Let $R subseteq S$ be two local PIDs with the same field of fractions, then $R=S$.












1












$begingroup$


Let $R$ and $S$ be two local principal ideal domains with the same field of fractions $K$. I want to show that if $Rsubseteq S$ then $R=S$.



I will denote as $mathfrak{m}_R=(m_R)$ and $mathfrak{m}_S=(m_S)$ the unique maximal ideal of the local rings $R$ and $S$ respectively. Then every non-unit of $R$ is of the form $x=m_R^nu$ for some $ninmathbb{N} $ and some unit $uin R$. Same goes for $S$. I though of using the following Lemma to attack this problem.



$textbf{Lemma}$ $colon$ Let $R$ be a local PID with field of fractions $K$. Let $S$ be any local domain with $Rsubseteq Ssubseteq K$. If $mathfrak{m}_R subseteq mathfrak{m}_S$, then $R=S$



I know that for local $R$ it is true that its maximal ideal consists of all the non-units. Also, if $R$ is a PID then, $xin K$ implies that $xin R$ or $x^{-1} in R$ (or both). Let $xin mathfrak{m}_R$, thus $x^{-1} notin R$. Now $x^{-1}$ cannot be in $mathfrak{m}_S$ because then, since $xin mathfrak{m}_RRightarrow x in R Rightarrow xin S$ we would have that $1=xx^{-1} in mathfrak{m}_S$ which is a contradiction. How can i prove that in fact $x^{-1}$ can't be in $S$?






abstract-algebra ring-theory commutative-algebra localization







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $R$ and $S$ be two local principal ideal domains with the same field of fractions $K$. I want to show that if $Rsubseteq S$ then $R=S$.



    I will denote as $mathfrak{m}_R=(m_R)$ and $mathfrak{m}_S=(m_S)$ the unique maximal ideal of the local rings $R$ and $S$ respectively. Then every non-unit of $R$ is of the form $x=m_R^nu$ for some $ninmathbb{N} $ and some unit $uin R$. Same goes for $S$. I though of using the following Lemma to attack this problem.



    $textbf{Lemma}$ $colon$ Let $R$ be a local PID with field of fractions $K$. Let $S$ be any local domain with $Rsubseteq Ssubseteq K$. If $mathfrak{m}_R subseteq mathfrak{m}_S$, then $R=S$



    I know that for local $R$ it is true that its maximal ideal consists of all the non-units. Also, if $R$ is a PID then, $xin K$ implies that $xin R$ or $x^{-1} in R$ (or both). Let $xin mathfrak{m}_R$, thus $x^{-1} notin R$. Now $x^{-1}$ cannot be in $mathfrak{m}_S$ because then, since $xin mathfrak{m}_RRightarrow x in R Rightarrow xin S$ we would have that $1=xx^{-1} in mathfrak{m}_S$ which is a contradiction. How can i prove that in fact $x^{-1}$ can't be in $S$?






    abstract-algebra ring-theory commutative-algebra localization







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $R$ and $S$ be two local principal ideal domains with the same field of fractions $K$. I want to show that if $Rsubseteq S$ then $R=S$.



      I will denote as $mathfrak{m}_R=(m_R)$ and $mathfrak{m}_S=(m_S)$ the unique maximal ideal of the local rings $R$ and $S$ respectively. Then every non-unit of $R$ is of the form $x=m_R^nu$ for some $ninmathbb{N} $ and some unit $uin R$. Same goes for $S$. I though of using the following Lemma to attack this problem.



      $textbf{Lemma}$ $colon$ Let $R$ be a local PID with field of fractions $K$. Let $S$ be any local domain with $Rsubseteq Ssubseteq K$. If $mathfrak{m}_R subseteq mathfrak{m}_S$, then $R=S$



      I know that for local $R$ it is true that its maximal ideal consists of all the non-units. Also, if $R$ is a PID then, $xin K$ implies that $xin R$ or $x^{-1} in R$ (or both). Let $xin mathfrak{m}_R$, thus $x^{-1} notin R$. Now $x^{-1}$ cannot be in $mathfrak{m}_S$ because then, since $xin mathfrak{m}_RRightarrow x in R Rightarrow xin S$ we would have that $1=xx^{-1} in mathfrak{m}_S$ which is a contradiction. How can i prove that in fact $x^{-1}$ can't be in $S$?






      abstract-algebra ring-theory commutative-algebra localization







      share|cite|improve this question











      $endgroup$




      Let $R$ and $S$ be two local principal ideal domains with the same field of fractions $K$. I want to show that if $Rsubseteq S$ then $R=S$.



