When a polynomial quotient algebra is semi-simple












3












$begingroup$


Let F be a field, and $p(x) in F[x]$ an irreducible polynomial. Consider $n in mathbb{N}$ and let's define an ideal $J = F[x]p(x)^n$ in polynomials algebra $F[x]$, and now we shall define the quotient algebra $A = F[x]/J$.



Prove that $A$ is semi-simple iff $n=1$. (when we define $A$ to be semi-simple if it can be written as a sum of minimal left ideals - which is equivalent to the definition that Jacobson radical is 0).



Now, I believe that i have an argument to the first part: if $A$ is semi-simple means $A = sum_{alpha in Gamma}J_{alpha}$ (where $J_{alpha}$ are left minimal ideals) we can assume by contradiction that $n>1$. let's define $I_k:= ${$[q(x)] in A: p(x)^k | q(x)$} for $k=1,...,n-1$ - which are all left ideals ($I_n = ${$[0]$} and $I_{n-1}$ is minimal and also $I_{n-1} subseteq I_1$). now since we know that there exists $alpha_0 in Gamma$ such that $p(x) in J_{alpha _ 0}$ we deduce $I_1 subseteq J_{alpha _0}$ which means contradiction to minimalism of $J_{alpha _0}$ or $I_{n-1} = J_{alpha_0}$ which means $I_{n-1} = I_{1}$ also a contradiction.



I'm not sure in my solution and also i seek for a proof in the other direction.



thanks ahead






abstract-algebra polynomials ideals







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  • $begingroup$
    “Which is equivalent to the definition that the Jacobson radical is 0”. I sure hope you mean “artinian and Jacobson radical is zero”
    $endgroup$
    – rschwieb
    Dec 22 '18 at 17:30
















3












$begingroup$


Let F be a field, and $p(x) in F[x]$ an irreducible polynomial. Consider $n in mathbb{N}$ and let's define an ideal $J = F[x]p(x)^n$ in polynomials algebra $F[x]$, and now we shall define the quotient algebra $A = F[x]/J$.



Prove that $A$ is semi-simple iff $n=1$. (when we define $A$ to be semi-simple if it can be written as a sum of minimal left ideals - which is equivalent to the definition that Jacobson radical is 0).



Now, I believe that i have an argument to the first part: if $A$ is semi-simple means $A = sum_{alpha in Gamma}J_{alpha}$ (where $J_{alpha}$ are left minimal ideals) we can assume by contradiction that $n>1$. let's define $I_k:= ${$[q(x)] in A: p(x)^k | q(x)$} for $k=1,...,n-1$ - which are all left ideals ($I_n = ${$[0]$} and $I_{n-1}$ is minimal and also $I_{n-1} subseteq I_1$). now since we know that there exists $alpha_0 in Gamma$ such that $p(x) in J_{alpha _ 0}$ we deduce $I_1 subseteq J_{alpha _0}$ which means contradiction to minimalism of $J_{alpha _0}$ or $I_{n-1} = J_{alpha_0}$ which means $I_{n-1} = I_{1}$ also a contradiction.



I'm not sure in my solution and also i seek for a proof in the other direction.



thanks ahead






abstract-algebra polynomials ideals







share|cite|improve this question









$endgroup$












  • $begingroup$
    “Which is equivalent to the definition that the Jacobson radical is 0”. I sure hope you mean “artinian and Jacobson radical is zero”
    $endgroup$
    – rschwieb
    Dec 22 '18 at 17:30














3












3








3


0



$begingroup$


Let F be a field, and $p(x) in F[x]$ an irreducible polynomial. Consider $n in mathbb{N}$ and let's define an ideal $J = F[x]p(x)^n$ in polynomials algebra $F[x]$, and now we shall define the quotient algebra $A = F[x]/J$.



Prove that $A$ is semi-simple iff $n=1$. (when we define $A$ to be semi-simple if it can be written as a sum of minimal left ideals - which is equivalent to the definition that Jacobson radical is 0).



