Factorization in the field of fractions of a local PID.











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Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.



I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.






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    up vote
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    favorite












    Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.



    I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.






    abstract-algebra ring-theory commutative-algebra







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.



      I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.






      abstract-algebra ring-theory commutative-algebra







      share|cite|improve this question













      Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.



      I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.






      abstract-algebra ring-theory commutative-algebra




      abstract-algebra ring-theory commutative-algebra






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      asked 12 hours ago









      tangentbundle

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          Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.



          If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
          $$
          frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
          $$






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            accepted










            Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.



            If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
            $$
            frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
            $$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.



              If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
              $$
              frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
              $$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.



                If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
                $$
                frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
                $$






                share|cite|improve this answer












                Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.



                If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
                $$
                frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 12 hours ago









                egreg

                173k1383197




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