A ring which contains a nonprime maximal ideal












4














I found this observation on my book, but I did not understand it well.



Let $R$ a commutative ring without identity, but having a single generator, then $R$ contains nonprime maximal ideal.



We suppose that $R=(a)$. First observe that the principal ideal $(a^2)$ is a proper ideal of $R$, since the generator $anotin (a^2)$. Indeed, were $a$ in $(a^2)$, we could write $a=ra^2+na^2$ for same $rin R$ and $nin mathbb{Z}$;



Why, in this case, the element $e=ra+na$ is a multiplicative identity for $R$? Therefore, since $e$ is a identity of $R$, they are violated the hypothesis.



Now, since $Rne (a^2)$, exist a maximal ideal $M$ of $R$ with $(a^2)subseteq M$. However $M$ is not a prime ideal, as can be seen by considering the product of elements in te complement of $M$ ($r,snotin M$), the product $rsin (a^2)subseteq M$. (Why this is the negation of a definition of prime ideal?)



Anyone could explain to me how things are?



Thanks!










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    4














    I found this observation on my book, but I did not understand it well.



    Let $R$ a commutative ring without identity, but having a single generator, then $R$ contains nonprime maximal ideal.



    We suppose that $R=(a)$. First observe that the principal ideal $(a^2)$ is a proper ideal of $R$, since the generator $anotin (a^2)$. Indeed, were $a$ in $(a^2)$, we could write $a=ra^2+na^2$ for same $rin R$ and $nin mathbb{Z}$;



    Why, in this case, the element $e=ra+na$ is a multiplicative identity for $R$? Therefore, since $e$ is a identity of $R$, they are violated the hypothesis.



    Now, since $Rne (a^2)$, exist a maximal ideal $M$ of $R$ with $(a^2)subseteq M$. However $M$ is not a prime ideal, as can be seen by considering the product of elements in te complement of $M$ ($r,snotin M$), the product $rsin (a^2)subseteq M$. (Why this is the negation of a definition of prime ideal?)



    Anyone could explain to me how things are?



    Thanks!










    share|cite|improve this question

























      4












      4








      4







      I found this observation on my book, but I did not understand it well.



      Let $R$ a commutative ring without identity, but having a single generator, then $R$ contains nonprime maximal ideal.



      We suppose that $R=(a)$. First observe that the principal ideal $(a^2)$ is a proper ideal of $R$, since the generator $anotin (a^2)$. Indeed, were $a$ in $(a^2)$, we could write $a=ra^2+na^2$ for same $rin R$ and $nin mathbb{Z}$;



      Why, in this case, the element $e=ra+na$ is a multiplicative identity for $R$? Therefore, since $e$ is a identity of $R$, they are violated the hypothesis.



      Now, since $Rne (a^2)$, exist a maximal ideal $M$ of $R$ with $(a^2)subseteq M$. However $M$ is not a prime ideal, as can be seen by considering the product of elements in te complement of $M$ ($r,snotin M$), the product $rsin (a^2)subseteq M$. (Why this is the negation of a definition of prime ideal?)



      Anyone could explain to me how things are?



      Thanks!










      share|cite|improve this question













      I found this observation on my book, but I did not understand it well.



      Let $R$ a commutative ring without identity, but having a single generator, then $R$ contains nonprime maximal ideal.



      We suppose that $R=(a)$. First observe that the principal ideal $(a^2)$ is a proper ideal of $R$, since the generator $anotin (a^2)$. Indeed, were $a$ in $(a^2)$, we could write $a=ra^2+na^2$ for same $rin R$ and $nin mathbb{Z}$;



      Why, in this case, the element $e=ra+na$ is a multiplicative identity for $R$? Therefore, since $e$ is a identity of $R$, they are violated the hypothesis.



      Now, since $Rne (a^2)$, exist a maximal ideal $M$ of $R$ with $(a^2)subseteq M$. However $M$ is not a prime ideal, as can be seen by considering the product of elements in te complement of $M$ ($r,snotin M$), the product $rsin (a^2)subseteq M$. (Why this is the negation of a definition of prime ideal?)



      Anyone could explain to me how things are?



      Thanks!







      abstract-algebra proof-explanation






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      asked Dec 3 '18 at 15:00









      Jack J.

      4421419




      4421419






















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          Since every element of $R$ can be written as $ra+na$ for some $rin R$ and $ninmathbb{Z}$, if we find $ein R$ such that $ae=a$, then $e$ will be the identity of $R$.