Now, I believe that i have an argument to the first part: if $A$ is semi-simple means $A = sum_{alpha in Gamma}J_{alpha}$ (where $J_{alpha}$ are left minimal ideals) we can assume by contradiction that $n>1$. let's define $I_k:= ${$[q(x)] in A: p(x)^k | q(x)$} for $k=1,...,n-1$ - which are all left ideals ($I_n = ${$[0]$} and $I_{n-1}$ is minimal and also $I_{n-1} subseteq I_1$). now since we know that there exists $alpha_0 in Gamma$ such that $p(x) in J_{alpha _ 0}$ we deduce $I_1 subseteq J_{alpha _0}$ which means contradiction to minimalism of $J_{alpha _0}$ or $I_{n-1} = J_{alpha_0}$ which means $I_{n-1} = I_{1}$ also a contradiction.



I'm not sure in my solution and also i seek for a proof in the other direction.



thanks ahead






abstract-algebra polynomials ideals







share|cite|improve this question









$endgroup$




Let F be a field, and $p(x) in F[x]$ an irreducible polynomial. Consider $n in mathbb{N}$ and let's define an ideal $J = F[x]p(x)^n$ in polynomials algebra $F[x]$, and now we shall define the quotient algebra $A = F[x]/J$.



Prove that $A$ is semi-simple iff $n=1$. (when we define $A$ to be semi-simple if it can be written as a sum of minimal left ideals - which is equivalent to the definition that Jacobson radical is 0).



Now, I believe that i have an argument to the first part: if $A$ is semi-simple means $A = sum_{alpha in Gamma}J_{alpha}$ (where $J_{alpha}$ are left minimal ideals) we can assume by contradiction that $n>1$. let's define $I_k:= ${$[q(x)] in A: p(x)^k | q(x)$} for $k=1,...,n-1$ - which are all left ideals ($I_n = ${$[0]$} and $I_{n-1}$ is minimal and also $I_{n-1} subseteq I_1$). now since we know that there exists $alpha_0 in Gamma$ such that $p(x) in J_{alpha _ 0}$ we deduce $I_1 subseteq J_{alpha _0}$ which means contradiction to minimalism of $J_{alpha _0}$ or $I_{n-1} = J_{alpha_0}$ which means $I_{n-1} = I_{1}$ also a contradiction.



I'm not sure in my solution and also i seek for a proof in the other direction.



thanks ahead






abstract-algebra polynomials ideals




abstract-algebra polynomials ideals






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share|cite|improve this question











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asked Dec 22 '18 at 15:53









ned grekerzbergned grekerzberg

482318




482318












  • $begingroup$
    “Which is equivalent to the definition that the Jacobson radical is 0”. I sure hope you mean “artinian and Jacobson radical is zero”
    $endgroup$
    – rschwieb
    Dec 22 '18 at 17:30


















  • $begingroup$
    “Which is equivalent to the definition that the Jacobson radical is 0”. I sure hope you mean “artinian and Jacobson radical is zero”
    $endgroup$
    – rschwieb
    Dec 22 '18 at 17:30
















$begingroup$
“Which is equivalent to the definition that the Jacobson radical is 0”. I sure hope you mean “artinian and Jacobson radical is zero”
$endgroup$
– rschwieb
Dec 22 '18 at 17:30




$begingroup$
“Which is equivalent to the definition that the Jacobson radical is 0”. I sure hope you mean “artinian and Jacobson radical is zero”
$endgroup$
– rschwieb
Dec 22 '18 at 17:30










2 Answers
2






active

oldest

votes


















0












$begingroup$

In the reverse direction, if n $>$ 1, then there exists a non-zero polynomial, namely $p(x)$ that is a nilpotent. The nil radical has a non-zero element therefore, so does the Jacobson radical.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
    $endgroup$
    – ned grekerzberg
    Dec 22 '18 at 18:09





















0












$begingroup$

If $n>1$, $p(x)$ is obviously a nonzero nilpotent, so the quotient would not be semisimple.