      I will denote as $mathfrak{m}_R=(m_R)$ and $mathfrak{m}_S=(m_S)$ the unique maximal ideal of the local rings $R$ and $S$ respectively. Then every non-unit of $R$ is of the form $x=m_R^nu$ for some $ninmathbb{N} $ and some unit $uin R$. Same goes for $S$. I though of using the following Lemma to attack this problem.



      $textbf{Lemma}$ $colon$ Let $R$ be a local PID with field of fractions $K$. Let $S$ be any local domain with $Rsubseteq Ssubseteq K$. If $mathfrak{m}_R subseteq mathfrak{m}_S$, then $R=S$



      I know that for local $R$ it is true that its maximal ideal consists of all the non-units. Also, if $R$ is a PID then, $xin K$ implies that $xin R$ or $x^{-1} in R$ (or both). Let $xin mathfrak{m}_R$, thus $x^{-1} notin R$. Now $x^{-1}$ cannot be in $mathfrak{m}_S$ because then, since $xin mathfrak{m}_RRightarrow x in R Rightarrow xin S$ we would have that $1=xx^{-1} in mathfrak{m}_S$ which is a contradiction. How can i prove that in fact $x^{-1}$ can't be in $S$?






      abstract-algebra ring-theory commutative-algebra localization




      abstract-algebra ring-theory commutative-algebra localization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 16:00









      Badam Baplan

      4,531722




      4,531722










      asked Dec 8 '18 at 12:52









      CorneliusCornelius

      1957




      1957






















          2 Answers
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          1












          $begingroup$

          First show that $S=T^{-1}R$ where $T$ is a multiplicative subset of $R$. This is true whenever $R$ is a PID. Now use the fact that R is local PID to show that $T^{-1}R = R $ or $K$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 10:28





















          0












          $begingroup$

          This is not true: Take $S = K$ the field of fractions of $R$.



          This is the only counter-example: if $R subseteq S subseteq K$ and $S$ contains an element of the form $u m_{R}^{n}$ with $u in R^{times}$ and $n < 0$, it follows that $S = K$. Thus either $S = R$ or $S = K$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
            $endgroup$
            – Cornelius
            Dec 8 '18 at 13:28












          • $begingroup$
            I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
            $endgroup$
            – Cornelius
            Dec 8 '18 at 17:11












          • $begingroup$
            Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
            $endgroup$
            – barto
            Dec 8 '18 at 17:15













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          2 Answers
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          2 Answers
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          active

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          active

          oldest

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          1












          $begingroup$

          First show that $S=T^{-1}R$ where $T$ is a multiplicative subset of $R$. This is true whenever $R$ is a PID. Now use the fact that R is local PID to show that $T^{-1}R = R $ or $K$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 10:28


















          1












          $begingroup$

          First show that $S=T^{-1}R$ where $T$ is a multiplicative subset of $R$. This is true whenever $R$ is a PID. Now use the fact that R is local PID to show that $T^{-1}R = R $ or $K$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 10:28
















          1












          1








          1





          $begingroup$

          First show that $S=T^{-1}R$ where $T$ is a multiplicative subset of $R$. This is true whenever $R$ is a PID. Now use the fact that R is local PID to show that $T^{-1}R = R $ or $K$






          share|cite|improve this answer









          $endgroup$



          First show that $S=T^{-1}R$ where $T$ is a multiplicative subset of $R$. This is true whenever $R$ is a PID. Now use the fact that R is local PID to show that $T^{-1}R = R $ or $K$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 10:08









          Soumik GhoshSoumik Ghosh

          19610




          19610












          • $begingroup$
            Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 10:28




















          • $begingroup$
            Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 10:28


















          $begingroup$
          Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
          $endgroup$
          – Cornelius
          Dec 11 '18 at 10:28






          $begingroup$
          Thanks! Your answer was enlightening. The intuition i thought of is that if an element $xin mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$.
          $endgroup$
          – Cornelius
          Dec 11 '18 at 10:28













          0












          $begingroup$

          This is not true: Take $S = K$ the field of fractions of $R$.



