Proving if k-algebras are finitely generated and noetherian












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Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.



For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.



Can somebody give me a hint as how to proceed ? Many thanks.










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    Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.



    For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.



    Can somebody give me a hint as how to proceed ? Many thanks.










    share|cite|improve this question

























      1












      1








      1







      Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.



      For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.



      Can somebody give me a hint as how to proceed ? Many thanks.










      share|cite|improve this question













      Given two $k$-algebras $R:=koplus x^2 , k[x] subseteq k[x]$ and $S:= koplus xy ,, k[x,y] subseteq k[x,y]$ I need to show if they are finitely generated and if they are noetherian.



      For $R$ to be finitely generated k-Algebra, we need to show that there exist finitely many elements, $a_1, a_2,....a_n$ such that $R=k[a_1,a_2,...a_n].$ But we know that it is contained in $k[x]$. For the second Question I Need to show if there is an ascending chain of ideals.



      Can somebody give me a hint as how to proceed ? Many thanks.







      abstract-algebra






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      asked Nov 25 at 14:15









      user249018

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          $R = k oplus x^2 k[x] = k [ x^2, x^3 ]$



          Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
          None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.






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            1 Answer
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            1 Answer
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            $R = k oplus x^2 k[x] = k [ x^2, x^3 ]$



            Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
            None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.






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              0














              $R = k oplus x^2 k[x] = k [ x^2, x^3 ]$



              Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
              None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.






              share|cite|improve this answer
























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                0






                $R = k oplus x^2 k[x] = k [ x^2, x^3 ]$



                Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
                None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.






                share|cite|improve this answer












                $R = k oplus x^2 k[x] = k [ x^2, x^3 ]$



                Also, $S$ is isomorphic to the semi-group algebra $k [ mathbb N times mathbb N ]$, which is not a finitely generated semi-group.
                None of the $(n, 1)$ can be written in terms of others. So $S$ is not Noetherian.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 15:39









                R.C.Cowsik

                31514




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