Why is $frac{1}{x+2} = sum_k^infty (-1)^k(1+x)^k in mathbb{Z} [[x]]$ not the inverse of $x+2$ in $mathbb{Z}$?












0












$begingroup$


As also stated in here, a formal power series is a unit in $R[[x]]$ iff it is constant coefficient is a unit in the ring $R$. However, for example, we can find the inverse of $1+x$ by observing that $frac{1}{1+x} = sum_k^infty (-x)^k$, so to find the inverse of $x+2$, I simply observed that
$$frac{1}{x+2} = frac{1}{1- (-1-x) } = sum_k^infty (-1)^k(1+x)^k in mathbb{Z} [[x]].$$



However, this contradicts with the stated theorem that since $2$ is not unit in $mathbb{Z}$, $x+2$ is a non unit, so what is wrong in my argument ?






abstract-algebra polynomials ring-theory irreducible-polynomials







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Why do you assert that complicated sum is an element of $Bbb Z[[x]]$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 8 at 17:20












  • $begingroup$
    Let $|sum_{n=N}^{N+m} a_n x^n| = 2^{-N}$ for $a_n in mathbb{Z}, a_Nne 0$. It is a non-archimedian absolute value on $mathbb{Z}[x]$ and $mathbb{Z}[[x]],|.|$ is the completion of $mathbb{Z}[x],|.|$. The sequence $sum_{k=0}^K (-1)^k (1+x)^k$ is not Cauchy in $mathbb{Z}[x],|.|$ so it diverges. Replacing $|.|$ by $|sum_{n=N}^{N+m} a_n (1+x)^n|_2 = 2^{-N}$ then $sum_{k=0}^K (-1)^k (1+x)^k$ is Cauchy and its limit belongs to $mathbb{Z}[[1+x]],|.|_2$
    $endgroup$
    – reuns
    Jan 8 at 17:32












  • $begingroup$
    @reuns Thanks your explanation, but I just took first year graduate course on Algebra, and I don't know what do you mean by "completion" of a polynomial ring. In fact, I just did a quick research, but I think that beyond what I know allows.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:59












  • $begingroup$
    @onurcanbektas $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to its usual absolute value : adding to the space the limits of Cauchy sequences en.wikipedia.org/wiki/Complete_metric_space An absolute value is a metric compatible with the addition and the multiplication which guaranties those extend to the completed metric space.
    $endgroup$
    – reuns
    Jan 9 at 15:26


















0












$begingroup$


As also stated in here, a formal power series is a unit in $R[[x]]$ iff it is constant coefficient is a unit in the ring $R$. However, for example, we can find the inverse of $1+x$ by observing that $frac{1}{1+x} = sum_k^infty (-x)^k$, so to find the inverse of $x+2$, I simply observed that
$$frac{1}{x+2} = frac{1}{1- (-1-x) } = sum_k^infty (-1)^k(1+x)^k in mathbb{Z} [[x]].$$



However, this contradicts with the stated theorem that since $2$ is not unit in $mathbb{Z}$, $x+2$ is a non unit, so what is wrong in my argument ?






abstract-algebra polynomials ring-theory irreducible-polynomials







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Why do you assert that complicated sum is an element of $Bbb Z[[x]]$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 8 at 17:20












  • $begingroup$
    Let $|sum_{n=N}^{N+m} a_n x^n| = 2^{-N}$ for $a_n in mathbb{Z}, a_Nne 0$. It is a non-archimedian absolute value on $mathbb{Z}[x]$ and $mathbb{Z}[[x]],|.|$ is the completion of $mathbb{Z}[x],|.|$. The sequence $sum_{k=0}^K (-1)^k (1+x)^k$ is not Cauchy in $mathbb{Z}[x],|.|$ so it diverges. Replacing $|.|$ by $|sum_{n=N}^{N+m} a_n (1+x)^n|_2 = 2^{-N}$ then $sum_{k=0}^K (-1)^k (1+x)^k$ is Cauchy and its limit belongs to $mathbb{Z}[[1+x]],|.|_2$
    $endgroup$
    – reuns
    Jan 8 at 17:32












  • $begingroup$
    @reuns Thanks your explanation, but I just took first year graduate course on Algebra, and I don't know what do you mean by "completion" of a polynomial ring. In fact, I just did a quick research, but I think that beyond what I know allows.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:59












  • $begingroup$
    @onurcanbektas $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to its usual absolute value : adding to the space the limits of Cauchy sequences en.wikipedia.org/wiki/Complete_metric_space An absolute value is a metric compatible with the addition and the multiplication which guaranties those extend to the completed metric space.
    $endgroup$
    – reuns
    Jan 9 at 15:26
















