Let $R$ be an integral domain. If $I cap J = IJ$ for all ideals $I,J$ of $R,$ then $R$ is a field.











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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







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    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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






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      asked yesterday









      Alexy Vincenzo

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          2 Answers
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          up vote
          4
          down vote



          accepted










          Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.






          share|cite|improve this answer





















          • 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


















          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.






          share|cite|improve this answer























          • 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













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          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 $I = J = Ra$.






          share|cite|improve this answer





















          • 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















          up vote
          4
          down vote



          accepted










          Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.






          share|cite|improve this answer





















          • 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













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.






          share|cite|improve this answer












          Hint: Let $ain Rsetminus{0}$ and consider $I = J = Ra$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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










          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.






          share|cite|improve this answer























          • 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

















          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.






          share|cite|improve this answer























          • 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















          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 20 hours ago









          user26857

          39.1k123882




          39.1k123882










          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




















          • 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




















           

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