Is it possible for an integral domain to have an ideal that cannot be generated by a countable set?












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This question came up when I was working on the following problem



Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:



i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.



ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.



I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.






ring-theory principal-ideal-domains integral-domain







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    0












    $begingroup$


    This question came up when I was working on the following problem



    Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:



    i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.



    ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.



    I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.






    ring-theory principal-ideal-domains integral-domain







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      This question came up when I was working on the following problem



      Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:



      i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.



      ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.



      I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.






      ring-theory principal-ideal-domains integral-domain







      share|cite|improve this question









      $endgroup$




      This question came up when I was working on the following problem



      Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:



      i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.



      ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.



      I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.






      ring-theory principal-ideal-domains integral-domain




      ring-theory principal-ideal-domains integral-domain






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 3 '18 at 10:12









      DanielDaniel

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      162






















          1 Answer
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          $begingroup$

          The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.



          It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:14










          • $begingroup$
            Sorry I meant Ideal* at the beginning
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:24











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          1 Answer
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          1 Answer
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          active

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          0












          $begingroup$

          The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.



          It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:14










          • $begingroup$
            Sorry I meant Ideal* at the beginning
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:24
















          0












          $begingroup$

          The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.



          It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:14










          • $begingroup$
            Sorry I meant Ideal* at the beginning
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:24














          0












          0








          0





          $begingroup$

          The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.



          It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.






          share|cite|improve this answer









          $endgroup$



          The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.



          It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 11:58









          rschwiebrschwieb

          107k12102250




          107k12102250












          • $begingroup$
            I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:14










          • $begingroup$
            Sorry I meant Ideal* at the beginning
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:24


















          • $begingroup$
            I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:14










          • $begingroup$
            Sorry I meant Ideal* at the beginning
            $endgroup$
            – Daniel
            Dec 3 '18 at 12:24
















          $begingroup$
          I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
          $endgroup$
          – Daniel
          Dec 3 '18 at 12:14




          $begingroup$
          I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
          $endgroup$
          – Daniel
          Dec 3 '18 at 12:14












          $begingroup$
          Sorry I meant Ideal* at the beginning
          $endgroup$
          – Daniel
          Dec 3 '18 at 12:24




          $begingroup$
          Sorry I meant Ideal* at the beginning
          $endgroup$
          – Daniel
          Dec 3 '18 at 12:24


















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