Are simple commutative monoids monogeneous?












3












$begingroup$


Let $M$ be a simple commutative monoid.



Is there a surjective monoid morphism $mathbb Nto M$?



If non-monogeneous simple commutative monoids do exist, what's known about them?



Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:



$bullet$ $M$ admits exactly two congruences,



$bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,



$bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.






abstract-algebra monoid







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $M$ be a simple commutative monoid.



    Is there a surjective monoid morphism $mathbb Nto M$?



    If non-monogeneous simple commutative monoids do exist, what's known about them?



    Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:



    $bullet$ $M$ admits exactly two congruences,



    $bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,



    $bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.






    abstract-algebra monoid







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Let $M$ be a simple commutative monoid.



      Is there a surjective monoid morphism $mathbb Nto M$?



      If non-monogeneous simple commutative monoids do exist, what's known about them?



      Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:



      $bullet$ $M$ admits exactly two congruences,



      $bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,



      $bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.






      abstract-algebra monoid







      share|cite|improve this question











      $endgroup$




      Let $M$ be a simple commutative monoid.



      Is there a surjective monoid morphism $mathbb Nto M$?



      If non-monogeneous simple commutative monoids do exist, what's known about them?



      Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:



      $bullet$ $M$ admits exactly two congruences,



      $bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,



      $bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.






      abstract-algebra monoid




      abstract-algebra monoid






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 20:54







      Pierre-Yves Gaillard

















      asked Jan 7 at 13:39









      Pierre-Yves GaillardPierre-Yves Gaillard

      13.3k23181




      13.3k23181






















          1 Answer
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          1












          $begingroup$

          Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
          If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.



          So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.






          share|cite|improve this answer









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            $begingroup$

            Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
            If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.



            So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
              If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.



              So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
                If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.



                So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.






                share|cite|improve this answer









                $endgroup$



                Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
                If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.



                So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 20:55









                Eric WofseyEric Wofsey

                183k13211338




                183k13211338






























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