Valuation of the p adic logarithm












0












$begingroup$


I'm stuck in some propertie about the $p$-adic logarithm. The propertie comes from a proposition in a Book by Dwork which I'm studying. The proposition says:



If $v_{p}(x)>frac{1}{p-1}$, then $v_{p}(log(1+x))=v_{p}(x)$. And if
$$frac{1}{p^{s}(p-1)}<v_{p}(x)<frac{1}{p^{s-1}(p-1)}$$
for some $sgeq 1$, then
$$v_{p}(log(1+x))=p^{s}v_{p}(x)-stext{.}$$



Here $v_{p}$ denote the $p-$adic valuation and we define
$$log(1+x)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n}x^{n}$$
which converges for $|x|_{p}<1$, or, equivalently, for $v_{p}(x)>0$.



I have the first part, and the reasoning suggest that we must verify that for every $n<p^{s}$ we have $p^{s}v_{p}(x)-s<p^{n}v_{p}(x)-v_{p}(n)$. But I don't have succeed.
I was trying to play with the inequalities, but the most closely to my objective is that $nv_{p}(x)<p^{s}v_{p}(x)$, and since $n<p^{s}$ we have $v_{p}(n)<s$ and so $nv_{p}(x)-s<p^{s}v_{p}(x)-v_{p}(x)$.



Can anyone give me some hint? I will appreciate.
Thanks






algebraic-number-theory absolute-value p-adic-number-theory valuation-theory







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  • $begingroup$
    What is the relation between the $,log(1+x),$ and $,log(1+x^p),$ series?
    $endgroup$
    – Somos
    Dec 8 '18 at 19:50


















0












$begingroup$


I'm stuck in some propertie about the $p$-adic logarithm. The propertie comes from a proposition in a Book by Dwork which I'm studying. The proposition says:



If $v_{p}(x)>frac{1}{p-1}$, then $v_{p}(log(1+x))=v_{p}(x)$. And if
$$frac{1}{p^{s}(p-1)}<v_{p}(x)<frac{1}{p^{s-1}(p-1)}$$
for some $sgeq 1$, then
$$v_{p}(log(1+x))=p^{s}v_{p}(x)-stext{.}$$



Here $v_{p}$ denote the $p-$adic valuation and we define
$$log(1+x)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n}x^{n}$$
which converges for $|x|_{p}<1$, or, equivalently, for $v_{p}(x)>0$.



I have the first part, and the reasoning suggest that we must verify that for every $n<p^{s}$ we have $p^{s}v_{p}(x)-s<p^{n}v_{p}(x)-v_{p}(n)$. But I don't have succeed.
I was trying to play with the inequalities, but the most closely to my objective is that $nv_{p}(x)<p^{s}v_{p}(x)$, and since $n<p^{s}$ we have $v_{p}(n)<s$ and so $nv_{p}(x)-s<p^{s}v_{p}(x)-v_{p}(x)$.



Can anyone give me some hint? I will appreciate.
Thanks






algebraic-number-theory absolute-value p-adic-number-theory valuation-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the relation between the $,log(1+x),$ and $,log(1+x^p),$ series?
    $endgroup$
    – Somos
    Dec 8 '18 at 19:50
















0












0








0





$begingroup$


I'm stuck in some propertie about the $p$-adic logarithm. The propertie comes from a proposition in a Book by Dwork which I'm studying. The proposition says:



If $v_{p}(x)>frac{1}{p-1}$, then $v_{p}(log(1+x))=v_{p}(x)$. And if
$$frac{1}{p^{s}(p-1)}<v_{p}(x)<frac{1}{p^{s-1}(p-1)}$$
for some $sgeq 1$, then
$$v_{p}(log(1+x))=p^{s}v_{p}(x)-stext{.}$$



Here $v_{p}$ denote the $p-$adic valuation and we define
$$log(1+x)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n}x^{n}$$
which converges for $|x|_{p}<1$, or, equivalently, for $v_{p}(x)>0$.



I have the first part, and the reasoning suggest that we must verify that for every $n<p^{s}$ we have $p^{s}v_{p}(x)-s<p^{n}v_{p}(x)-v_{p}(n)$. But I don't have succeed.
I was trying to play with the inequalities, but the most closely to my objective is that $nv_{p}(x)<p^{s}v_{p}(x)$, and since $n<p^{s}$ we have $v_{p}(n)<s$ and so $nv_{p}(x)-s<p^{s}v_{p}(x)-v_{p}(x)$.