          If $ain(a^2)$, then $a=ra^2+na^2$, for some $rin R$ and $ninmathbb{Z}$. Set $e=ra+na$; then $ae=a$.



          Now take an ideal $M$ of $R$ maximal with respect to the property that





          1. $(a^2)subseteq M$;


          2. $anotin M$.


          Such a maximal element exists by Zorn's lemma: the union of a chain of ideals containing $(a^2)$ and not containing $a$ is again an ideal satisfying the same property. Clearly such an ideal is not prime, because $anotin M$, but $aa=a^2in M$.



          Let's prove that $M$ is maximal. If $I$ is an ideal of $R$ properly containing $M$, then $I$ contains $(a^2)$, so by the choice of $M$ we must have $ain I$. Hence $I=R$.



          Note: the fact that $R=(a)$ is crucial. A ring without an identity can fail to have maximal ideals at all.






          share|cite|improve this answer





























            3














            Suppose that $a=ra^2+na^2$. Then, $a(ra+na)=a$, so $ae=a$. Then, for any $xin R$, $x=ta+ma$ for some $tin R$, $min mathbb{Z}$. So $xe= (ta+ma)e= tae+mae= ta+ma=x$. Therefore, $e$ is the identity element.



            To prove that $M$ is not a prime ideal, it is enough to check that there are two ideals $A$, $B$ with $ABsubseteq M$, and $A$ and $B$ are not contained in $M$. By taking $A=B=(a)$, $AB=(a)(a)subseteq (a^2)subseteq M$, but $(a)=R$ is not contained in $M$ because it is a maximal ideal, so by definition $Mneq R$.






            share|cite|improve this answer































              3














              $4mathbb{Z}$ in $2mathbb{Z}$. Note that $4 = 2 cdot 2 in 4mathbb{Z}$, but $2 not in 4 mathbb{Z}$. There is no ideal between $4mathbb{Z}$ and $2 mathbb{Z}$.



              (Note that this is a concrete example of your more abstract example.)






              share|cite|improve this answer





























                2














                For the first part, you have the idea, but if $ain (a^2)$, you can write $a=P(a^2) $ where $P$ is a polynomial. Let $M$ be a maximal ideal containing $(a^2)$, $a.a=a^2$ is in $M$ but $a$ is not in $ M$. This implies that $M$ is not prime.






                share|cite|improve this answer































                  2














                  For the second question, the definition of a prime ideal is that $I$ is prime if for any $a, b in R$ such that $ab in I$ then either $a in I$ or $b in I$. So if we can find $a, b in R setminus I$ with $ab in I$ then this would show that $I$ was not prime.



                  Now in the situation of the question we have a maximal ideal $M$ of $R = (a)$ with $(a^2) subseteq M$. Let $r, s in R setminus M$. Then $r = ta + ma$ and $s = ua + na$ for some $t, u in R$, $m, n in Bbb Z$, so $rs = (tu + tn + mu + mn)a^2 in (a^2) subset M$. Thus $rs in M$ but $r notin M$, $s notin M$, as required to show that $M$ was not prime.






                  share|cite|improve this answer





















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                    5 Answers
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                    5 Answers
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                    3














                    Since every element of $R$ can be written as $ra+na$ for some $rin R$ and $ninmathbb{Z}$, if we find $ein R$ such that $ae=a$, then $e$ will be the identity of $R$.



                    If $ain(a^2)$, then $a=ra^2+na^2$, for some $rin R$ and $ninmathbb{Z}$. Set $e=ra+na$; then $ae=a$.



                    Now take an ideal $M$ of $R$ maximal with respect to the property that





                    1. $(a^2)subseteq M$;


                    2. $anotin M$.


                    Such a maximal element exists by Zorn's lemma: the union of a chain of ideals containing $(a^2)$ and not containing $a$ is again an ideal satisfying the same property. Clearly such an ideal is not prime, because $anotin M$, but $aa=a^2in M$.



                    Let's prove that $M$ is maximal. If $I$ is an ideal of $R$ properly containing $M$, then $I$ contains $(a^2)$, so by the choice of $M$ we must have $ain I$. Hence $I=R$.



