Characterizing von Neumann regularity












0












$begingroup$


Let $R$ be a commutative unital ring. I want to show the following equivalences:



(1) $R$ is zero-dimensional and reduced,



(2) every ideal in $R$ is radical,



(3) $R$ is von Neumann regular.



See wikipedia for the definitions. The only step I am missing is (1) to (2). I only see that the zero ideal (0) is radical. Any idea would be welcome. Or a reference to a textbook containing this proof. Thanks.






commutative-algebra







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $R$ be a commutative unital ring. I want to show the following equivalences:



    (1) $R$ is zero-dimensional and reduced,



    (2) every ideal in $R$ is radical,



    (3) $R$ is von Neumann regular.



    See wikipedia for the definitions. The only step I am missing is (1) to (2). I only see that the zero ideal (0) is radical. Any idea would be welcome. Or a reference to a textbook containing this proof. Thanks.






    commutative-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $R$ be a commutative unital ring. I want to show the following equivalences:



      (1) $R$ is zero-dimensional and reduced,



      (2) every ideal in $R$ is radical,



      (3) $R$ is von Neumann regular.



      See wikipedia for the definitions. The only step I am missing is (1) to (2). I only see that the zero ideal (0) is radical. Any idea would be welcome. Or a reference to a textbook containing this proof. Thanks.






      commutative-algebra







      share|cite|improve this question









      $endgroup$




      Let $R$ be a commutative unital ring. I want to show the following equivalences:



      (1) $R$ is zero-dimensional and reduced,



      (2) every ideal in $R$ is radical,



      (3) $R$ is von Neumann regular.



      See wikipedia for the definitions. The only step I am missing is (1) to (2). I only see that the zero ideal (0) is radical. Any idea would be welcome. Or a reference to a textbook containing this proof. Thanks.






      commutative-algebra




      commutative-algebra






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 15:45









      rayray

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          1 Answer
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          $begingroup$

          (2) <-> (3) is pretty much trivial. The key observations are that $x^2R$ being radical implies that $x$ is a Von Neumann Regular (VNR) element, and that $x$ being a VNR element implies that $x^n$ divides $x$.



          So if you're good with (3) -> (1), we might as well just show (1) -> (3) directly.



          For good measure we'll do it two ways.



          First Way: (This is the way I've seen most often). We start by considering the module $xR / x^2R$. If we can show that this is the zero module, then we'll have $xR = x^2R$ — end of proof. Being zero is a local property, so in fact we only need to show that $(xR / x^2R)_mathfrak{m} = 0$ for each maximal ideal of $R$. Since localization distributes over quotients, we're actually looking at $xR_mathfrak{m}/x^2R_mathfrak{m}$. By assumption that $R$ is $0$-dimensional and reduced, $R_mathfrak{m}$ is a field. Hence the image of $x$ in $R_mathfrak{m}$ is either $0$ or a unit. In either case, certainly $xR_mathfrak{m}/x^2R_mathfrak{m} = 0$.



          Second Way: Note that $x in R$ is a VNR element iff $xR$ is a direct summand of $R$. Indeed, if $xR$ is a direct summand of $R$, then $xR + J = R$ with $xR cap J = 0$, so we can write $xa + j = 1$ for some $a in R, j in J$. Then $xj = 0$ and $x^2a = x$ follows. Conversely, if $x^2a = x$, then $axR = xR$ and $ax$ is an idempotent. When $e in R$ is an idempotent, one always has $eR oplus (1-e)R = R$.



          With this in mind, our approach will be to show that $xR oplus Ann(x) = R$ for any element $x$ of a reduced $0$-dimensional ring. First check that $xR cap Ann(x) = 0$ because $R$ is reduced (indeed this characterizes reduced rings). Then note $xR + Ann(x)$ is never contained in a minimal prime of a reduced ring, as $PR_P = 0$. Since $R$ is assumed to be $0$-dimensional, moreover $xR + Ann(x)$ is not contained in any prime ideal, and must be the entire ring.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
            $endgroup$
            – ray
            Dec 31 '18 at 9:28










          • $begingroup$
            I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
            $endgroup$
            – ray
            Dec 31 '18 at 12:14













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          $begingroup$

          (2) <-> (3) is pretty much trivial. The key observations are that $x^2R$ being radical implies that $x$ is a Von Neumann Regular (VNR) element, and that $x$ being a VNR element implies that $x^n$ divides $x$.



