Graded and Filtered Algebras












1












$begingroup$


This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.



The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.



Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.



His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).



I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!






abstract-algebra graded-rings quantum-groups filtrations







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$endgroup$

















    1












    $begingroup$


    This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.



    The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.



    Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.



    His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).



    I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!






    abstract-algebra graded-rings quantum-groups filtrations







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.



      The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.



      Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.



      His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).



      I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!






      abstract-algebra graded-rings quantum-groups filtrations







      share|cite|improve this question











      $endgroup$




      This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.



      The general set-up is this. Let $A$ be a filtered algebra with filtration $F_0(A) subset F_1(A) subset cdots subset A$, and suppose that $I$ is a two-sided ideal of $A$. The quotient algebra is then filtered with filtration $F_i(A/I) = F_i(A)/F_i(A) cap I$. We know that for a filtered algebra, $A$, there exists an associated graded algebra, called $mathrm{gr}(A) = oplus S_i$ where $S_i = F_i(A)/F_{i-1}(A)$.



      Define $M(2)$ as the polynomial algebra $k[a,b,c,d]$ and define $SL(2) = M(2)/(ad - bc -1)$.



      His first claim is that $mathrm{gr}(A/I) = oplus_{i in mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) cap I)$. Following this, he claims that $mathrm{gr}(SL(2)) cong k[a,b,c,d]/(ad-bc)$ (note that the ideal $(ad-bc-1)$ is not generated by homogeneous elements so that $SL(2)$ is not graded).



      I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!






      abstract-algebra graded-rings quantum-groups filtrations




      abstract-algebra graded-rings quantum-groups filtrations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 10:55









      user26857

      39.3k124183




      39.3k124183










      asked Mar 21 '12 at 22:12









      Elden ElmantoElden Elmanto

      684315




      684315






















          1 Answer
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          $begingroup$

          We have
          $$begin{align*}
          text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
          &cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
          &cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$



          which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.



          With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yup. Thank you very much!
            $endgroup$
            – Elden Elmanto
            Mar 21 '12 at 23:08











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          $begingroup$

          We have
          $$begin{align*}
          text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
          &cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
          &cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$



          which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.



          With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yup. Thank you very much!
            $endgroup$
            – Elden Elmanto
            Mar 21 '12 at 23:08
















          2












          $begingroup$

          We have
          $$begin{align*}
          text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
          &cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
          &cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$



          which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.



          With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yup. Thank you very much!
            $endgroup$
            – Elden Elmanto
            Mar 21 '12 at 23:08














          2












          2








          2





          $begingroup$

          We have
          $$begin{align*}
          text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
          &cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
          &cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$



          which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.



          With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?






          share|cite|improve this answer









          $endgroup$



          We have
          $$begin{align*}
          text{gr}(A/I) &=& bigoplus F_i(A/I)/F_{i-1}(A/I) \
          &cong& bigoplus (F_i(A)/F_i(A) cap I)/(F_{i-1}(A)/F_{i-1}(A) cap I) \
          &cong& bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) cap I) end{align*}$$



          which follows from the fact that $F_{i-1}(A) cap I subseteq F_i(A) cap I$ and I suppose the third isomorphism theorem.



          With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 21 '12 at 22:38









          Qiaochu YuanQiaochu Yuan

          279k32586931




          279k32586931












          • $begingroup$
            Yup. Thank you very much!
            $endgroup$
            – Elden Elmanto
            Mar 21 '12 at 23:08


















          • $begingroup$
            Yup. Thank you very much!
            $endgroup$
            – Elden Elmanto
            Mar 21 '12 at 23:08
















          $begingroup$
          Yup. Thank you very much!
          $endgroup$
          – Elden Elmanto
          Mar 21 '12 at 23:08




          $begingroup$
          Yup. Thank you very much!
          $endgroup$
          – Elden Elmanto
          Mar 21 '12 at 23:08


















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