                    Note: the fact that $R=(a)$ is crucial. A ring without an identity can fail to have maximal ideals at all.






                    share|cite|improve this answer


























                      3














                      Since every element of $R$ can be written as $ra+na$ for some $rin R$ and $ninmathbb{Z}$, if we find $ein R$ such that $ae=a$, then $e$ will be the identity of $R$.



                      If $ain(a^2)$, then $a=ra^2+na^2$, for some $rin R$ and $ninmathbb{Z}$. Set $e=ra+na$; then $ae=a$.



                      Now take an ideal $M$ of $R$ maximal with respect to the property that





                      1. $(a^2)subseteq M$;


                      2. $anotin M$.


                      Such a maximal element exists by Zorn's lemma: the union of a chain of ideals containing $(a^2)$ and not containing $a$ is again an ideal satisfying the same property. Clearly such an ideal is not prime, because $anotin M$, but $aa=a^2in M$.



                      Let's prove that $M$ is maximal. If $I$ is an ideal of $R$ properly containing $M$, then $I$ contains $(a^2)$, so by the choice of $M$ we must have $ain I$. Hence $I=R$.



                      Note: the fact that $R=(a)$ is crucial. A ring without an identity can fail to have maximal ideals at all.






                      share|cite|improve this answer
























                        3












                        3








                        3






                        Since every element of $R$ can be written as $ra+na$ for some $rin R$ and $ninmathbb{Z}$, if we find $ein R$ such that $ae=a$, then $e$ will be the identity of $R$.



                        If $ain(a^2)$, then $a=ra^2+na^2$, for some $rin R$ and $ninmathbb{Z}$. Set $e=ra+na$; then $ae=a$.



                        Now take an ideal $M$ of $R$ maximal with respect to the property that





                        1. $(a^2)subseteq M$;


                        2. $anotin M$.


                        Such a maximal element exists by Zorn's lemma: the union of a chain of ideals containing $(a^2)$ and not containing $a$ is again an ideal satisfying the same property. Clearly such an ideal is not prime, because $anotin M$, but $aa=a^2in M$.



                        Let's prove that $M$ is maximal. If $I$ is an ideal of $R$ properly containing $M$, then $I$ contains $(a^2)$, so by the choice of $M$ we must have $ain I$. Hence $I=R$.



                        Note: the fact that $R=(a)$ is crucial. A ring without an identity can fail to have maximal ideals at all.






                        share|cite|improve this answer












                        Since every element of $R$ can be written as $ra+na$ for some $rin R$ and $ninmathbb{Z}$, if we find $ein R$ such that $ae=a$, then $e$ will be the identity of $R$.



                        If $ain(a^2)$, then $a=ra^2+na^2$, for some $rin R$ and $ninmathbb{Z}$. Set $e=ra+na$; then $ae=a$.



                        Now take an ideal $M$ of $R$ maximal with respect to the property that





                        1. $(a^2)subseteq M$;


                        2. $anotin M$.


                        Such a maximal element exists by Zorn's lemma: the union of a chain of ideals containing $(a^2)$ and not containing $a$ is again an ideal satisfying the same property. Clearly such an ideal is not prime, because $anotin M$, but $aa=a^2in M$.



                        Let's prove that $M$ is maximal. If $I$ is an ideal of $R$ properly containing $M$, then $I$ contains $(a^2)$, so by the choice of $M$ we must have $ain I$. Hence $I=R$.



                        Note: the fact that $R=(a)$ is crucial. A ring without an identity can fail to have maximal ideals at all.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 3 '18 at 15:47









                        egreg

                        178k1484201




                        178k1484201























                            3














                            Suppose that $a=ra^2+na^2$. Then, $a(ra+na)=a$, so $ae=a$. Then, for any $xin R$, $x=ta+ma$ for some $tin R$, $min mathbb{Z}$. So $xe= (ta+ma)e= tae+mae= ta+ma=x$. Therefore, $e$ is the identity element.



                            To prove that $M$ is not a prime ideal, it is enough to check that there are two ideals $A$, $B$ with $ABsubseteq M$, and $A$ and $B$ are not contained in $M$. By taking $A=B=(a)$, $AB=(a)(a)subseteq (a^2)subseteq M$, but $(a)=R$ is not contained in $M$ because it is a maximal ideal, so by definition $Mneq R$.






                            share|cite|improve this answer




























                              3














                              Suppose that $a=ra^2+na^2$. Then, $a(ra+na)=a$, so $ae=a$. Then, for any $xin R$, $x=ta+ma$ for some $tin R$, $min mathbb{Z}$. So $xe= (ta+ma)e= tae+mae= ta+ma=x$. Therefore, $e$ is the identity element.