0












0








0





$begingroup$


As also stated in here, a formal power series is a unit in $R[[x]]$ iff it is constant coefficient is a unit in the ring $R$. However, for example, we can find the inverse of $1+x$ by observing that $frac{1}{1+x} = sum_k^infty (-x)^k$, so to find the inverse of $x+2$, I simply observed that
$$frac{1}{x+2} = frac{1}{1- (-1-x) } = sum_k^infty (-1)^k(1+x)^k in mathbb{Z} [[x]].$$



However, this contradicts with the stated theorem that since $2$ is not unit in $mathbb{Z}$, $x+2$ is a non unit, so what is wrong in my argument ?






abstract-algebra polynomials ring-theory irreducible-polynomials







share|cite|improve this question









$endgroup$




As also stated in here, a formal power series is a unit in $R[[x]]$ iff it is constant coefficient is a unit in the ring $R$. However, for example, we can find the inverse of $1+x$ by observing that $frac{1}{1+x} = sum_k^infty (-x)^k$, so to find the inverse of $x+2$, I simply observed that
$$frac{1}{x+2} = frac{1}{1- (-1-x) } = sum_k^infty (-1)^k(1+x)^k in mathbb{Z} [[x]].$$



However, this contradicts with the stated theorem that since $2$ is not unit in $mathbb{Z}$, $x+2$ is a non unit, so what is wrong in my argument ?






abstract-algebra polynomials ring-theory irreducible-polynomials




abstract-algebra polynomials ring-theory irreducible-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 17:11









onurcanbektasonurcanbektas

3,47911037




3,47911037








  • 4




    $begingroup$
    Why do you assert that complicated sum is an element of $Bbb Z[[x]]$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 8 at 17:20












  • $begingroup$
    Let $|sum_{n=N}^{N+m} a_n x^n| = 2^{-N}$ for $a_n in mathbb{Z}, a_Nne 0$. It is a non-archimedian absolute value on $mathbb{Z}[x]$ and $mathbb{Z}[[x]],|.|$ is the completion of $mathbb{Z}[x],|.|$. The sequence $sum_{k=0}^K (-1)^k (1+x)^k$ is not Cauchy in $mathbb{Z}[x],|.|$ so it diverges. Replacing $|.|$ by $|sum_{n=N}^{N+m} a_n (1+x)^n|_2 = 2^{-N}$ then $sum_{k=0}^K (-1)^k (1+x)^k$ is Cauchy and its limit belongs to $mathbb{Z}[[1+x]],|.|_2$
    $endgroup$
    – reuns
    Jan 8 at 17:32












  • $begingroup$
    @reuns Thanks your explanation, but I just took first year graduate course on Algebra, and I don't know what do you mean by "completion" of a polynomial ring. In fact, I just did a quick research, but I think that beyond what I know allows.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:59












  • $begingroup$
    @onurcanbektas $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to its usual absolute value : adding to the space the limits of Cauchy sequences en.wikipedia.org/wiki/Complete_metric_space An absolute value is a metric compatible with the addition and the multiplication which guaranties those extend to the completed metric space.
    $endgroup$
    – reuns
    Jan 9 at 15:26
















  • 4




    $begingroup$
    Why do you assert that complicated sum is an element of $Bbb Z[[x]]$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 8 at 17:20












  • $begingroup$
    Let $|sum_{n=N}^{N+m} a_n x^n| = 2^{-N}$ for $a_n in mathbb{Z}, a_Nne 0$. It is a non-archimedian absolute value on $mathbb{Z}[x]$ and $mathbb{Z}[[x]],|.|$ is the completion of $mathbb{Z}[x],|.|$. The sequence $sum_{k=0}^K (-1)^k (1+x)^k$ is not Cauchy in $mathbb{Z}[x],|.|$ so it diverges. Replacing $|.|$ by $|sum_{n=N}^{N+m} a_n (1+x)^n|_2 = 2^{-N}$ then $sum_{k=0}^K (-1)^k (1+x)^k$ is Cauchy and its limit belongs to $mathbb{Z}[[1+x]],|.|_2$
    $endgroup$
    – reuns
    Jan 8 at 17:32












  • $begingroup$
    @reuns Thanks your explanation, but I just took first year graduate course on Algebra, and I don't know what do you mean by "completion" of a polynomial ring. In fact, I just did a quick research, but I think that beyond what I know allows.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:59












  • $begingroup$
    @onurcanbektas $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to its usual absolute value : adding to the space the limits of Cauchy sequences en.wikipedia.org/wiki/Complete_metric_space An absolute value is a metric compatible with the addition and the multiplication which guaranties those extend to the completed metric space.
    $endgroup$
    – reuns
    Jan 9 at 15:26










4




4




$begingroup$
Why do you assert that complicated sum is an element of $Bbb Z[[x]]$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 17:20






$begingroup$
Why do you assert that complicated sum is an element of $Bbb Z[[x]]$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 17:20














