Passing from a set of generators to a set of homogeneous generators












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Consider a graded ring $R=R_0oplus R_1oplusdots$. Suppose the irrelevant ideal $R_1oplus R_2oplus dots$ is finitely generated. How come we can assume that it is generated by a finite number of homogeneous elements? I'm not satisfied with the answer "by taking homogeneous parts". I have this example in mind: $(x^2+x)$ is strictly contained in the ideal generated by its homogeneous parts $(x,x^2)$. Of course in this case the irrelevant ideal of the polynomial ring isn't finitely generated, but why can't the same situation be true for finitely generated irrelevant ideals?






abstract-algebra ring-theory ideals graded-rings







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    $begingroup$


    Consider a graded ring $R=R_0oplus R_1oplusdots$. Suppose the irrelevant ideal $R_1oplus R_2oplus dots$ is finitely generated. How come we can assume that it is generated by a finite number of homogeneous elements? I'm not satisfied with the answer "by taking homogeneous parts". I have this example in mind: $(x^2+x)$ is strictly contained in the ideal generated by its homogeneous parts $(x,x^2)$. Of course in this case the irrelevant ideal of the polynomial ring isn't finitely generated, but why can't the same situation be true for finitely generated irrelevant ideals?






    abstract-algebra ring-theory ideals graded-rings







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Consider a graded ring $R=R_0oplus R_1oplusdots$. Suppose the irrelevant ideal $R_1oplus R_2oplus dots$ is finitely generated. How come we can assume that it is generated by a finite number of homogeneous elements? I'm not satisfied with the answer "by taking homogeneous parts". I have this example in mind: $(x^2+x)$ is strictly contained in the ideal generated by its homogeneous parts $(x,x^2)$. Of course in this case the irrelevant ideal of the polynomial ring isn't finitely generated, but why can't the same situation be true for finitely generated irrelevant ideals?






      abstract-algebra ring-theory ideals graded-rings







      share|cite|improve this question









      $endgroup$




      Consider a graded ring $R=R_0oplus R_1oplusdots$. Suppose the irrelevant ideal $R_1oplus R_2oplus dots$ is finitely generated. How come we can assume that it is generated by a finite number of homogeneous elements? I'm not satisfied with the answer "by taking homogeneous parts". I have this example in mind: $(x^2+x)$ is strictly contained in the ideal generated by its homogeneous parts $(x,x^2)$. Of course in this case the irrelevant ideal of the polynomial ring isn't finitely generated, but why can't the same situation be true for finitely generated irrelevant ideals?






      abstract-algebra ring-theory ideals graded-rings




      abstract-algebra ring-theory ideals graded-rings






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      asked Jan 17 at 3:06









      user437309user437309

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          Let $I$ be the irrelevant ideal. Notice that if $xin I$, then every homogeneous part of $x$ is in $I$. Indeed, by definition of $I$, every nonzero homogeneous part of $x$ must have positive degree, and thus be in $I$. So, we can indeed just take the homogeneous parts of a set of generators for $I$.



          So to be clear, the claim here is not that you can always just take the homogeneous parts of a set of generators for an ideal to get a homogeneous set of generators for the same ideal. Rather, we can do this specifically for the irrelevant ideal because it has the special property that if $xin I$, every homogeneous part of $x$ is also in $I$.






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            $begingroup$

            Let $I$ be the irrelevant ideal. Notice that if $xin I$, then every homogeneous part of $x$ is in $I$. Indeed, by definition of $I$, every nonzero homogeneous part of $x$ must have positive degree, and thus be in $I$. So, we can indeed just take the homogeneous parts of a set of generators for $I$.



            So to be clear, the claim here is not that you can always just take the homogeneous parts of a set of generators for an ideal to get a homogeneous set of generators for the same ideal. Rather, we can do this specifically for the irrelevant ideal because it has the special property that if $xin I$, every homogeneous part of $x$ is also in $I$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let $I$ be the irrelevant ideal. Notice that if $xin I$, then every homogeneous part of $x$ is in $I$. Indeed, by definition of $I$, every nonzero homogeneous part of $x$ must have positive degree, and thus be in $I$. So, we can indeed just take the homogeneous parts of a set of generators for $I$.



              So to be clear, the claim here is not that you can always just take the homogeneous parts of a set of generators for an ideal to get a homogeneous set of generators for the same ideal. Rather, we can do this specifically for the irrelevant ideal because it has the special property that if $xin I$, every homogeneous part of $x$ is also in $I$.






              share|cite|improve this answer









              $endgroup$
















                1












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                1





                $begingroup$

                Let $I$ be the irrelevant ideal. Notice that if $xin I$, then every homogeneous part of $x$ is in $I$. Indeed, by definition of $I$, every nonzero homogeneous part of $x$ must have positive degree, and thus be in $I$. So, we can indeed just take the homogeneous parts of a set of generators for $I$.



                So to be clear, the claim here is not that you can always just take the homogeneous parts of a set of generators for an ideal to get a homogeneous set of generators for the same ideal. Rather, we can do this specifically for the irrelevant ideal because it has the special property that if $xin I$, every homogeneous part of $x$ is also in $I$.






                share|cite|improve this answer









                $endgroup$



                Let $I$ be the irrelevant ideal. Notice that if $xin I$, then every homogeneous part of $x$ is in $I$. Indeed, by definition of $I$, every nonzero homogeneous part of $x$ must have positive degree, and thus be in $I$. So, we can indeed just take the homogeneous parts of a set of generators for $I$.



                So to be clear, the claim here is not that you can always just take the homogeneous parts of a set of generators for an ideal to get a homogeneous set of generators for the same ideal. Rather, we can do this specifically for the irrelevant ideal because it has the special property that if $xin I$, every homogeneous part of $x$ is also in $I$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 3:12









                Eric WofseyEric Wofsey

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