + (b)$ a non-principal proper ideal in $L$. Consider equation of the form $xa + yb = 0$. Is it true that then there exists a proper ideal $J$ such that any $x, y$ that satisfy the equation belong in $J$?
commutative-algebra ideals dedekind-domain
asked Jul 20 at 18:06
UnrealVillager
256
Let $L$ be a Dedekind domain and $I = (a) + (b)$ a non-principal proper ideal in $L$. Consider equation of the form $xa + yb = 0$. Is it true that then there exists a proper ideal $J$ such that any $x, y$ that satisfy the equation belong in $J$?
commutative-algebra ideals dedekind-domain
asked Jul 20 at 18:06
UnrealVillager
256
edited Jul 20 at 18:28
asked Jul 20 at 18:06
UnrealVillager
256
asked Jul 20 at 18:06
UnrealVillager
256
asked Jul 20 at 18:06
UnrealVillager
256
256
I have edited the question to make it clearer.
– UnrealVillager
Jul 20 at 18:29
Is it clear to you that such a proper ideal does not exist if $I$ is principal? If yes, can you reduce to that case by localizing?
– Mohan
Jul 20 at 19:21
Well, looks like I just found a counterexample to that logic, Mohan. For $I = (5, 2sqrt-5)$ in $mathbbZ[sqrt-5]$ the equation seems to imply that $x, y in (sqrt-5)$. Unless I made a mistake somewhere in calculations. Could you please expand a little on your logic? Maybe I'm missing something... Thank you for your answer.
– UnrealVillager
Jul 20 at 19:33
add a comment |Â
I have edited the question to make it clearer.
– UnrealVillager
Jul 20 at 18:29
Is it clear to you that such a proper ideal does not exist if $I$ is principal? If yes, can you reduce to that case by localizing?
– Mohan
Jul 20 at 19:21
Well, looks like I just found a counterexample to that logic, Mohan. For $I = (5, 2sqrt-5)$ in $mathbbZ[sqrt-5]$ the equation seems to imply that $x, y in (sqrt-5)$. Unless I made a mistake somewhere in calculations. Could you please expand a little on your logic? Maybe I'm missing something... Thank you for your answer.
– UnrealVillager
Jul 20 at 19:33
I have edited the question to make it clearer.
– UnrealVillager
Jul 20 at 18:29
I have edited the question to make it clearer.
– UnrealVillager
Jul 20 at 18:29
Is it clear to you that such a proper ideal does not exist if $I$ is principal? If yes, can you reduce to that case by localizing?
– Mohan
Jul 20 at 19:21
Is it clear to you that such a proper ideal does not exist if $I$ is principal? If yes, can you reduce to that case by localizing?
– Mohan
Jul 20 at 19:21
Well, looks like I just found a counterexample to that logic, Mohan. For $I = (5, 2sqrt-5)$ in $mathbbZ[sqrt-5]$ the equation seems to imply that $x, y in (sqrt-5)$. Unless I made a mistake somewhere in calculations. Could you please expand a little on your logic? Maybe I'm missing something... Thank you for your answer.
– UnrealVillager
Jul 20 at 19:33
Well, looks like I just found a counterexample to that logic, Mohan. For $I = (5, 2sqrt-5)$ in $mathbbZ[sqrt-5]$ the equation seems to imply that $x, y in (sqrt-5)$. Unless I made a mistake somewhere in calculations. Could you please expand a little on your logic? Maybe I'm missing something... Thank you for your answer.
– UnrealVillager
Jul 20 at 19:33
add a comment |Â
1 Answer
1
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0
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Let $Msubset L$ be any maximal ideal. We will show that there exists $x,yin L$ with $xa+yb=0$ and $(x,y)$ is not contained in $M$, which will show that no such proper ideal $J$ as in your question exists.
If you localize at $M$, then $L_M$ is a pid and thus $(a,b)$ is principal, generated by either $a$ or $b$. Wlog, assume $a$ generates it and then $b=pa$ for some $pin L_M$, which we can write as $pa-b=0$. Now, we can find $snotin M$ such that $sp=qin L$ and we get an equation, $qa-sb=0$. Since $snotin M$, we have proved our claim.
answered Jul 20 at 19:43
Mohan
11k1816
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $Msubset L$ be any maximal ideal. We will show that there exists $x,yin L$ with $xa+yb=0$ and $(x,y)$ is not contained in $M$, which will show that no such proper ideal $J$ as in your question exists.