$begingroup$
Let $|sum_{n=N}^{N+m} a_n x^n| = 2^{-N}$ for $a_n in mathbb{Z}, a_Nne 0$. It is a non-archimedian absolute value on $mathbb{Z}[x]$ and $mathbb{Z}[[x]],|.|$ is the completion of $mathbb{Z}[x],|.|$. The sequence $sum_{k=0}^K (-1)^k (1+x)^k$ is not Cauchy in $mathbb{Z}[x],|.|$ so it diverges. Replacing $|.|$ by $|sum_{n=N}^{N+m} a_n (1+x)^n|_2 = 2^{-N}$ then $sum_{k=0}^K (-1)^k (1+x)^k$ is Cauchy and its limit belongs to $mathbb{Z}[[1+x]],|.|_2$
$endgroup$
– reuns
Jan 8 at 17:32






$begingroup$
Let $|sum_{n=N}^{N+m} a_n x^n| = 2^{-N}$ for $a_n in mathbb{Z}, a_Nne 0$. It is a non-archimedian absolute value on $mathbb{Z}[x]$ and $mathbb{Z}[[x]],|.|$ is the completion of $mathbb{Z}[x],|.|$. The sequence $sum_{k=0}^K (-1)^k (1+x)^k$ is not Cauchy in $mathbb{Z}[x],|.|$ so it diverges. Replacing $|.|$ by $|sum_{n=N}^{N+m} a_n (1+x)^n|_2 = 2^{-N}$ then $sum_{k=0}^K (-1)^k (1+x)^k$ is Cauchy and its limit belongs to $mathbb{Z}[[1+x]],|.|_2$
$endgroup$
– reuns
Jan 8 at 17:32














$begingroup$
@reuns Thanks your explanation, but I just took first year graduate course on Algebra, and I don't know what do you mean by "completion" of a polynomial ring. In fact, I just did a quick research, but I think that beyond what I know allows.
$endgroup$
– onurcanbektas
Jan 8 at 17:59






$begingroup$
@reuns Thanks your explanation, but I just took first year graduate course on Algebra, and I don't know what do you mean by "completion" of a polynomial ring. In fact, I just did a quick research, but I think that beyond what I know allows.
$endgroup$
– onurcanbektas
Jan 8 at 17:59














$begingroup$
@onurcanbektas $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to its usual absolute value : adding to the space the limits of Cauchy sequences en.wikipedia.org/wiki/Complete_metric_space An absolute value is a metric compatible with the addition and the multiplication which guaranties those extend to the completed metric space.
$endgroup$
– reuns
Jan 9 at 15:26






$begingroup$
@onurcanbektas $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to its usual absolute value : adding to the space the limits of Cauchy sequences en.wikipedia.org/wiki/Complete_metric_space An absolute value is a metric compatible with the addition and the multiplication which guaranties those extend to the completed metric space.
$endgroup$
– reuns
Jan 9 at 15:26












2 Answers
2






active

oldest

votes


















2












$begingroup$

Because
$$sum_k(-1)^k(1+x)^k=sum_{kgeq 0}(-1)^k + sum_{k geq 0}sum_{1leq rleq k}{kchoose r}x^r$$
is not convergent in $mathbb{Z}[[x]]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:26








  • 1




    $begingroup$
    What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
    $endgroup$
    – David Hill
    Jan 8 at 17:27












  • $begingroup$
    "What is the constant term ?": a good question :). Thanks for you answer.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:55












  • $begingroup$
    $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
    $endgroup$
    – reuns
    Jan 9 at 15:29










  • $begingroup$
    @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
    $endgroup$
    – David Hill
    Jan 9 at 16:04



















4












$begingroup$

Not that $ sum_k^infty (-1)^k(1+x)^k$ is not an element of $ mathbb{Z} [[x]]$. For example the constant term of $sum_k^infty (-1)^k(1+x)^k $ is not an integer numbr, indeed, $ sum_k^infty (-1)^k(1+0)^k $ is divergence.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Because
    $$sum_k(-1)^k(1+x)^k=sum_{kgeq 0}(-1)^k + sum_{k geq 0}sum_{1leq rleq k}{kchoose r}x^r$$
    is not convergent in $mathbb{Z}[[x]]$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:26








    • 1




      $begingroup$
      What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
      $endgroup$
      – David Hill
      Jan 8 at 17:27












    • $begingroup$
      "What is the constant term ?": a good question :). Thanks for you answer.
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:55












    • $begingroup$
      $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
      $endgroup$
      – reuns
      Jan 9 at 15:29










    • $begingroup$
      @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
      $endgroup$
      – David Hill
      Jan 9 at 16:04
















    2












    $begingroup$

    Because
    $$sum_k(-1)^k(1+x)^k=sum_{kgeq 0}(-1)^k + sum_{k geq 0}sum_{1leq rleq k}{kchoose r}x^r$$
    is not convergent in $mathbb{Z}[[x]]$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:26








