If $I$ is a maximal ideal and $ain R -I$, then the assumption that $I + (a) = R$ gives a contradiction












1












$begingroup$


While I'm trying to prove that




Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.




.I stuck at showing that




For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.




My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$



But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?






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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
    $endgroup$
    – rschwieb
    Jan 8 at 14:50








  • 1




    $begingroup$
    @rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
    $endgroup$
    – onurcanbektas
    Jan 8 at 16:08










  • $begingroup$
    Yeah: that completely illuminates where your misunderstanding was.
    $endgroup$
    – rschwieb
    Jan 8 at 16:24


















1












$begingroup$


While I'm trying to prove that




Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.




.I stuck at showing that




For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.




My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$



But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?






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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
    $endgroup$
    – rschwieb
    Jan 8 at 14:50








  • 1




    $begingroup$
    @rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
    $endgroup$
    – onurcanbektas
    Jan 8 at 16:08










  • $begingroup$
    Yeah: that completely illuminates where your misunderstanding was.
    $endgroup$
    – rschwieb
    Jan 8 at 16:24
















1












1








1





$begingroup$


While I'm trying to prove that




Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.




.I stuck at showing that




For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.




My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$



But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?






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







share|cite|improve this question











$endgroup$




While I'm trying to prove that




Let $S$ be a multiplicative set in the commutative ring $R$ with
identity s.t $0 not in S$.Let $I$ be a maximal ideal in $S^c = R -
S$. Then show that $I$ is a prime ideal.




.I stuck at showing that




For $ab in I$ and $a,b not in I$, if $a in S^c - I$, then $$I +
(a) = R$$ gives a contradiction.




My first (and in fact only) attempt was observing that $exists (j in I, r in R)$ s.t
$$j + ra = 1_R.$$



But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?






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




abstract-algebra ring-theory ideals 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 Jan 8 at 16:07







onurcanbektas

















asked Jan 8 at 8:12









onurcanbektasonurcanbektas

3,40711036




3,40711036












  • $begingroup$
    Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
    $endgroup$
    – rschwieb
    Jan 8 at 14:50








  • 1




    $begingroup$
    @rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
    $endgroup$
    – onurcanbektas
    Jan 8 at 16:08










  • $begingroup$
    Yeah: that completely illuminates where your misunderstanding was.
    $endgroup$
    – rschwieb
    Jan 8 at 16:24




















  • $begingroup$
    Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
    $endgroup$
    – rschwieb
    Jan 8 at 14:50








  • 1




    $begingroup$
    @rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
    $endgroup$
    – onurcanbektas
    Jan 8 at 16:08










  • $begingroup$
    Yeah: that completely illuminates where your misunderstanding was.
    $endgroup$
    – rschwieb
    Jan 8 at 16:24


















$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50






$begingroup$
Sounds like you're just leaving something out about what you're proving, or else misunderstood what the intended contradiction is.
$endgroup$
– rschwieb
Jan 8 at 14:50






1




1




$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08




$begingroup$
@rschwieb Yes, I left out some part intentionally because I didn't want someone to give the answer to the main thing that I was trying to prove; I just wanted help at a particular point; see my edit.
$endgroup$
– onurcanbektas
Jan 8 at 16:08












$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24






$begingroup$
Yeah: that completely illuminates where your misunderstanding was.
$endgroup$
– rschwieb
Jan 8 at 16:24












4 Answers
4






active

oldest

votes


















2












$begingroup$

With the added context, we can now see where your problem lies:




While I'm trying to prove that



Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.




The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.



That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    A contradiction? No.



    Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.



    Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.



    It is not a contradiction but a correct conclusion.




    P.S.



    This answer was posted before the illuminating edit of the OP.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
      $endgroup$
      – onurcanbektas
      Jan 8 at 16:13



















    1












    $begingroup$

    This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)



    In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      As mentioned above, this is not a contradiction.



      If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        With the added context, we can now see where your problem lies:




        While I'm trying to prove that



        Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.




        The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.



        That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          With the added context, we can now see where your problem lies:




          While I'm trying to prove that



          Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.




          The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.



          That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            With the added context, we can now see where your problem lies:




            While I'm trying to prove that



            Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.




            The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.



            That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.






            share|cite|improve this answer









            $endgroup$



            With the added context, we can now see where your problem lies:




            While I'm trying to prove that



            Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.




            The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime. In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.



            That seems to be your misunderstanding. You need to proceed instead with $(a,I)cap Sneqemptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)cap Sneqemptyset$ and see what you can do with the two statements.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 16:19









            rschwiebrschwieb

            107k12102250




            107k12102250























                2












                $begingroup$

                A contradiction? No.



                Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.



                Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.



                It is not a contradiction but a correct conclusion.




                P.S.



                This answer was posted before the illuminating edit of the OP.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
                  $endgroup$
                  – onurcanbektas
                  Jan 8 at 16:13
















                2












                $begingroup$

                A contradiction? No.



                Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.



                Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.



                It is not a contradiction but a correct conclusion.




                P.S.



                This answer was posted before the illuminating edit of the OP.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
                  $endgroup$
                  – onurcanbektas
                  Jan 8 at 16:13














                2












                2








                2





                $begingroup$

                A contradiction? No.



                Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.



                Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.



                It is not a contradiction but a correct conclusion.




                P.S.



                This answer was posted before the illuminating edit of the OP.






                share|cite|improve this answer











                $endgroup$



                A contradiction? No.



                Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $anotin I$ as an element.



                Then $I+(a)=R$ is the only possibility, because $I+(a)neq R$ would violate the maximality of $I$.



                It is not a contradiction but a correct conclusion.




                P.S.



                This answer was posted before the illuminating edit of the OP.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 8 at 16:42

























                answered Jan 8 at 8:19









                drhabdrhab

                102k545136




                102k545136












                • $begingroup$
                  I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
                  $endgroup$
                  – onurcanbektas
                  Jan 8 at 16:13


















                • $begingroup$
                  I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
                  $endgroup$
                  – onurcanbektas
                  Jan 8 at 16:13
















                $begingroup$
                I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
                $endgroup$
                – onurcanbektas
                Jan 8 at 16:13




                $begingroup$
                I hate doing controversial edits to the question after I've asked, but I think I missed a lot of point in the question s.t a reader cannot post a helpful answer, so see my edit please.
                $endgroup$
                – onurcanbektas
                Jan 8 at 16:13











                1












                $begingroup$

                This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)



                In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)



                  In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)



                    In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?






                    share|cite|improve this answer









                    $endgroup$



                    This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $mathbb Z$ and that $(2)+(3)=mathbb Z$ (show that $1$ is in the sum.)



                    In fact, if $I subset A$ is a maximal ideal and $I+(a) neq A$ for $a notin I$, we have that $I subset I+(a) subsetneq A$ so how could $I$ have been maximal?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 8 at 8:18









                    Andres MejiaAndres Mejia

                    16.2k21548




                    16.2k21548























                        0












                        $begingroup$

                        As mentioned above, this is not a contradiction.



                        If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          As mentioned above, this is not a contradiction.



                          If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            As mentioned above, this is not a contradiction.



                            If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.






                            share|cite|improve this answer









                            $endgroup$



                            As mentioned above, this is not a contradiction.



                            If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a in R-I$ such that $I+(a) subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 8 at 8:38









                            Yanger MaYanger Ma

                            1014




                            1014






























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                                張江高科駅

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                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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