Propostion 1.6 on Atiyah's commutative algebra text












0












$begingroup$



Let $A$ be a ring (commutative with 1) and $M$ a maximal ideal of $A$.
Let $x in A-M $. Since $M$ is maximal, the ideal generated by $x$ and
$M$ is $(1)$, i.e. the entire ring.




I don't understand how that is true. If $x$ were a unit, certainly. However, $x$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $x$ and $M$ can generate $A$?






abstract-algebra commutative-algebra







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$endgroup$








  • 2




    $begingroup$
    the ideal generated by $x$ and $M$ strictly contains $M$, hence it must be the whole thing.
    $endgroup$
    – dezdichado
    Jan 6 at 6:42
















0












$begingroup$



Let $A$ be a ring (commutative with 1) and $M$ a maximal ideal of $A$.
Let $x in A-M $. Since $M$ is maximal, the ideal generated by $x$ and
$M$ is $(1)$, i.e. the entire ring.




I don't understand how that is true. If $x$ were a unit, certainly. However, $x$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $x$ and $M$ can generate $A$?






abstract-algebra commutative-algebra







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    the ideal generated by $x$ and $M$ strictly contains $M$, hence it must be the whole thing.
    $endgroup$
    – dezdichado
    Jan 6 at 6:42














0












0








0





$begingroup$



Let $A$ be a ring (commutative with 1) and $M$ a maximal ideal of $A$.
Let $x in A-M $. Since $M$ is maximal, the ideal generated by $x$ and
$M$ is $(1)$, i.e. the entire ring.




I don't understand how that is true. If $x$ were a unit, certainly. However, $x$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $x$ and $M$ can generate $A$?






abstract-algebra commutative-algebra







share|cite|improve this question









$endgroup$





Let $A$ be a ring (commutative with 1) and $M$ a maximal ideal of $A$.
Let $x in A-M $. Since $M$ is maximal, the ideal generated by $x$ and
$M$ is $(1)$, i.e. the entire ring.




I don't understand how that is true. If $x$ were a unit, certainly. However, $x$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $x$ and $M$ can generate $A$?






abstract-algebra commutative-algebra




abstract-algebra commutative-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 6:38









NawajNawaj

283




283








  • 2




    $begingroup$
    the ideal generated by $x$ and $M$ strictly contains $M$, hence it must be the whole thing.
    $endgroup$
    – dezdichado
    Jan 6 at 6:42














  • 2




    $begingroup$
    the ideal generated by $x$ and $M$ strictly contains $M$, hence it must be the whole thing.
    $endgroup$
    – dezdichado
    Jan 6 at 6:42








2




2




$begingroup$
the ideal generated by $x$ and $M$ strictly contains $M$, hence it must be the whole thing.
$endgroup$
– dezdichado
Jan 6 at 6:42




$begingroup$
the ideal generated by $x$ and $M$ strictly contains $M$, hence it must be the whole thing.
$endgroup$
– dezdichado
Jan 6 at 6:42










2 Answers
2






active

oldest

votes


















1












$begingroup$

The ideal generated by $x$ and $M$ strictly contains $M$ (as $x notin M$ by assumption). Then, since $M$ is maximal, by definition, any ideal containing $M$ must be the whole ring; thus, $(x,M)$ is the entire ring (as dezdichado said in the comments).






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $mathscr Msubset A$ a maximal ideal implies that for any ideal $mathscr Jsupsetneqmathscr M$, $mathscr J=(1)$.



    But if $xin Asetminus mathscr M$, then $(x,mathscr M)supsetneq mathscr M$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
      $endgroup$
      – mathworker21
      Jan 6 at 7:12










    • $begingroup$
      @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
      $endgroup$
      – Chris Custer
      Jan 6 at 7:22










    • $begingroup$
      $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
      $endgroup$
      – mathworker21
      Jan 6 at 7:33












    • $begingroup$
      @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
      $endgroup$
      – Chris Custer
      Jan 6 at 7:56











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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

