Mixing
Let $R$ be an integral domain. If $I cap J = IJ$ for all ideals $I,J$ of $R,$ then $R$ is a field.
up vote
-1
down vote
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Let $R$ be an integral domain. If $I cap J = IJ$ for all ideals $I,J$ of $R,$ how do I show $R$ is a field? Hints will suffice. Thank you.
commutative-algebra
asked yesterday
Alexy Vincenzo
2,1673923
add a comment |
up vote
-1
down vote
favorite
Let $R$ be an integral domain. If $I cap J = IJ$ for all ideals $I,J$ of $R,$ how do I show $R$ is a field? Hints will suffice. Thank you.
commutative-algebra
asked yesterday
Alexy Vincenzo
2,1673923
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $R$ be an integral domain. If $I cap J = IJ$ for all ideals $I,J$ of $R,$ how do I show $R$ is a field? Hints will suffice. Thank you.
commutative-algebra
asked yesterday
Alexy Vincenzo
2,1673923
Let $R$ be an integral domain. If $I cap J = IJ$ for all ideals $I,J$ of $R,$ how do I show $R$ is a field? Hints will suffice. Thank you.
commutative-algebra
commutative-algebra
asked yesterday
Alexy Vincenzo
2,1673923
asked yesterday
Alexy Vincenzo
2,1673923
asked yesterday
Alexy Vincenzo
2,1673923
asked yesterday
Alexy Vincenzo
2,1673923
asked yesterday
Alexy Vincenzo
2,1673923
2,1673923
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.
answered yesterday
Claudius
3,8541516
Haha I got it from that thanks
– Richard Martin
yesterday
1
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
1
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
|
show 1 more comment
up vote
1
down vote
Arithmetically: since ideal intersection $= rm lcm$ for principal ideals, the hypothesis implies for all $,a,bneq 0,$ we have $,{rm lcm}(a,b) = ab,,$ so $,gcd(a,b)=1,,$ i.e. $,a,b$ have only unit common factors. Hence for $,b=a,$ the common factor $,a,$ of $,a,b$ is a unit, so $,aneq 0,Rightarrow, a,$ is a unit, so $R$ is a field.
edited 20 hours ago
user26857
39.1k123882
answered yesterday
Bill Dubuque
206k29189621
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.
answered yesterday
Claudius
3,8541516
Haha I got it from that thanks
– Richard Martin
yesterday
1
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
1
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
|
show 1 more comment
up vote
4
down vote
accepted
Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.
answered yesterday
Claudius
3,8541516
Haha I got it from that thanks
– Richard Martin
yesterday
1
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
1
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
|
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.
answered yesterday
Claudius
3,8541516
Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.
answered yesterday
Claudius
3,8541516
answered yesterday
Claudius
3,8541516
answered yesterday
Claudius
3,8541516
answered yesterday
Claudius
3,8541516
3,8541516
Haha I got it from that thanks
– Richard Martin
yesterday
1
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
1
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
|
show 1 more comment
Haha I got it from that thanks
– Richard Martin
yesterday
1
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
1
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
Haha I got it from that thanks
– Richard Martin
yesterday
Haha I got it from that thanks
– Richard Martin
yesterday
1
1
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Let $r in R.$ Then $r'a^{2} = ra, $ for some $r' in R.$ Since $R$ is integral domain, $r = r'a in Ra.$ So $1 in Ra$ and hence $a$ is unit.
– Alexy Vincenzo
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$?
– John Douma
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
@JohnDouma If $J=R$, then the product will be $Ra$, not $R$.
– Claudius
yesterday
1
1
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with.
– Claudius
yesterday
|
show 1 more comment
up vote
1
down vote
Arithmetically: since ideal intersection $= rm lcm$ for principal ideals, the hypothesis implies for all $,a,bneq 0,$ we have $,{rm lcm}(a,b) = ab,,$ so $,gcd(a,b)=1,,$ i.e. $,a,b$ have only unit common factors. Hence for $,b=a,$ the common factor $,a,$ of $,a,b$ is a unit, so $,aneq 0,Rightarrow, a,$ is a unit, so $R$ is a field.
edited 20 hours ago
user26857
39.1k123882
answered yesterday
Bill Dubuque
206k29189621
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
add a comment |
up vote
1
down vote
Arithmetically: since ideal intersection $= rm lcm$ for principal ideals, the hypothesis implies for all $,a,bneq 0,$ we have $,{rm lcm}(a,b) = ab,,$ so $,gcd(a,b)=1,,$ i.e. $,a,b$ have only unit common factors. Hence for $,b=a,$ the common factor $,a,$ of $,a,b$ is a unit, so $,aneq 0,Rightarrow, a,$ is a unit, so $R$ is a field.
edited 20 hours ago
user26857
39.1k123882
answered yesterday
Bill Dubuque
206k29189621
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
Arithmetically: since ideal intersection $= rm lcm$ for principal ideals, the hypothesis implies for all $,a,bneq 0,$ we have $,{rm lcm}(a,b) = ab,,$ so $,gcd(a,b)=1,,$ i.e. $,a,b$ have only unit common factors. Hence for $,b=a,$ the common factor $,a,$ of $,a,b$ is a unit, so $,aneq 0,Rightarrow, a,$ is a unit, so $R$ is a field.
edited 20 hours ago
user26857
39.1k123882
answered yesterday
Bill Dubuque
206k29189621
Arithmetically: since ideal intersection $= rm lcm$ for principal ideals, the hypothesis implies for all $,a,bneq 0,$ we have $,{rm lcm}(a,b) = ab,,$ so $,gcd(a,b)=1,,$ i.e. $,a,b$ have only unit common factors. Hence for $,b=a,$ the common factor $,a,$ of $,a,b$ is a unit, so $,aneq 0,Rightarrow, a,$ is a unit, so $R$ is a field.
edited 20 hours ago
user26857
39.1k123882
answered yesterday
Bill Dubuque
206k29189621
edited 20 hours ago
user26857
39.1k123882
edited 20 hours ago
user26857
39.1k123882
edited 20 hours ago
user26857
39.1k123882
39.1k123882
answered yesterday
Bill Dubuque
206k29189621
answered yesterday
Bill Dubuque
206k29189621
answered yesterday
Bill Dubuque
206k29189621
206k29189621
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
add a comment |
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
Essentially we used: $,0neq a$ is a unit $!iff! gcd(a,x) = 1!iff! {rm lcm}(a,x) = ax, $ for all $,xneq 0. $ Recall that in general domains $gcd$ and $rm lcm$ are defined only up to unit factors (associates).
– Bill Dubuque
yesterday
add a comment |
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