Application of p-adic valuation












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Happy new year, I need help with solving the following problem



Let $p$ be a prime number, $f(X)=sumlimits_{i=0}^na_{i}X^{i}in mathbb{Q}[X]$ and $c_{p}(f):=min{v_{p}(a_{i})|i=0,...,n}$, $v_{p}:mathbb{Q}to mathbb{Z}cup{ infty}$ is the unique extension of the p-adic valuation.



i) For $g,hin mathbb{Q}[X]$ the following holds: $c_{p}(g*h)=c_{p}(g)+c_{p}(h)$



ii) $f(X)in mathbb{Q}[X]$ is in $mathbb{Z}[X]$ if and only if $c_p(f)geq0$ for all prime numbers $p$.



i) I know that I have to find a $r$ such that $v_{p}(d_r)=c_{p}(g)+c_{p}(h)$ holds but I'm not sure how to do that. $d_{j}=sumlimits_{i=0}^ja_{i}b_{j-i}$, this is obtained by multiplying two elements $g,hin mathbb{Q}[X]$, $g(X)=sumlimits_{i=0}^na_{i}X^{i}$, $h(X)=sumlimits_{i=0}^mb_{i}X^{i}$, and we get $(g*h)(X)=sumlimits_{i=0}^{n+m}d_{i}X^{i}.$



Thanks in advance for any help.






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    $begingroup$


    Happy new year, I need help with solving the following problem



    Let $p$ be a prime number, $f(X)=sumlimits_{i=0}^na_{i}X^{i}in mathbb{Q}[X]$ and $c_{p}(f):=min{v_{p}(a_{i})|i=0,...,n}$, $v_{p}:mathbb{Q}to mathbb{Z}cup{ infty}$ is the unique extension of the p-adic valuation.



    i) For $g,hin mathbb{Q}[X]$ the following holds: $c_{p}(g*h)=c_{p}(g)+c_{p}(h)$



    ii) $f(X)in mathbb{Q}[X]$ is in $mathbb{Z}[X]$ if and only if $c_p(f)geq0$ for all prime numbers $p$.



    i) I know that I have to find a $r$ such that $v_{p}(d_r)=c_{p}(g)+c_{p}(h)$ holds but I'm not sure how to do that. $d_{j}=sumlimits_{i=0}^ja_{i}b_{j-i}$, this is obtained by multiplying two elements $g,hin mathbb{Q}[X]$, $g(X)=sumlimits_{i=0}^na_{i}X^{i}$, $h(X)=sumlimits_{i=0}^mb_{i}X^{i}$, and we get $(g*h)(X)=sumlimits_{i=0}^{n+m}d_{i}X^{i}.$



    Thanks in advance for any help.






    abstract-algebra







    share|cite|improve this question











    $endgroup$















      0












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      0





      $begingroup$


      Happy new year, I need help with solving the following problem



      Let $p$ be a prime number, $f(X)=sumlimits_{i=0}^na_{i}X^{i}in mathbb{Q}[X]$ and $c_{p}(f):=min{v_{p}(a_{i})|i=0,...,n}$, $v_{p}:mathbb{Q}to mathbb{Z}cup{ infty}$ is the unique extension of the p-adic valuation.



      i) For $g,hin mathbb{Q}[X]$ the following holds: $c_{p}(g*h)=c_{p}(g)+c_{p}(h)$



      ii) $f(X)in mathbb{Q}[X]$ is in $mathbb{Z}[X]$ if and only if $c_p(f)geq0$ for all prime numbers $p$.



      i) I know that I have to find a $r$ such that $v_{p}(d_r)=c_{p}(g)+c_{p}(h)$ holds but I'm not sure how to do that. $d_{j}=sumlimits_{i=0}^ja_{i}b_{j-i}$, this is obtained by multiplying two elements $g,hin mathbb{Q}[X]$, $g(X)=sumlimits_{i=0}^na_{i}X^{i}$, $h(X)=sumlimits_{i=0}^mb_{i}X^{i}$, and we get $(g*h)(X)=sumlimits_{i=0}^{n+m}d_{i}X^{i}.$