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    1












    $begingroup$

    The ideal generated by $x$ and $M$ strictly contains $M$ (as $x notin M$ by assumption). Then, since $M$ is maximal, by definition, any ideal containing $M$ must be the whole ring; thus, $(x,M)$ is the entire ring (as dezdichado said in the comments).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The ideal generated by $x$ and $M$ strictly contains $M$ (as $x notin M$ by assumption). Then, since $M$ is maximal, by definition, any ideal containing $M$ must be the whole ring; thus, $(x,M)$ is the entire ring (as dezdichado said in the comments).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The ideal generated by $x$ and $M$ strictly contains $M$ (as $x notin M$ by assumption). Then, since $M$ is maximal, by definition, any ideal containing $M$ must be the whole ring; thus, $(x,M)$ is the entire ring (as dezdichado said in the comments).






        share|cite|improve this answer











        $endgroup$



        The ideal generated by $x$ and $M$ strictly contains $M$ (as $x notin M$ by assumption). Then, since $M$ is maximal, by definition, any ideal containing $M$ must be the whole ring; thus, $(x,M)$ is the entire ring (as dezdichado said in the comments).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 21:24









        user26857

        39.4k124183




        39.4k124183










        answered Jan 6 at 7:37









        JJC94JJC94

        1549




        1549























            0












            $begingroup$

            $mathscr Msubset A$ a maximal ideal implies that for any ideal $mathscr Jsupsetneqmathscr M$, $mathscr J=(1)$.



            But if $xin Asetminus mathscr M$, then $(x,mathscr M)supsetneq mathscr M$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:12










            • $begingroup$
              @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:22










            • $begingroup$
              $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:33












            • $begingroup$
              @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:56
















            0












            $begingroup$

            $mathscr Msubset A$ a maximal ideal implies that for any ideal $mathscr Jsupsetneqmathscr M$, $mathscr J=(1)$.



            But if $xin Asetminus mathscr M$, then $(x,mathscr M)supsetneq mathscr M$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:12










            • $begingroup$
              @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:22










            • $begingroup$
              $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:33












            • $begingroup$
              @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:56














            0












            0








            0





            $begingroup$

            $mathscr Msubset A$ a maximal ideal implies that for any ideal $mathscr Jsupsetneqmathscr M$, $mathscr J=(1)$.



            But if $xin Asetminus mathscr M$, then $(x,mathscr M)supsetneq mathscr M$.






            share|cite|improve this answer











            $endgroup$



            $mathscr Msubset A$ a maximal ideal implies that for any ideal $mathscr Jsupsetneqmathscr M$, $mathscr J=(1)$.



            But if $xin Asetminus mathscr M$, then $(x,mathscr M)supsetneq mathscr M$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 6 at 7:57

























            answered Jan 6 at 6:49









            Chris CusterChris Custer

            13.1k3827




            13.1k3827












            • $begingroup$
              I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:12










            • $begingroup$
              @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:22










            • $begingroup$
              $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:33












            • $begingroup$
              @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:56


















            • $begingroup$
              I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:12










            • $begingroup$
              @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:22










            • $begingroup$
              $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
              $endgroup$
              – mathworker21
              Jan 6 at 7:33












            • $begingroup$
              @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
              $endgroup$
              – Chris Custer
              Jan 6 at 7:56
















            $begingroup$
            I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
            $endgroup$
            – mathworker21
            Jan 6 at 7:12




            $begingroup$
            I would say "implies that" instead of "means that", since $A$ is not a maximal ideal.
            $endgroup$
            – mathworker21
            Jan 6 at 7:12












            $begingroup$
            @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
            $endgroup$
            – Chris Custer
            Jan 6 at 7:22




            $begingroup$
            @mathworker21thanks for the suggestion. I'm not quite sure I see your point, though. I chose "means that" because it really is the very definition.
            $endgroup$
            – Chris Custer
            Jan 6 at 7:22












            $begingroup$
            $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
            $endgroup$
            – mathworker21
            Jan 6 at 7:33






            $begingroup$
            $A$ satisfies the property that for any ideal $F supsetneq A$, $F = (1)$, but $A$ is not a maximal ideal.
            $endgroup$
            – mathworker21
            Jan 6 at 7:33














            $begingroup$
            @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
            $endgroup$
            – Chris Custer
            Jan 6 at 7:56




            $begingroup$
            @mathworker21now I think I see your point. You mean "implies" as opposed to "if and only if ". See my edit.
            $endgroup$
            – Chris Custer
            Jan 6 at 7:56


















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