What is meant by $langle X_1,…,X_nrangle$ in algebra?












1












$begingroup$


I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.






abstract-algebra







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  • 2




    $begingroup$
    As a tip, you can use langle and rangle, respectively, to produce $langle,rangle$.
    $endgroup$
    – Clayton
    Jan 22 at 15:26
















1












$begingroup$


I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.






abstract-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    As a tip, you can use langle and rangle, respectively, to produce $langle,rangle$.
    $endgroup$
    – Clayton
    Jan 22 at 15:26














1












1








1





$begingroup$


I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.






abstract-algebra







share|cite|improve this question











$endgroup$




I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 15:25









Clayton

19.2k33286




19.2k33286










asked Jan 22 at 15:14









DoleDole

926514




926514








  • 2




    $begingroup$
    As a tip, you can use langle and rangle, respectively, to produce $langle,rangle$.
    $endgroup$
    – Clayton
    Jan 22 at 15:26














  • 2




    $begingroup$
    As a tip, you can use langle and rangle, respectively, to produce $langle,rangle$.
    $endgroup$
    – Clayton
    Jan 22 at 15:26








2




2




$begingroup$
As a tip, you can use langle and rangle, respectively, to produce $langle,rangle$.
$endgroup$
– Clayton
Jan 22 at 15:26




$begingroup$
As a tip, you can use langle and rangle, respectively, to produce $langle,rangle$.
$endgroup$
– Clayton
Jan 22 at 15:26










2 Answers
2






active

oldest

votes


















3












$begingroup$

In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.



It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.



    1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.



    2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.



    3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.



    4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      +1 very thorough
      $endgroup$
      – rschwieb
      Jan 22 at 15:46











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.



    It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.



      It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.



        It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.






        share|cite|improve this answer









        $endgroup$



        In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.



        It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 15:21









        rschwiebrschwieb

        107k12102251




        107k12102251























            3












            $begingroup$

            There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.



            1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.



            2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.



            3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.



            4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              +1 very thorough
              $endgroup$
              – rschwieb
              Jan 22 at 15:46
















            3












            $begingroup$

            There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.



            1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.



            2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.



            3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.



            4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              +1 very thorough
              $endgroup$
              – rschwieb
              Jan 22 at 15:46














            3












            3








            3





            $begingroup$

            There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.



            1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.



            2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.



            3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.



            4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).






            share|cite|improve this answer









            $endgroup$



            There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.



            1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.



            2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.



            3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.



            4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 22 at 15:31









            Jose BroxJose Brox

            3,15711128




            3,15711128








            • 1




              $begingroup$
              +1 very thorough
              $endgroup$
              – rschwieb
              Jan 22 at 15:46














            • 1




              $begingroup$
              +1 very thorough
              $endgroup$
              – rschwieb
              Jan 22 at 15:46








            1




            1




            $begingroup$
            +1 very thorough
            $endgroup$
            – rschwieb
            Jan 22 at 15:46




            $begingroup$
            +1 very thorough
            $endgroup$
            – rschwieb
            Jan 22 at 15:46


















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