Can anyone give me some hint? I will appreciate.
Thanks






algebraic-number-theory absolute-value p-adic-number-theory valuation-theory







share|cite|improve this question











$endgroup$




I'm stuck in some propertie about the $p$-adic logarithm. The propertie comes from a proposition in a Book by Dwork which I'm studying. The proposition says:



If $v_{p}(x)>frac{1}{p-1}$, then $v_{p}(log(1+x))=v_{p}(x)$. And if
$$frac{1}{p^{s}(p-1)}<v_{p}(x)<frac{1}{p^{s-1}(p-1)}$$
for some $sgeq 1$, then
$$v_{p}(log(1+x))=p^{s}v_{p}(x)-stext{.}$$



Here $v_{p}$ denote the $p-$adic valuation and we define
$$log(1+x)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n}x^{n}$$
which converges for $|x|_{p}<1$, or, equivalently, for $v_{p}(x)>0$.



I have the first part, and the reasoning suggest that we must verify that for every $n<p^{s}$ we have $p^{s}v_{p}(x)-s<p^{n}v_{p}(x)-v_{p}(n)$. But I don't have succeed.
I was trying to play with the inequalities, but the most closely to my objective is that $nv_{p}(x)<p^{s}v_{p}(x)$, and since $n<p^{s}$ we have $v_{p}(n)<s$ and so $nv_{p}(x)-s<p^{s}v_{p}(x)-v_{p}(x)$.



Can anyone give me some hint? I will appreciate.
Thanks






algebraic-number-theory absolute-value p-adic-number-theory valuation-theory




algebraic-number-theory absolute-value p-adic-number-theory valuation-theory






share|cite|improve this question















share|cite|improve this question













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edited Dec 8 '18 at 19:51









Somos

13.2k11034




13.2k11034










asked Dec 8 '18 at 18:38









Camacho CamachitoCamacho Camachito

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  • $begingroup$
    What is the relation between the $,log(1+x),$ and $,log(1+x^p),$ series?
    $endgroup$
    – Somos
    Dec 8 '18 at 19:50




















  • $begingroup$
    What is the relation between the $,log(1+x),$ and $,log(1+x^p),$ series?
    $endgroup$
    – Somos
    Dec 8 '18 at 19:50


















$begingroup$
What is the relation between the $,log(1+x),$ and $,log(1+x^p),$ series?
$endgroup$
– Somos
Dec 8 '18 at 19:50






$begingroup$
What is the relation between the $,log(1+x),$ and $,log(1+x^p),$ series?
$endgroup$
– Somos
Dec 8 '18 at 19:50












1 Answer
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$begingroup$

I’ll give but a hint. Look at the actual series for $log(1+x)$. Now, given a number $x$ for which $v_p(x)$ is in one of the specified open intervals, see what the value of each monomial in the series turns out to be. You’ll see that only one monomial takes on a minimum value. Now use the fact that even in an infinite sum, if one term has a $v_p$-value (say $upsilon$) less than all the others, then the $v_p$-value of the sum is $upsilon$.






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    $begingroup$

    I’ll give but a hint. Look at the actual series for $log(1+x)$. Now, given a number $x$ for which $v_p(x)$ is in one of the specified open intervals, see what the value of each monomial in the series turns out to be. You’ll see that only one monomial takes on a minimum value. Now use the fact that even in an infinite sum, if one term has a $v_p$-value (say $upsilon$) less than all the others, then the $v_p$-value of the sum is $upsilon$.






    share|cite|improve this answer









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      3












      $begingroup$

      I’ll give but a hint. Look at the actual series for $log(1+x)$. Now, given a number $x$ for which $v_p(x)$ is in one of the specified open intervals, see what the value of each monomial in the series turns out to be. You’ll see that only one monomial takes on a minimum value. Now use the fact that even in an infinite sum, if one term has a $v_p$-value (say $upsilon$) less than all the others, then the $v_p$-value of the sum is $upsilon$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        I’ll give but a hint. Look at the actual series for $log(1+x)$. Now, given a number $x$ for which $v_p(x)$ is in one of the specified open intervals, see what the value of each monomial in the series turns out to be. You’ll see that only one monomial takes on a minimum value. Now use the fact that even in an infinite sum, if one term has a $v_p$-value (say $upsilon$) less than all the others, then the $v_p$-value of the sum is $upsilon$.






        share|cite|improve this answer









        $endgroup$



        I’ll give but a hint. Look at the actual series for $log(1+x)$. Now, given a number $x$ for which $v_p(x)$ is in one of the specified open intervals, see what the value of each monomial in the series turns out to be. You’ll see that only one monomial takes on a minimum value. Now use the fact that even in an infinite sum, if one term has a $v_p$-value (say $upsilon$) less than all the others, then the $v_p$-value of the sum is $upsilon$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 2:56









        LubinLubin

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