Is a polynomial $f$ zero at $(a_1,ldots,a_n)$ iff $f$ lies in the ideal $(X_1-a_1,ldots,X_n-a_n)$?












4












$begingroup$


This is probably a very silly question:



If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
$$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.




Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
$$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
If this does not work in general, let $R=K$ be a field.







abstract-algebra polynomials ideals roots







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$endgroup$

















    4












    $begingroup$


    This is probably a very silly question:



    If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
    $$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
    where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.




    Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
    $$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
    If this does not work in general, let $R=K$ be a field.







    abstract-algebra polynomials ideals roots







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      This is probably a very silly question:



      If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
      $$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
      where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.




      Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
      $$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
      If this does not work in general, let $R=K$ be a field.







      abstract-algebra polynomials ideals roots







      share|cite|improve this question











      $endgroup$




      This is probably a very silly question:



      If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
      $$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
      where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.




      Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
      $$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
      If this does not work in general, let $R=K$ be a field.







      abstract-algebra polynomials ideals roots




      abstract-algebra polynomials ideals roots






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 1 '17 at 17:49









      coproc

      977514




      977514










      asked Jul 7 '15 at 13:54









      user8463524user8463524

      577315




      577315






















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          $begingroup$

          Yes. Another way to see this is by the following:



          We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
          for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!






          share|cite|improve this answer











          $endgroup$





















            4












            $begingroup$

            Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
            $$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
            and
            $$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
            where
            $$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$






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              2 Answers
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              2 Answers
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              1












              $begingroup$

              Yes. Another way to see this is by the following:



              We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
              for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Yes. Another way to see this is by the following:



                We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
                for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Yes. Another way to see this is by the following:



                  We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
                  for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!






                  share|cite|improve this answer











                  $endgroup$



                  Yes. Another way to see this is by the following:



                  We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
                  for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 18 '18 at 20:39

























                  answered Jul 7 '15 at 14:10









                  EoinEoin

                  4,5211927




                  4,5211927























                      4












                      $begingroup$

                      Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
                      $$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
                      and
                      $$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
                      where
                      $$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$






                      share|cite|improve this answer









                      $endgroup$


















                        4












                        $begingroup$

                        Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
                        $$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
                        and
                        $$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
                        where
                        $$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$






                        share|cite|improve this answer









                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
                          $$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
                          and
                          $$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
                          where
                          $$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$






                          share|cite|improve this answer









                          $endgroup$



                          Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
                          $$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
                          and
                          $$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
                          where
                          $$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jul 7 '15 at 14:02









                          Zev ChonolesZev Chonoles

                          110k16231430




                          110k16231430






























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