Show that any prime ideal from such a ring is maximal.












9















Let R be a commutative ring with an identity such that for all $rin$ R, there exists some $n>1$ such that $r^n = r$. Show that any prime ideal is maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.)




Any hints?






abstract-algebra ring-theory maximal-and-prime-ideals







share|cite|improve this question




















  • 3




    You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want.
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 7:42






  • 1




    Yes, you are correct, I apologize.
    – user41916
    Oct 11 '12 at 8:04






  • 1




    Oh, there is no need to apologize! Just keeping things precise is good for the universe :-)
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 8:06
















9















Let R be a commutative ring with an identity such that for all $rin$ R, there exists some $n>1$ such that $r^n = r$. Show that any prime ideal is maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.)




Any hints?






abstract-algebra ring-theory maximal-and-prime-ideals







share|cite|improve this question




















  • 3




    You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want.
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 7:42






  • 1




    Yes, you are correct, I apologize.
    – user41916
    Oct 11 '12 at 8:04






  • 1




    Oh, there is no need to apologize! Just keeping things precise is good for the universe :-)
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 8:06














9












9








9


2






Let R be a commutative ring with an identity such that for all $rin$ R, there exists some $n>1$ such that $r^n = r$. Show that any prime ideal is maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.)




Any hints?






abstract-algebra ring-theory maximal-and-prime-ideals







share|cite|improve this question
















Let R be a commutative ring with an identity such that for all $rin$ R, there exists some $n>1$ such that $r^n = r$. Show that any prime ideal is maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.)




Any hints?






abstract-algebra ring-theory maximal-and-prime-ideals




abstract-algebra ring-theory maximal-and-prime-ideals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '15 at 22:06









user26857

39.2k123882




39.2k123882










asked Oct 11 '12 at 7:06







user41916















  • 3




    You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want.
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 7:42






  • 1




    Yes, you are correct, I apologize.
    – user41916
    Oct 11 '12 at 8:04






  • 1




    Oh, there is no need to apologize! Just keeping things precise is good for the universe :-)
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 8:06














  • 3




    You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want.
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 7:42






  • 1




    Yes, you are correct, I apologize.
    – user41916
    Oct 11 '12 at 8:04






  • 1




    Oh, there is no need to apologize! Just keeping things precise is good for the universe :-)
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 8:06








3




3




You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want.
– Mariano Suárez-Álvarez
Oct 11 '12 at 7:42




You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want.
– Mariano Suárez-Álvarez
Oct 11 '12 at 7:42




1




1




Yes, you are correct, I apologize.
– user41916
Oct 11 '12 at 8:04




Yes, you are correct, I apologize.
– user41916
Oct 11 '12 at 8:04




1




1




Oh, there is no need to apologize! Just keeping things precise is good for the universe :-)
– Mariano Suárez-Álvarez
Oct 11 '12 at 8:06




Oh, there is no need to apologize! Just keeping things precise is good for the universe :-)
– Mariano Suárez-Álvarez
Oct 11 '12 at 8:06










3 Answers
3






active

oldest

votes


















5














Hint: Reduce to the case that $R$ is an integral domain satisfying $forall r exists n (r^n=r)$, and show that $R$ is a field.






share|cite|improve this answer



















  • 4




    Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
    – Mariano Suárez-Álvarez
    Oct 11 '12 at 7:41



















3














Let $,D,$ be any integer domain and let $,din D,$ be s.t. $,d^n=d,,,,,1<ninBbb N,$ , then:



$$d^n=dLongrightarrow d(d^{n-1}-1)=0Longleftrightarrow d=0,,,text{or},,,d^{n-1}=1,$$



so if $,d,$ is not zero then it must be a unit.



$$-------o----------o---------o---$$



In our case: let $,Ileq R,$ be a prime ideal and let $, rin Rsetminus I,$, then:



$$exists,ninBbb N,,s.t.,,r^n=rLongrightarrow left(r+Iright)^n=r^n+I=r+Iin R/I$$



Now use the first part with $,D:=R/I,,,,,,d=r+I,$ and deduce $,R/I,$ is actually a field...






share|cite|improve this answer































    0














    Proof: Let $mathfrak{p}$ be a prime ideal in $A$. Note that since $mathfrak{p}$ is a prime ideal we have that $A / mathfrak{p}$ is an integral domain.