      Thanks in advance for any help.






      abstract-algebra







      share|cite|improve this question











      $endgroup$




      Happy new year, I need help with solving the following problem



      Let $p$ be a prime number, $f(X)=sumlimits_{i=0}^na_{i}X^{i}in mathbb{Q}[X]$ and $c_{p}(f):=min{v_{p}(a_{i})|i=0,...,n}$, $v_{p}:mathbb{Q}to mathbb{Z}cup{ infty}$ is the unique extension of the p-adic valuation.



      i) For $g,hin mathbb{Q}[X]$ the following holds: $c_{p}(g*h)=c_{p}(g)+c_{p}(h)$



      ii) $f(X)in mathbb{Q}[X]$ is in $mathbb{Z}[X]$ if and only if $c_p(f)geq0$ for all prime numbers $p$.



      i) I know that I have to find a $r$ such that $v_{p}(d_r)=c_{p}(g)+c_{p}(h)$ holds but I'm not sure how to do that. $d_{j}=sumlimits_{i=0}^ja_{i}b_{j-i}$, this is obtained by multiplying two elements $g,hin mathbb{Q}[X]$, $g(X)=sumlimits_{i=0}^na_{i}X^{i}$, $h(X)=sumlimits_{i=0}^mb_{i}X^{i}$, and we get $(g*h)(X)=sumlimits_{i=0}^{n+m}d_{i}X^{i}.$



      Thanks in advance for any help.






      abstract-algebra




      abstract-algebra






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      share|cite|improve this question













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      edited Jan 4 at 22:09







      Manwell

















      asked Jan 4 at 20:38









      ManwellManwell

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          Let $U_1,U_2 in mathbb{Z}[X]$. Iff $c_p(U_j) = k_j$ then $$U_j(X) = p^{k_j} (a_j X^{n_j}+ V_j(X))+p^{k_j+1} W_j(X)$$ with $V_j,W_j in mathbb{Z}[X],deg(V_j) < n_j, p nmid a_j$.



          Then $$U_1(X)U_2(X) = p^{k_1+k_2}(a_1 a_2 X^{n_1+n_2}+V_3(X)) + p^{k_1+k_2+1} W_3(X)$$
          with $V_3,W_3 in mathbb{Z}[X],deg(V_3) < n_1+n_2, p nmid a_1a_2$.



          thus $$c_p(U_1U_2) = k_1+k_2 = c_p(U_1)+c_p(U_2)$$




          Note we can state it with the big-O notation : if $U_j(X) = p^{k_j} a_j X^{n_j} + O(p^{k_j} X^{n_j-1})+O(p^{k_j+1})$ then $U_1(X)U_2(X) = p^{k_1+k_2} a_1a_2 X^{n_1+n_2} + O(p^{k_1+k_2} X^{n_1+n_2-1})+O(p^{k_1+k_2+1})$






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            $begingroup$

            Let $U_1,U_2 in mathbb{Z}[X]$. Iff $c_p(U_j) = k_j$ then $$U_j(X) = p^{k_j} (a_j X^{n_j}+ V_j(X))+p^{k_j+1} W_j(X)$$ with $V_j,W_j in mathbb{Z}[X],deg(V_j) < n_j, p nmid a_j$.



            Then $$U_1(X)U_2(X) = p^{k_1+k_2}(a_1 a_2 X^{n_1+n_2}+V_3(X)) + p^{k_1+k_2+1} W_3(X)$$
            with $V_3,W_3 in mathbb{Z}[X],deg(V_3) < n_1+n_2, p nmid a_1a_2$.



            thus $$c_p(U_1U_2) = k_1+k_2 = c_p(U_1)+c_p(U_2)$$




            Note we can state it with the big-O notation : if $U_j(X) = p^{k_j} a_j X^{n_j} + O(p^{k_j} X^{n_j-1})+O(p^{k_j+1})$ then $U_1(X)U_2(X) = p^{k_1+k_2} a_1a_2 X^{n_1+n_2} + O(p^{k_1+k_2} X^{n_1+n_2-1})+O(p^{k_1+k_2+1})$






            share|cite|improve this answer









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              0












              $begingroup$

              Let $U_1,U_2 in mathbb{Z}[X]$. Iff $c_p(U_j) = k_j$ then $$U_j(X) = p^{k_j} (a_j X^{n_j}+ V_j(X))+p^{k_j+1} W_j(X)$$ with $V_j,W_j in mathbb{Z}[X],deg(V_j) < n_j, p nmid a_j$.