If $n=1$, then $p(x)$’s irreducibility means the ideal it generated is maximal, so the quotient is a field, which is certainly semisimple.



In general, a quotient of a polynomial ring over a field is semisimple iff the polynomial has a square free factorization.






share|cite|improve this answer











$endgroup$













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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

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    active

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    votes









    0












    $begingroup$

    In the reverse direction, if n $>$ 1, then there exists a non-zero polynomial, namely $p(x)$ that is a nilpotent. The nil radical has a non-zero element therefore, so does the Jacobson radical.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
      $endgroup$
      – ned grekerzberg
      Dec 22 '18 at 18:09


















    0












    $begingroup$

    In the reverse direction, if n $>$ 1, then there exists a non-zero polynomial, namely $p(x)$ that is a nilpotent. The nil radical has a non-zero element therefore, so does the Jacobson radical.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
      $endgroup$
      – ned grekerzberg
      Dec 22 '18 at 18:09
















    0












    0








    0





    $begingroup$

    In the reverse direction, if n $>$ 1, then there exists a non-zero polynomial, namely $p(x)$ that is a nilpotent. The nil radical has a non-zero element therefore, so does the Jacobson radical.






    share|cite|improve this answer









    $endgroup$



    In the reverse direction, if n $>$ 1, then there exists a non-zero polynomial, namely $p(x)$ that is a nilpotent. The nil radical has a non-zero element therefore, so does the Jacobson radical.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 22 '18 at 16:06









    Joel PereiraJoel Pereira

    78719




    78719












    • $begingroup$
      First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
      $endgroup$
      – ned grekerzberg
      Dec 22 '18 at 18:09




















    • $begingroup$
      First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
      $endgroup$
      – ned grekerzberg
      Dec 22 '18 at 18:09


















    $begingroup$
    First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
    $endgroup$
    – ned grekerzberg
    Dec 22 '18 at 18:09






    $begingroup$
    First, thank you Joel and rschwieb for answering my question. Although i just realized that if A (algebra with unit) is indeed a sum of minimal left ideals then it's Jacobson radical is zero, but not the other way around!!! can you find an answer using my definition for semi-simple algebra?
    $endgroup$
    – ned grekerzberg
    Dec 22 '18 at 18:09













    0












    $begingroup$

    If $n>1$, $p(x)$ is obviously a nonzero nilpotent, so the quotient would not be semisimple.



    If $n=1$, then $p(x)$’s irreducibility means the ideal it generated is maximal, so the quotient is a field, which is certainly semisimple.



    In general, a quotient of a polynomial ring over a field is semisimple iff the polynomial has a square free factorization.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If $n>1$, $p(x)$ is obviously a nonzero nilpotent, so the quotient would not be semisimple.



      If $n=1$, then $p(x)$’s irreducibility means the ideal it generated is maximal, so the quotient is a field, which is certainly semisimple.



      In general, a quotient of a polynomial ring over a field is semisimple iff the polynomial has a square free factorization.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If $n>1$, $p(x)$ is obviously a nonzero nilpotent, so the quotient would not be semisimple.



        If $n=1$, then $p(x)$’s irreducibility means the ideal it generated is maximal, so the quotient is a field, which is certainly semisimple.



        In general, a quotient of a polynomial ring over a field is semisimple iff the polynomial has a square free factorization.






        share|cite|improve this answer











        $endgroup$



        If $n>1$, $p(x)$ is obviously a nonzero nilpotent, so the quotient would not be semisimple.



        If $n=1$, then $p(x)$’s irreducibility means the ideal it generated is maximal, so the quotient is a field, which is certainly semisimple.



        In general, a quotient of a polynomial ring over a field is semisimple iff the polynomial has a square free factorization.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 22 '18 at 17:39

























        answered Dec 22 '18 at 16:43









        rschwiebrschwieb

        107k12102251




        107k12102251






























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