Counter-example: Radical of a k-algebra












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$begingroup$


I have the following question. Let $k$ a field, $A$ a $k$-algebra, and $khookrightarrow K$ a field extension. It´s well known that $$Rad(A)otimes_{k}Ksubset Rad(Aotimes_{k}K)$$



($Rad(A)$=radical of $A$).



When is it $Rad(A)otimes_{k}K= Rad(Aotimes_{k}K)$?



Is it always true?






commutative-algebra







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$endgroup$

















    2












    $begingroup$


    I have the following question. Let $k$ a field, $A$ a $k$-algebra, and $khookrightarrow K$ a field extension. It´s well known that $$Rad(A)otimes_{k}Ksubset Rad(Aotimes_{k}K)$$



    ($Rad(A)$=radical of $A$).



    When is it $Rad(A)otimes_{k}K= Rad(Aotimes_{k}K)$?



    Is it always true?






    commutative-algebra







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I have the following question. Let $k$ a field, $A$ a $k$-algebra, and $khookrightarrow K$ a field extension. It´s well known that $$Rad(A)otimes_{k}Ksubset Rad(Aotimes_{k}K)$$



      ($Rad(A)$=radical of $A$).



      When is it $Rad(A)otimes_{k}K= Rad(Aotimes_{k}K)$?



      Is it always true?






      commutative-algebra







      share|cite|improve this question









      $endgroup$




      I have the following question. Let $k$ a field, $A$ a $k$-algebra, and $khookrightarrow K$ a field extension. It´s well known that $$Rad(A)otimes_{k}Ksubset Rad(Aotimes_{k}K)$$



      ($Rad(A)$=radical of $A$).



      When is it $Rad(A)otimes_{k}K= Rad(Aotimes_{k}K)$?



      Is it always true?






      commutative-algebra




      commutative-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jul 30 '17 at 10:24









      matemagreekmatemagreek

      16413




      16413






















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          $begingroup$

          The equality does not always hold. A counter-example is given by $k = mathbb{F}_2(t^2)$ and $A=K=mathbb{F}_2(t)$. Then





          • $Rad(A) otimes_k K = 0$;


          • $Rad(A otimes_k K)$ is not zero, since $A otimes_k K = mathbb{F}_2(t) otimes_{mathbb{F}_2(t^2)} mathbb{F}_2(t)$ contains a non-zero nilpotent element, namely $1otimes t + totimes 1$.


          Thus the inclusion $Rad(A) otimes_k K subset Rad(A otimes_k K)$ is strict in this case.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            The equality does not always hold. A counter-example is given by $k = mathbb{F}_2(t^2)$ and $A=K=mathbb{F}_2(t)$. Then





            • $Rad(A) otimes_k K = 0$;


            • $Rad(A otimes_k K)$ is not zero, since $A otimes_k K = mathbb{F}_2(t) otimes_{mathbb{F}_2(t^2)} mathbb{F}_2(t)$ contains a non-zero nilpotent element, namely $1otimes t + totimes 1$.


            Thus the inclusion $Rad(A) otimes_k K subset Rad(A otimes_k K)$ is strict in this case.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The equality does not always hold. A counter-example is given by $k = mathbb{F}_2(t^2)$ and $A=K=mathbb{F}_2(t)$. Then





              • $Rad(A) otimes_k K = 0$;


              • $Rad(A otimes_k K)$ is not zero, since $A otimes_k K = mathbb{F}_2(t) otimes_{mathbb{F}_2(t^2)} mathbb{F}_2(t)$ contains a non-zero nilpotent element, namely $1otimes t + totimes 1$.


              Thus the inclusion $Rad(A) otimes_k K subset Rad(A otimes_k K)$ is strict in this case.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The equality does not always hold. A counter-example is given by $k = mathbb{F}_2(t^2)$ and $A=K=mathbb{F}_2(t)$. Then





                • $Rad(A) otimes_k K = 0$;


                • $Rad(A otimes_k K)$ is not zero, since $A otimes_k K = mathbb{F}_2(t) otimes_{mathbb{F}_2(t^2)} mathbb{F}_2(t)$ contains a non-zero nilpotent element, namely $1otimes t + totimes 1$.


                Thus the inclusion $Rad(A) otimes_k K subset Rad(A otimes_k K)$ is strict in this case.






                share|cite|improve this answer









                $endgroup$



                The equality does not always hold. A counter-example is given by $k = mathbb{F}_2(t^2)$ and $A=K=mathbb{F}_2(t)$. Then





                • $Rad(A) otimes_k K = 0$;


                • $Rad(A otimes_k K)$ is not zero, since $A otimes_k K = mathbb{F}_2(t) otimes_{mathbb{F}_2(t^2)} mathbb{F}_2(t)$ contains a non-zero nilpotent element, namely $1otimes t + totimes 1$.


                Thus the inclusion $Rad(A) otimes_k K subset Rad(A otimes_k K)$ is strict in this case.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 19:52









                Pierre-Guy PlamondonPierre-Guy Plamondon

                8,88511739




                8,88511739






























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