Determine the maximal ideals of $mathbb{R}^2$.












1















Determine the maximal ideals of $mathbb{R}^2$.




Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?










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  • I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
    – Ian Coley
    Oct 20 '13 at 5:12










  • No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
    – anon
    Oct 20 '13 at 5:14












  • I guess OP means the product ring structure where addition and multiplication are defined componentwise.
    – user43208
    Oct 20 '13 at 5:29










  • I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
    – DonAntonio
    Oct 20 '13 at 10:55


















1















Determine the maximal ideals of $mathbb{R}^2$.




Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?










share|cite|improve this question
























  • I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
    – Ian Coley
    Oct 20 '13 at 5:12










  • No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
    – anon
    Oct 20 '13 at 5:14












  • I guess OP means the product ring structure where addition and multiplication are defined componentwise.
    – user43208
    Oct 20 '13 at 5:29










  • I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
    – DonAntonio
    Oct 20 '13 at 10:55
















1












1








1








Determine the maximal ideals of $mathbb{R}^2$.




Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?










share|cite|improve this question
















Determine the maximal ideals of $mathbb{R}^2$.




Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?







abstract-algebra ring-theory






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share|cite|improve this question













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share|cite|improve this question








edited Dec 26 '18 at 21:09









user26857

39.2k123983




39.2k123983










asked Oct 20 '13 at 5:09









sarah

2111834




2111834












  • I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
    – Ian Coley
    Oct 20 '13 at 5:12










  • No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
    – anon
    Oct 20 '13 at 5:14












  • I guess OP means the product ring structure where addition and multiplication are defined componentwise.
    – user43208
    Oct 20 '13 at 5:29










  • I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
    – DonAntonio
    Oct 20 '13 at 10:55




















  • I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
    – Ian Coley
    Oct 20 '13 at 5:12










  • No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
    – anon
    Oct 20 '13 at 5:14












  • I guess OP means the product ring structure where addition and multiplication are defined componentwise.
    – user43208
    Oct 20 '13 at 5:29










  • I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
    – DonAntonio
    Oct 20 '13 at 10:55


















I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12




I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12












No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14






No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14














I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29




I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29












I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55






I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55












1 Answer
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Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then



$$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$



And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.





On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:




  • If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.


  • If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.


  • Likewise for $(a, 0)$.







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    1 Answer
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    1 Answer
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    Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then



    $$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$



    And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.





    On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:




    • If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.


    • If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.


    • Likewise for $(a, 0)$.







    share|cite|improve this answer


























      3














      Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then



      $$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$



      And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.





      On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:




      • If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.


      • If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.


      • Likewise for $(a, 0)$.







      share|cite|improve this answer
























        3












        3








        3






        Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then



        $$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$



        And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.





        On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:




        • If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.


        • If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.


        • Likewise for $(a, 0)$.







        share|cite|improve this answer












        Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then



        $$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$



        And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.





        On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:




        • If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.


        • If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.


        • Likewise for $(a, 0)$.








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        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 20 '13 at 5:14







        user61527





































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