          This is the only counter-example: if $R subseteq S subseteq K$ and $S$ contains an element of the form $u m_{R}^{n}$ with $u in R^{times}$ and $n < 0$, it follows that $S = K$. Thus either $S = R$ or $S = K$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
            $endgroup$
            – Cornelius
            Dec 8 '18 at 13:28












          • $begingroup$
            I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
            $endgroup$
            – Cornelius
            Dec 8 '18 at 17:11












          • $begingroup$
            Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
            $endgroup$
            – barto
            Dec 8 '18 at 17:15


















          0












          $begingroup$

          This is not true: Take $S = K$ the field of fractions of $R$.



          This is the only counter-example: if $R subseteq S subseteq K$ and $S$ contains an element of the form $u m_{R}^{n}$ with $u in R^{times}$ and $n < 0$, it follows that $S = K$. Thus either $S = R$ or $S = K$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
            $endgroup$
            – Cornelius
            Dec 8 '18 at 13:28












          • $begingroup$
            I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
            $endgroup$
            – Cornelius
            Dec 8 '18 at 17:11












          • $begingroup$
            Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
            $endgroup$
            – barto
            Dec 8 '18 at 17:15
















          0












          0








          0





          $begingroup$

          This is not true: Take $S = K$ the field of fractions of $R$.



          This is the only counter-example: if $R subseteq S subseteq K$ and $S$ contains an element of the form $u m_{R}^{n}$ with $u in R^{times}$ and $n < 0$, it follows that $S = K$. Thus either $S = R$ or $S = K$.






          share|cite|improve this answer











          $endgroup$



          This is not true: Take $S = K$ the field of fractions of $R$.



          This is the only counter-example: if $R subseteq S subseteq K$ and $S$ contains an element of the form $u m_{R}^{n}$ with $u in R^{times}$ and $n < 0$, it follows that $S = K$. Thus either $S = R$ or $S = K$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 13:38

























          answered Dec 8 '18 at 13:13









          bartobarto

          13.7k32682




          13.7k32682








          • 1




            $begingroup$
            The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
            $endgroup$
            – Cornelius
            Dec 8 '18 at 13:28












          • $begingroup$
            I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
            $endgroup$
            – Cornelius
            Dec 8 '18 at 17:11












          • $begingroup$
            Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
            $endgroup$
            – barto
            Dec 8 '18 at 17:15
















          • 1




            $begingroup$
            The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
            $endgroup$
            – Cornelius
            Dec 8 '18 at 13:28












          • $begingroup$
            I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
            $endgroup$
            – Cornelius
            Dec 8 '18 at 17:11












          • $begingroup$
            Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
            $endgroup$
            – barto
            Dec 8 '18 at 17:15










          1




          1




          $begingroup$
          The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
          $endgroup$
          – Cornelius
          Dec 8 '18 at 13:28






          $begingroup$
          The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $Rsubseteq S subsetneq K$
          $endgroup$
          – Cornelius
          Dec 8 '18 at 13:28














          $begingroup$
          I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
          $endgroup$
          – Cornelius
          Dec 8 '18 at 17:11






          $begingroup$
          I can't see why an element of the form $um_R^n$ with $u in R^{times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ?
          $endgroup$
          – Cornelius
          Dec 8 '18 at 17:11














          $begingroup$
          Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
          $endgroup$
          – barto
          Dec 8 '18 at 17:15






          $begingroup$
          Every element of $K$ is of the form $u m_R^n$ with $n in mathbb Z$. If $m_R^{n_0} in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} cdot (m_R^{n_0})^{N} in R cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 geq 0$.
          $endgroup$
          – barto
          Dec 8 '18 at 17:15




















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