    We will show first that $A / mathfrak{p}$ is a field. Choose $x + mathfrak{p} in A / mathfrak{p}$ such that $x + p neq 0_{A/ mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+mathfrak{p})^n = x^n + mathfrak{p} = x + mathfrak{p}.$$
    We then have that
    begin{align*}
    (x+mathfrak{p})^n = x + mathfrak{p} &implies (x+mathfrak{p})^n - (x+mathfrak{p}) = 0_{A/mathfrak{p}} \
    &implies (x + mathfrak{p})left((x+mathfrak{p})^{n-1} - 1_{A/mathfrak{p}}right) = 0_{A/mathfrak{p}} \
    &implies x+ mathfrak{p} = 0_{A/mathfrak{p}} text{or} (x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}
    end{align*}
    where the last implication follows from the fact that $A / mathfrak{p}$ is an integral domain. Now since $x + mathfrak{p} neq 0_{A/mathfrak{p}}$ we must have that $(x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}$. But then we have $(x+ mathfrak{p})(x+mathfrak{p})^{n-2} = 1_{A/mathfrak{p}}$ and so $(x+ mathfrak{p})^{n-2}$ is an inverse of $x + mathfrak{p}$.



    Note that since $n > 1$ it follows that $(x+ mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ mathfrak{p}$ has an inverse and so $A / mathfrak{p}$ is a field. Hence $mathfrak{p}$ is maximal in $A$. $square$






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      Hint: Reduce to the case that $R$ is an integral domain satisfying $forall r exists n (r^n=r)$, and show that $R$ is a field.






      share|cite|improve this answer



















      • 4




        Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
        – Mariano Suárez-Álvarez
        Oct 11 '12 at 7:41
















      5














      Hint: Reduce to the case that $R$ is an integral domain satisfying $forall r exists n (r^n=r)$, and show that $R$ is a field.






      share|cite|improve this answer



















      • 4




        Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
        – Mariano Suárez-Álvarez
        Oct 11 '12 at 7:41














      5












      5








      5






      Hint: Reduce to the case that $R$ is an integral domain satisfying $forall r exists n (r^n=r)$, and show that $R$ is a field.






      share|cite|improve this answer














      Hint: Reduce to the case that $R$ is an integral domain satisfying $forall r exists n (r^n=r)$, and show that $R$ is a field.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Oct 11 '12 at 7:28

























      answered Oct 11 '12 at 7:08









      Martin Brandenburg

      107k13157326




      107k13157326








      • 4




        Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
        – Mariano Suárez-Álvarez
        Oct 11 '12 at 7:41














      • 4




        Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
        – Mariano Suárez-Álvarez
        Oct 11 '12 at 7:41








      4




      4




      Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
      – Mariano Suárez-Álvarez
      Oct 11 '12 at 7:41




      Slightly more verbosely: in order to prove that a prime ideal $mathfrak psubseteq R$ is maximal, it is enough to show that the quotient $R/mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c
      – Mariano Suárez-Álvarez
      Oct 11 '12 at 7:41











      3














      Let $,D,$ be any integer domain and let $,din D,$ be s.t. $,d^n=d,,,,,1<ninBbb N,$ , then:



      $$d^n=dLongrightarrow d(d^{n-1}-1)=0Longleftrightarrow d=0,,,text{or},,,d^{n-1}=1,$$



      so if $,d,$ is not zero then it must be a unit.



      $$-------o----------o---------o---$$



      In our case: let $,Ileq R,$ be a prime ideal and let $, rin Rsetminus I,$, then:



      $$exists,ninBbb N,,s.t.,,r^n=rLongrightarrow left(r+Iright)^n=r^n+I=r+Iin R/I$$



      Now use the first part with $,D:=R/I,,,,,,d=r+I,$ and deduce $,R/I,$ is actually a field...






      share|cite|improve this answer




























        3














        Let $,D,$ be any integer domain and let $,din D,$ be s.t. $,d^n=d,,,,,1<ninBbb N,$ , then:



        $$d^n=dLongrightarrow d(d^{n-1}-1)=0Longleftrightarrow d=0,,,text{or},,,d^{n-1}=1,$$



        so if $,d,$ is not zero then it must be a unit.



        $$-------o----------o---------o---$$



        In our case: let $,Ileq R,$ be a prime ideal and let $, rin Rsetminus I,$, then:



        $$exists,ninBbb N,,s.t.,,r^n=rLongrightarrow left(r+Iright)^n=r^n+I=r+Iin R/I$$



        Now use the first part with $,D:=R/I,,,,,,d=r+I,$ and deduce $,R/I,$ is actually a field...






