Show that the ring of polynomials with coefficients in a field, and in infinitely many variables, is not Noetherian

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Show that the ring of polynomials with coefficients in a
field, and in infinitely many variables, is not Noetherian, that is, $R = k [x_i: igeq1]$ is not Noetherian.



I know that I need to exhibit an ideal of the ring that is not finitely generated, what could this ideal be? Could it be $(x_1,x_2,...,)$?
Or could I give the following chain of ideals that do not have a maximal element $(x_1)subset(x_1,x_2)subset(x_1,x_2,x_3)subset...$?How can all ideals that are not finitely generated be classified? What to do in the case where the number of variables is non-countable?










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    up vote
    3
    down vote

    favorite
    1












    Show that the ring of polynomials with coefficients in a
    field, and in infinitely many variables, is not Noetherian, that is, $R = k [x_i: igeq1]$ is not Noetherian.



    I know that I need to exhibit an ideal of the ring that is not finitely generated, what could this ideal be? Could it be $(x_1,x_2,...,)$?
    Or could I give the following chain of ideals that do not have a maximal element $(x_1)subset(x_1,x_2)subset(x_1,x_2,x_3)subset...$?How can all ideals that are not finitely generated be classified? What to do in the case where the number of variables is non-countable?










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      up vote
      3
      down vote

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      up vote
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      Show that the ring of polynomials with coefficients in a
      field, and in infinitely many variables, is not Noetherian, that is, $R = k [x_i: igeq1]$ is not Noetherian.



      I know that I need to exhibit an ideal of the ring that is not finitely generated, what could this ideal be? Could it be $(x_1,x_2,...,)$?
      Or could I give the following chain of ideals that do not have a maximal element $(x_1)subset(x_1,x_2)subset(x_1,x_2,x_3)subset...$?How can all ideals that are not finitely generated be classified? What to do in the case where the number of variables is non-countable?










      share|cite|improve this question















      Show that the ring of polynomials with coefficients in a
      field, and in infinitely many variables, is not Noetherian, that is, $R = k [x_i: igeq1]$ is not Noetherian.



      I know that I need to exhibit an ideal of the ring that is not finitely generated, what could this ideal be? Could it be $(x_1,x_2,...,)$?
      Or could I give the following chain of ideals that do not have a maximal element $(x_1)subset(x_1,x_2)subset(x_1,x_2,x_3)subset...$?How can all ideals that are not finitely generated be classified? What to do in the case where the number of variables is non-countable?







      abstract-algebra ring-theory commutative-algebra noetherian finitely-generated






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      edited Aug 21 at 11:02









      user26857

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      asked Aug 21 at 2:58









      Nash

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          A ring is Noetherian if and only if it satisfies the ascending chain condition, i.e. every increasing chain of ideals terminates. Now you have a chain
          $$(x_1)subsetneq(x_1,x_2)subsetneq(x_1,x_2,x_3)subsetneqcdots$$
          that never terminates, so $k[x_i: ige 1]$ is not Noetherian.






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            You've got all the right ideas. You don't need to fully classify the non-finitely-generated ideals to prove that $(x_1,x_2,dots)$ is not finitely generated. One way to do it is to reason:



            Suppose $I = (x_1,x_2,dots)$ were finitely generated. Then there would be a finite list of generators $f_1,dots,f_n$. Each $f_i$ would be a polynomial in finitely many variables (because that's what elements of this ring are). Thus, there would be some $N$ that is the maximum $N$ for which $x_N$ appears in any of the $f_i$'s. Then, supposedly, every element of $I$ would be expressible as a linear combination of the $f_i$'s with coefficients in the ring of polynomials. But how would you ever express $x_k$ this way when $k>N$?






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              A ring is Noetherian if and only if it satisfies the ascending chain condition, i.e. every increasing chain of ideals terminates. Now you have a chain
              $$(x_1)subsetneq(x_1,x_2)subsetneq(x_1,x_2,x_3)subsetneqcdots$$
              that never terminates, so $k[x_i: ige 1]$ is not Noetherian.