              Then $$U_1(X)U_2(X) = p^{k_1+k_2}(a_1 a_2 X^{n_1+n_2}+V_3(X)) + p^{k_1+k_2+1} W_3(X)$$
              with $V_3,W_3 in mathbb{Z}[X],deg(V_3) < n_1+n_2, p nmid a_1a_2$.



              thus $$c_p(U_1U_2) = k_1+k_2 = c_p(U_1)+c_p(U_2)$$




              Note we can state it with the big-O notation : if $U_j(X) = p^{k_j} a_j X^{n_j} + O(p^{k_j} X^{n_j-1})+O(p^{k_j+1})$ then $U_1(X)U_2(X) = p^{k_1+k_2} a_1a_2 X^{n_1+n_2} + O(p^{k_1+k_2} X^{n_1+n_2-1})+O(p^{k_1+k_2+1})$






              share|cite|improve this answer









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                0












                0








                0





                $begingroup$

                Let $U_1,U_2 in mathbb{Z}[X]$. Iff $c_p(U_j) = k_j$ then $$U_j(X) = p^{k_j} (a_j X^{n_j}+ V_j(X))+p^{k_j+1} W_j(X)$$ with $V_j,W_j in mathbb{Z}[X],deg(V_j) < n_j, p nmid a_j$.



                Then $$U_1(X)U_2(X) = p^{k_1+k_2}(a_1 a_2 X^{n_1+n_2}+V_3(X)) + p^{k_1+k_2+1} W_3(X)$$
                with $V_3,W_3 in mathbb{Z}[X],deg(V_3) < n_1+n_2, p nmid a_1a_2$.



                thus $$c_p(U_1U_2) = k_1+k_2 = c_p(U_1)+c_p(U_2)$$




                Note we can state it with the big-O notation : if $U_j(X) = p^{k_j} a_j X^{n_j} + O(p^{k_j} X^{n_j-1})+O(p^{k_j+1})$ then $U_1(X)U_2(X) = p^{k_1+k_2} a_1a_2 X^{n_1+n_2} + O(p^{k_1+k_2} X^{n_1+n_2-1})+O(p^{k_1+k_2+1})$






                share|cite|improve this answer









                $endgroup$



                Let $U_1,U_2 in mathbb{Z}[X]$. Iff $c_p(U_j) = k_j$ then $$U_j(X) = p^{k_j} (a_j X^{n_j}+ V_j(X))+p^{k_j+1} W_j(X)$$ with $V_j,W_j in mathbb{Z}[X],deg(V_j) < n_j, p nmid a_j$.



                Then $$U_1(X)U_2(X) = p^{k_1+k_2}(a_1 a_2 X^{n_1+n_2}+V_3(X)) + p^{k_1+k_2+1} W_3(X)$$
                with $V_3,W_3 in mathbb{Z}[X],deg(V_3) < n_1+n_2, p nmid a_1a_2$.



                thus $$c_p(U_1U_2) = k_1+k_2 = c_p(U_1)+c_p(U_2)$$




                Note we can state it with the big-O notation : if $U_j(X) = p^{k_j} a_j X^{n_j} + O(p^{k_j} X^{n_j-1})+O(p^{k_j+1})$ then $U_1(X)U_2(X) = p^{k_1+k_2} a_1a_2 X^{n_1+n_2} + O(p^{k_1+k_2} X^{n_1+n_2-1})+O(p^{k_1+k_2+1})$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 23:49









                reunsreuns

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                21k21250






























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