        share|cite|improve this answer


























          3












          3








          3






          Let $,D,$ be any integer domain and let $,din D,$ be s.t. $,d^n=d,,,,,1<ninBbb N,$ , then:



          $$d^n=dLongrightarrow d(d^{n-1}-1)=0Longleftrightarrow d=0,,,text{or},,,d^{n-1}=1,$$



          so if $,d,$ is not zero then it must be a unit.



          $$-------o----------o---------o---$$



          In our case: let $,Ileq R,$ be a prime ideal and let $, rin Rsetminus I,$, then:



          $$exists,ninBbb N,,s.t.,,r^n=rLongrightarrow left(r+Iright)^n=r^n+I=r+Iin R/I$$



          Now use the first part with $,D:=R/I,,,,,,d=r+I,$ and deduce $,R/I,$ is actually a field...






          share|cite|improve this answer














          Let $,D,$ be any integer domain and let $,din D,$ be s.t. $,d^n=d,,,,,1<ninBbb N,$ , then:



          $$d^n=dLongrightarrow d(d^{n-1}-1)=0Longleftrightarrow d=0,,,text{or},,,d^{n-1}=1,$$



          so if $,d,$ is not zero then it must be a unit.



          $$-------o----------o---------o---$$



          In our case: let $,Ileq R,$ be a prime ideal and let $, rin Rsetminus I,$, then:



          $$exists,ninBbb N,,s.t.,,r^n=rLongrightarrow left(r+Iright)^n=r^n+I=r+Iin R/I$$



          Now use the first part with $,D:=R/I,,,,,,d=r+I,$ and deduce $,R/I,$ is actually a field...







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '15 at 23:21









          user26857

          39.2k123882




          39.2k123882










          answered Oct 11 '12 at 11:19









          DonAntonio

          176k1491225




          176k1491225























              0














              Proof: Let $mathfrak{p}$ be a prime ideal in $A$. Note that since $mathfrak{p}$ is a prime ideal we have that $A / mathfrak{p}$ is an integral domain.



              We will show first that $A / mathfrak{p}$ is a field. Choose $x + mathfrak{p} in A / mathfrak{p}$ such that $x + p neq 0_{A/ mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+mathfrak{p})^n = x^n + mathfrak{p} = x + mathfrak{p}.$$
              We then have that
              begin{align*}
              (x+mathfrak{p})^n = x + mathfrak{p} &implies (x+mathfrak{p})^n - (x+mathfrak{p}) = 0_{A/mathfrak{p}} \
              &implies (x + mathfrak{p})left((x+mathfrak{p})^{n-1} - 1_{A/mathfrak{p}}right) = 0_{A/mathfrak{p}} \
              &implies x+ mathfrak{p} = 0_{A/mathfrak{p}} text{or} (x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}
              end{align*}
              where the last implication follows from the fact that $A / mathfrak{p}$ is an integral domain. Now since $x + mathfrak{p} neq 0_{A/mathfrak{p}}$ we must have that $(x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}$. But then we have $(x+ mathfrak{p})(x+mathfrak{p})^{n-2} = 1_{A/mathfrak{p}}$ and so $(x+ mathfrak{p})^{n-2}$ is an inverse of $x + mathfrak{p}$.



              Note that since $n > 1$ it follows that $(x+ mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ mathfrak{p}$ has an inverse and so $A / mathfrak{p}$ is a field. Hence $mathfrak{p}$ is maximal in $A$. $square$






              share|cite|improve this answer


























                0














                Proof: Let $mathfrak{p}$ be a prime ideal in $A$. Note that since $mathfrak{p}$ is a prime ideal we have that $A / mathfrak{p}$ is an integral domain.



                We will show first that $A / mathfrak{p}$ is a field. Choose $x + mathfrak{p} in A / mathfrak{p}$ such that $x + p neq 0_{A/ mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+mathfrak{p})^n = x^n + mathfrak{p} = x + mathfrak{p}.$$
                We then have that
                begin{align*}
                (x+mathfrak{p})^n = x + mathfrak{p} &implies (x+mathfrak{p})^n - (x+mathfrak{p}) = 0_{A/mathfrak{p}} \
                &implies (x + mathfrak{p})left((x+mathfrak{p})^{n-1} - 1_{A/mathfrak{p}}right) = 0_{A/mathfrak{p}} \
                &implies x+ mathfrak{p} = 0_{A/mathfrak{p}} text{or} (x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}
                end{align*}
                where the last implication follows from the fact that $A / mathfrak{p}$ is an integral domain. Now since $x + mathfrak{p} neq 0_{A/mathfrak{p}}$ we must have that $(x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}$. But then we have $(x+ mathfrak{p})(x+mathfrak{p})^{n-2} = 1_{A/mathfrak{p}}$ and so $(x+ mathfrak{p})^{n-2}$ is an inverse of $x + mathfrak{p}$.