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                up vote
                4
                down vote













                A ring is Noetherian if and only if it satisfies the ascending chain condition, i.e. every increasing chain of ideals terminates. Now you have a chain
                $$(x_1)subsetneq(x_1,x_2)subsetneq(x_1,x_2,x_3)subsetneqcdots$$
                that never terminates, so $k[x_i: ige 1]$ is not Noetherian.






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                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  A ring is Noetherian if and only if it satisfies the ascending chain condition, i.e. every increasing chain of ideals terminates. Now you have a chain
                  $$(x_1)subsetneq(x_1,x_2)subsetneq(x_1,x_2,x_3)subsetneqcdots$$
                  that never terminates, so $k[x_i: ige 1]$ is not Noetherian.






                  share|cite|improve this answer












                  A ring is Noetherian if and only if it satisfies the ascending chain condition, i.e. every increasing chain of ideals terminates. Now you have a chain
                  $$(x_1)subsetneq(x_1,x_2)subsetneq(x_1,x_2,x_3)subsetneqcdots$$
                  that never terminates, so $k[x_i: ige 1]$ is not Noetherian.







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                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 21 at 3:05









                  Eclipse Sun

                  6,6221336




                  6,6221336




















                      up vote
                      3
                      down vote













                      You've got all the right ideas. You don't need to fully classify the non-finitely-generated ideals to prove that $(x_1,x_2,dots)$ is not finitely generated. One way to do it is to reason:



                      Suppose $I = (x_1,x_2,dots)$ were finitely generated. Then there would be a finite list of generators $f_1,dots,f_n$. Each $f_i$ would be a polynomial in finitely many variables (because that's what elements of this ring are). Thus, there would be some $N$ that is the maximum $N$ for which $x_N$ appears in any of the $f_i$'s. Then, supposedly, every element of $I$ would be expressible as a linear combination of the $f_i$'s with coefficients in the ring of polynomials. But how would you ever express $x_k$ this way when $k>N$?






                      share|cite|improve this answer
























                        up vote
                        3
                        down vote













                        You've got all the right ideas. You don't need to fully classify the non-finitely-generated ideals to prove that $(x_1,x_2,dots)$ is not finitely generated. One way to do it is to reason:



                        Suppose $I = (x_1,x_2,dots)$ were finitely generated. Then there would be a finite list of generators $f_1,dots,f_n$. Each $f_i$ would be a polynomial in finitely many variables (because that's what elements of this ring are). Thus, there would be some $N$ that is the maximum $N$ for which $x_N$ appears in any of the $f_i$'s. Then, supposedly, every element of $I$ would be expressible as a linear combination of the $f_i$'s with coefficients in the ring of polynomials. But how would you ever express $x_k$ this way when $k>N$?






                        share|cite|improve this answer






















                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          You've got all the right ideas. You don't need to fully classify the non-finitely-generated ideals to prove that $(x_1,x_2,dots)$ is not finitely generated. One way to do it is to reason:



                          Suppose $I = (x_1,x_2,dots)$ were finitely generated. Then there would be a finite list of generators $f_1,dots,f_n$. Each $f_i$ would be a polynomial in finitely many variables (because that's what elements of this ring are). Thus, there would be some $N$ that is the maximum $N$ for which $x_N$ appears in any of the $f_i$'s. Then, supposedly, every element of $I$ would be expressible as a linear combination of the $f_i$'s with coefficients in the ring of polynomials. But how would you ever express $x_k$ this way when $k>N$?






                          share|cite|improve this answer












                          You've got all the right ideas. You don't need to fully classify the non-finitely-generated ideals to prove that $(x_1,x_2,dots)$ is not finitely generated. One way to do it is to reason:



                          Suppose $I = (x_1,x_2,dots)$ were finitely generated. Then there would be a finite list of generators $f_1,dots,f_n$. Each $f_i$ would be a polynomial in finitely many variables (because that's what elements of this ring are). Thus, there would be some $N$ that is the maximum $N$ for which $x_N$ appears in any of the $f_i$'s. Then, supposedly, every element of $I$ would be expressible as a linear combination of the $f_i$'s with coefficients in the ring of polynomials. But how would you ever express $x_k$ this way when $k>N$?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 21 at 4:46









                          Ben Blum-Smith

                          9,71522981




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