    • 1




      $begingroup$
      What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
      $endgroup$
      – David Hill
      Jan 8 at 17:27












    • $begingroup$
      "What is the constant term ?": a good question :). Thanks for you answer.
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:55












    • $begingroup$
      $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
      $endgroup$
      – reuns
      Jan 9 at 15:29










    • $begingroup$
      @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
      $endgroup$
      – David Hill
      Jan 9 at 16:04














    2












    2








    2





    $begingroup$

    Because
    $$sum_k(-1)^k(1+x)^k=sum_{kgeq 0}(-1)^k + sum_{k geq 0}sum_{1leq rleq k}{kchoose r}x^r$$
    is not convergent in $mathbb{Z}[[x]]$.






    share|cite|improve this answer









    $endgroup$



    Because
    $$sum_k(-1)^k(1+x)^k=sum_{kgeq 0}(-1)^k + sum_{k geq 0}sum_{1leq rleq k}{kchoose r}x^r$$
    is not convergent in $mathbb{Z}[[x]]$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 17:22









    David HillDavid Hill

    9,4761619




    9,4761619












    • $begingroup$
      What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:26








    • 1




      $begingroup$
      What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
      $endgroup$
      – David Hill
      Jan 8 at 17:27












    • $begingroup$
      "What is the constant term ?": a good question :). Thanks for you answer.
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:55












    • $begingroup$
      $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
      $endgroup$
      – reuns
      Jan 9 at 15:29










    • $begingroup$
      @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
      $endgroup$
      – David Hill
      Jan 9 at 16:04


















    • $begingroup$
      What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:26








    • 1




      $begingroup$
      What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
      $endgroup$
      – David Hill
      Jan 8 at 17:27












    • $begingroup$
      "What is the constant term ?": a good question :). Thanks for you answer.
      $endgroup$
      – onurcanbektas
      Jan 8 at 17:55












    • $begingroup$
      $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
      $endgroup$
      – reuns
      Jan 9 at 15:29










    • $begingroup$
      @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
      $endgroup$
      – David Hill
      Jan 9 at 16:04
















    $begingroup$
    What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:26






    $begingroup$
    What do you mean by convergent in this setup ? Isn't each coefficient of $x^r$ an integer ?
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:26






    1




    1




    $begingroup$
    What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
    $endgroup$
    – David Hill
    Jan 8 at 17:27






    $begingroup$
    What is the constant term? What is the coefficient of $x^k$? The sum is nonsense. @onurcanbektas
    $endgroup$
    – David Hill
    Jan 8 at 17:27














    $begingroup$
    "What is the constant term ?": a good question :). Thanks for you answer.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:55






    $begingroup$
    "What is the constant term ?": a good question :). Thanks for you answer.
    $endgroup$
    – onurcanbektas
    Jan 8 at 17:55














    $begingroup$
    $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
    $endgroup$
    – reuns
    Jan 9 at 15:29




    $begingroup$
    $sum_{k=0}^infty (-1)^k(1+x)^k in mathbb{Z}[[1+x]]$ @DavidHill
    $endgroup$
    – reuns
    Jan 9 at 15:29












    $begingroup$
    @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
    $endgroup$
    – David Hill
    Jan 9 at 16:04




    $begingroup$
    @reuns What is your point? The problem regards finding inverses in $mathbb{Z}[[x]].
    $endgroup$
    – David Hill
    Jan 9 at 16:04











    4












    $begingroup$

    Not that $ sum_k^infty (-1)^k(1+x)^k$ is not an element of $ mathbb{Z} [[x]]$. For example the constant term of $sum_k^infty (-1)^k(1+x)^k $ is not an integer numbr, indeed, $ sum_k^infty (-1)^k(1+0)^k $ is divergence.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Not that $ sum_k^infty (-1)^k(1+x)^k$ is not an element of $ mathbb{Z} [[x]]$. For example the constant term of $sum_k^infty (-1)^k(1+x)^k $ is not an integer numbr, indeed, $ sum_k^infty (-1)^k(1+0)^k $ is divergence.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Not that $ sum_k^infty (-1)^k(1+x)^k$ is not an element of $ mathbb{Z} [[x]]$. For example the constant term of $sum_k^infty (-1)^k(1+x)^k $ is not an integer numbr, indeed, $ sum_k^infty (-1)^k(1+0)^k $ is divergence.






        share|cite|improve this answer









        $endgroup$



        Not that $ sum_k^infty (-1)^k(1+x)^k$ is not an element of $ mathbb{Z} [[x]]$. For example the constant term of $sum_k^infty (-1)^k(1+x)^k $ is not an integer numbr, indeed, $ sum_k^infty (-1)^k(1+0)^k $ is divergence.







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        answered Jan 8 at 17:30









        Es.RoEs.Ro

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