                              To prove that $M$ is not a prime ideal, it is enough to check that there are two ideals $A$, $B$ with $ABsubseteq M$, and $A$ and $B$ are not contained in $M$. By taking $A=B=(a)$, $AB=(a)(a)subseteq (a^2)subseteq M$, but $(a)=R$ is not contained in $M$ because it is a maximal ideal, so by definition $Mneq R$.






                              share|cite|improve this answer


























                                3












                                3








                                3






                                Suppose that $a=ra^2+na^2$. Then, $a(ra+na)=a$, so $ae=a$. Then, for any $xin R$, $x=ta+ma$ for some $tin R$, $min mathbb{Z}$. So $xe= (ta+ma)e= tae+mae= ta+ma=x$. Therefore, $e$ is the identity element.



                                To prove that $M$ is not a prime ideal, it is enough to check that there are two ideals $A$, $B$ with $ABsubseteq M$, and $A$ and $B$ are not contained in $M$. By taking $A=B=(a)$, $AB=(a)(a)subseteq (a^2)subseteq M$, but $(a)=R$ is not contained in $M$ because it is a maximal ideal, so by definition $Mneq R$.






                                share|cite|improve this answer














                                Suppose that $a=ra^2+na^2$. Then, $a(ra+na)=a$, so $ae=a$. Then, for any $xin R$, $x=ta+ma$ for some $tin R$, $min mathbb{Z}$. So $xe= (ta+ma)e= tae+mae= ta+ma=x$. Therefore, $e$ is the identity element.



                                To prove that $M$ is not a prime ideal, it is enough to check that there are two ideals $A$, $B$ with $ABsubseteq M$, and $A$ and $B$ are not contained in $M$. By taking $A=B=(a)$, $AB=(a)(a)subseteq (a^2)subseteq M$, but $(a)=R$ is not contained in $M$ because it is a maximal ideal, so by definition $Mneq R$.







                                share|cite|improve this answer














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                                share|cite|improve this answer








                                edited Dec 3 '18 at 15:24

























                                answered Dec 3 '18 at 15:18









                                Karen

                                855




                                855























                                    3














                                    $4mathbb{Z}$ in $2mathbb{Z}$. Note that $4 = 2 cdot 2 in 4mathbb{Z}$, but $2 not in 4 mathbb{Z}$. There is no ideal between $4mathbb{Z}$ and $2 mathbb{Z}$.



                                    (Note that this is a concrete example of your more abstract example.)






                                    share|cite|improve this answer


























                                      3














                                      $4mathbb{Z}$ in $2mathbb{Z}$. Note that $4 = 2 cdot 2 in 4mathbb{Z}$, but $2 not in 4 mathbb{Z}$. There is no ideal between $4mathbb{Z}$ and $2 mathbb{Z}$.



                                      (Note that this is a concrete example of your more abstract example.)






                                      share|cite|improve this answer
























                                        3












                                        3








                                        3






                                        $4mathbb{Z}$ in $2mathbb{Z}$. Note that $4 = 2 cdot 2 in 4mathbb{Z}$, but $2 not in 4 mathbb{Z}$. There is no ideal between $4mathbb{Z}$ and $2 mathbb{Z}$.



                                        (Note that this is a concrete example of your more abstract example.)






                                        share|cite|improve this answer












                                        $4mathbb{Z}$ in $2mathbb{Z}$. Note that $4 = 2 cdot 2 in 4mathbb{Z}$, but $2 not in 4 mathbb{Z}$. There is no ideal between $4mathbb{Z}$ and $2 mathbb{Z}$.



                                        (Note that this is a concrete example of your more abstract example.)







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Dec 3 '18 at 20:46









                                        Eric Towers

                                        31.8k22265




                                        31.8k22265























                                            2














                                            For the first part, you have the idea, but if $ain (a^2)$, you can write $a=P(a^2) $ where $P$ is a polynomial. Let $M$ be a maximal ideal containing $(a^2)$, $a.a=a^2$ is in $M$ but $a$ is not in $ M$. This implies that $M$ is not prime.