If you localize at $M$, then $L_M$ is a pid and thus $(a,b)$ is principal, generated by either $a$ or $b$. Wlog, assume $a$ generates it and then $b=pa$ for some $pin L_M$, which we can write as $pa-b=0$. Now, we can find $snotin M$ such that $sp=qin L$ and we get an equation, $qa-sb=0$. Since $snotin M$, we have proved our claim.
answered Jul 20 at 19:43
Mohan
11k1816
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
add a comment |Â
up vote
0
down vote
accepted
Let $Msubset L$ be any maximal ideal. We will show that there exists $x,yin L$ with $xa+yb=0$ and $(x,y)$ is not contained in $M$, which will show that no such proper ideal $J$ as in your question exists.
If you localize at $M$, then $L_M$ is a pid and thus $(a,b)$ is principal, generated by either $a$ or $b$. Wlog, assume $a$ generates it and then $b=pa$ for some $pin L_M$, which we can write as $pa-b=0$. Now, we can find $snotin M$ such that $sp=qin L$ and we get an equation, $qa-sb=0$. Since $snotin M$, we have proved our claim.
answered Jul 20 at 19:43
Mohan
11k1816
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $Msubset L$ be any maximal ideal. We will show that there exists $x,yin L$ with $xa+yb=0$ and $(x,y)$ is not contained in $M$, which will show that no such proper ideal $J$ as in your question exists.
If you localize at $M$, then $L_M$ is a pid and thus $(a,b)$ is principal, generated by either $a$ or $b$. Wlog, assume $a$ generates it and then $b=pa$ for some $pin L_M$, which we can write as $pa-b=0$. Now, we can find $snotin M$ such that $sp=qin L$ and we get an equation, $qa-sb=0$. Since $snotin M$, we have proved our claim.
answered Jul 20 at 19:43
Mohan
11k1816
Let $Msubset L$ be any maximal ideal. We will show that there exists $x,yin L$ with $xa+yb=0$ and $(x,y)$ is not contained in $M$, which will show that no such proper ideal $J$ as in your question exists.
If you localize at $M$, then $L_M$ is a pid and thus $(a,b)$ is principal, generated by either $a$ or $b$. Wlog, assume $a$ generates it and then $b=pa$ for some $pin L_M$, which we can write as $pa-b=0$. Now, we can find $snotin M$ such that $sp=qin L$ and we get an equation, $qa-sb=0$. Since $snotin M$, we have proved our claim.
answered Jul 20 at 19:43
Mohan
11k1816
answered Jul 20 at 19:43
Mohan
11k1816
answered Jul 20 at 19:43
Mohan
11k1816
answered Jul 20 at 19:43
Mohan
11k1816
11k1816
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
add a comment |Â
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
Mohan, I understand now. Makes total sense and thank you. I also realized now that I formulated the initial question incorrectly. I'm wondering rather if the equation implies that $x$ and $y$ belong to some maximal ideal $M_x,y$? ($M_x,y$ has to depend on $x, y$, as you have just demonstrated).
– UnrealVillager
Jul 20 at 20:01
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
...in other words, can $x$ and $y$ be selected in such a way that $(x, y) = L$?
– UnrealVillager
Jul 20 at 20:03
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
@UnrealVillager You can easily check that if there are $x,y$ with $xa+yb=0$ and $(x,y)=L$, then $(a,b)$ is principal by using the relation $px+qy=1$ for some $p,qin L$.
– Mohan
Jul 20 at 20:21
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
Ok, that's because in that case both $a$ and $b$ sit in $(aq-bp)$. Got it. Thanks a lot for your help with this today, Mohan.
– UnrealVillager
Jul 20 at 20:49
add a comment |Â
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I have edited the question to make it clearer.
– UnrealVillager
Jul 20 at 18:29
Is it clear to you that such a proper ideal does not exist if $I$ is principal? If yes, can you reduce to that case by localizing?
– Mohan
Jul 20 at 19:21
Well, looks like I just found a counterexample to that logic, Mohan. For $I = (5, 2sqrt-5)$ in $mathbbZ[sqrt-5]$ the equation seems to imply that $x, y in (sqrt-5)$. Unless I made a mistake somewhere in calculations. Could you please expand a little on your logic? Maybe I'm missing something... Thank you for your answer.
– UnrealVillager
Jul 20 at 19:33