                Note that since $n > 1$ it follows that $(x+ mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ mathfrak{p}$ has an inverse and so $A / mathfrak{p}$ is a field. Hence $mathfrak{p}$ is maximal in $A$. $square$






                share|cite|improve this answer
























                  0












                  0








                  0






                  Proof: Let $mathfrak{p}$ be a prime ideal in $A$. Note that since $mathfrak{p}$ is a prime ideal we have that $A / mathfrak{p}$ is an integral domain.



                  We will show first that $A / mathfrak{p}$ is a field. Choose $x + mathfrak{p} in A / mathfrak{p}$ such that $x + p neq 0_{A/ mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+mathfrak{p})^n = x^n + mathfrak{p} = x + mathfrak{p}.$$
                  We then have that
                  begin{align*}
                  (x+mathfrak{p})^n = x + mathfrak{p} &implies (x+mathfrak{p})^n - (x+mathfrak{p}) = 0_{A/mathfrak{p}} \
                  &implies (x + mathfrak{p})left((x+mathfrak{p})^{n-1} - 1_{A/mathfrak{p}}right) = 0_{A/mathfrak{p}} \
                  &implies x+ mathfrak{p} = 0_{A/mathfrak{p}} text{or} (x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}
                  end{align*}
                  where the last implication follows from the fact that $A / mathfrak{p}$ is an integral domain. Now since $x + mathfrak{p} neq 0_{A/mathfrak{p}}$ we must have that $(x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}$. But then we have $(x+ mathfrak{p})(x+mathfrak{p})^{n-2} = 1_{A/mathfrak{p}}$ and so $(x+ mathfrak{p})^{n-2}$ is an inverse of $x + mathfrak{p}$.



                  Note that since $n > 1$ it follows that $(x+ mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ mathfrak{p}$ has an inverse and so $A / mathfrak{p}$ is a field. Hence $mathfrak{p}$ is maximal in $A$. $square$






                  share|cite|improve this answer












                  Proof: Let $mathfrak{p}$ be a prime ideal in $A$. Note that since $mathfrak{p}$ is a prime ideal we have that $A / mathfrak{p}$ is an integral domain.



                  We will show first that $A / mathfrak{p}$ is a field. Choose $x + mathfrak{p} in A / mathfrak{p}$ such that $x + p neq 0_{A/ mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+mathfrak{p})^n = x^n + mathfrak{p} = x + mathfrak{p}.$$
                  We then have that
                  begin{align*}
                  (x+mathfrak{p})^n = x + mathfrak{p} &implies (x+mathfrak{p})^n - (x+mathfrak{p}) = 0_{A/mathfrak{p}} \
                  &implies (x + mathfrak{p})left((x+mathfrak{p})^{n-1} - 1_{A/mathfrak{p}}right) = 0_{A/mathfrak{p}} \
                  &implies x+ mathfrak{p} = 0_{A/mathfrak{p}} text{or} (x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}
                  end{align*}
                  where the last implication follows from the fact that $A / mathfrak{p}$ is an integral domain. Now since $x + mathfrak{p} neq 0_{A/mathfrak{p}}$ we must have that $(x+mathfrak{p})^{n-1} = 1_{A/mathfrak{p}}$. But then we have $(x+ mathfrak{p})(x+mathfrak{p})^{n-2} = 1_{A/mathfrak{p}}$ and so $(x+ mathfrak{p})^{n-2}$ is an inverse of $x + mathfrak{p}$.



                  Note that since $n > 1$ it follows that $(x+ mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ mathfrak{p}$ has an inverse and so $A / mathfrak{p}$ is a field. Hence $mathfrak{p}$ is maximal in $A$. $square$







                  share|cite|improve this answer












                  share|cite|improve this answer



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                  answered Dec 13 at 2:33









                  Perturbative

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                  4,04511449






























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                      Aardman Animations

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                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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                      Are they similar matrix

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                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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