                                            share|cite|improve this answer




























                                              2














                                              For the first part, you have the idea, but if $ain (a^2)$, you can write $a=P(a^2) $ where $P$ is a polynomial. Let $M$ be a maximal ideal containing $(a^2)$, $a.a=a^2$ is in $M$ but $a$ is not in $ M$. This implies that $M$ is not prime.






                                              share|cite|improve this answer


























                                                2












                                                2








                                                2






                                                For the first part, you have the idea, but if $ain (a^2)$, you can write $a=P(a^2) $ where $P$ is a polynomial. Let $M$ be a maximal ideal containing $(a^2)$, $a.a=a^2$ is in $M$ but $a$ is not in $ M$. This implies that $M$ is not prime.






                                                share|cite|improve this answer














                                                For the first part, you have the idea, but if $ain (a^2)$, you can write $a=P(a^2) $ where $P$ is a polynomial. Let $M$ be a maximal ideal containing $(a^2)$, $a.a=a^2$ is in $M$ but $a$ is not in $ M$. This implies that $M$ is not prime.







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Dec 3 '18 at 15:20

























                                                answered Dec 3 '18 at 15:10









                                                Tsemo Aristide

                                                56k11444




                                                56k11444























                                                    2














                                                    For the second question, the definition of a prime ideal is that $I$ is prime if for any $a, b in R$ such that $ab in I$ then either $a in I$ or $b in I$. So if we can find $a, b in R setminus I$ with $ab in I$ then this would show that $I$ was not prime.



                                                    Now in the situation of the question we have a maximal ideal $M$ of $R = (a)$ with $(a^2) subseteq M$. Let $r, s in R setminus M$. Then $r = ta + ma$ and $s = ua + na$ for some $t, u in R$, $m, n in Bbb Z$, so $rs = (tu + tn + mu + mn)a^2 in (a^2) subset M$. Thus $rs in M$ but $r notin M$, $s notin M$, as required to show that $M$ was not prime.






                                                    share|cite|improve this answer


























                                                      2














                                                      For the second question, the definition of a prime ideal is that $I$ is prime if for any $a, b in R$ such that $ab in I$ then either $a in I$ or $b in I$. So if we can find $a, b in R setminus I$ with $ab in I$ then this would show that $I$ was not prime.



                                                      Now in the situation of the question we have a maximal ideal $M$ of $R = (a)$ with $(a^2) subseteq M$. Let $r, s in R setminus M$. Then $r = ta + ma$ and $s = ua + na$ for some $t, u in R$, $m, n in Bbb Z$, so $rs = (tu + tn + mu + mn)a^2 in (a^2) subset M$. Thus $rs in M$ but $r notin M$, $s notin M$, as required to show that $M$ was not prime.






                                                      share|cite|improve this answer
























                                                        2












                                                        2








                                                        2






                                                        For the second question, the definition of a prime ideal is that $I$ is prime if for any $a, b in R$ such that $ab in I$ then either $a in I$ or $b in I$. So if we can find $a, b in R setminus I$ with $ab in I$ then this would show that $I$ was not prime.



                                                        Now in the situation of the question we have a maximal ideal $M$ of $R = (a)$ with $(a^2) subseteq M$. Let $r, s in R setminus M$. Then $r = ta + ma$ and $s = ua + na$ for some $t, u in R$, $m, n in Bbb Z$, so $rs = (tu + tn + mu + mn)a^2 in (a^2) subset M$. Thus $rs in M$ but $r notin M$, $s notin M$, as required to show that $M$ was not prime.






                                                        share|cite|improve this answer












                                                        For the second question, the definition of a prime ideal is that $I$ is prime if for any $a, b in R$ such that $ab in I$ then either $a in I$ or $b in I$. So if we can find $a, b in R setminus I$ with $ab in I$ then this would show that $I$ was not prime.



                                                        Now in the situation of the question we have a maximal ideal $M$ of $R = (a)$ with $(a^2) subseteq M$. Let $r, s in R setminus M$. Then $r = ta + ma$ and $s = ua + na$ for some $t, u in R$, $m, n in Bbb Z$, so $rs = (tu + tn + mu + mn)a^2 in (a^2) subset M$. Thus $rs in M$ but $r notin M$, $s notin M$, as required to show that $M$ was not prime.







                                                        share|cite|improve this answer












                                                        share|cite|improve this answer



                                                        share|cite|improve this answer










                                                        answered Dec 3 '18 at 15:24









                                                        Christopher

                                                        6,44711628




                                                        6,44711628






























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