          So if you're good with (3) -> (1), we might as well just show (1) -> (3) directly.



          For good measure we'll do it two ways.



          First Way: (This is the way I've seen most often). We start by considering the module $xR / x^2R$. If we can show that this is the zero module, then we'll have $xR = x^2R$ — end of proof. Being zero is a local property, so in fact we only need to show that $(xR / x^2R)_mathfrak{m} = 0$ for each maximal ideal of $R$. Since localization distributes over quotients, we're actually looking at $xR_mathfrak{m}/x^2R_mathfrak{m}$. By assumption that $R$ is $0$-dimensional and reduced, $R_mathfrak{m}$ is a field. Hence the image of $x$ in $R_mathfrak{m}$ is either $0$ or a unit. In either case, certainly $xR_mathfrak{m}/x^2R_mathfrak{m} = 0$.



          Second Way: Note that $x in R$ is a VNR element iff $xR$ is a direct summand of $R$. Indeed, if $xR$ is a direct summand of $R$, then $xR + J = R$ with $xR cap J = 0$, so we can write $xa + j = 1$ for some $a in R, j in J$. Then $xj = 0$ and $x^2a = x$ follows. Conversely, if $x^2a = x$, then $axR = xR$ and $ax$ is an idempotent. When $e in R$ is an idempotent, one always has $eR oplus (1-e)R = R$.



          With this in mind, our approach will be to show that $xR oplus Ann(x) = R$ for any element $x$ of a reduced $0$-dimensional ring. First check that $xR cap Ann(x) = 0$ because $R$ is reduced (indeed this characterizes reduced rings). Then note $xR + Ann(x)$ is never contained in a minimal prime of a reduced ring, as $PR_P = 0$. Since $R$ is assumed to be $0$-dimensional, moreover $xR + Ann(x)$ is not contained in any prime ideal, and must be the entire ring.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
            $endgroup$
            – ray
            Dec 31 '18 at 9:28










          • $begingroup$
            I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
            $endgroup$
            – ray
            Dec 31 '18 at 12:14


















          1












          $begingroup$

          (2) <-> (3) is pretty much trivial. The key observations are that $x^2R$ being radical implies that $x$ is a Von Neumann Regular (VNR) element, and that $x$ being a VNR element implies that $x^n$ divides $x$.



          So if you're good with (3) -> (1), we might as well just show (1) -> (3) directly.



          For good measure we'll do it two ways.



          First Way: (This is the way I've seen most often). We start by considering the module $xR / x^2R$. If we can show that this is the zero module, then we'll have $xR = x^2R$ — end of proof. Being zero is a local property, so in fact we only need to show that $(xR / x^2R)_mathfrak{m} = 0$ for each maximal ideal of $R$. Since localization distributes over quotients, we're actually looking at $xR_mathfrak{m}/x^2R_mathfrak{m}$. By assumption that $R$ is $0$-dimensional and reduced, $R_mathfrak{m}$ is a field. Hence the image of $x$ in $R_mathfrak{m}$ is either $0$ or a unit. In either case, certainly $xR_mathfrak{m}/x^2R_mathfrak{m} = 0$.



          Second Way: Note that $x in R$ is a VNR element iff $xR$ is a direct summand of $R$. Indeed, if $xR$ is a direct summand of $R$, then $xR + J = R$ with $xR cap J = 0$, so we can write $xa + j = 1$ for some $a in R, j in J$. Then $xj = 0$ and $x^2a = x$ follows. Conversely, if $x^2a = x$, then $axR = xR$ and $ax$ is an idempotent. When $e in R$ is an idempotent, one always has $eR oplus (1-e)R = R$.



          With this in mind, our approach will be to show that $xR oplus Ann(x) = R$ for any element $x$ of a reduced $0$-dimensional ring. First check that $xR cap Ann(x) = 0$ because $R$ is reduced (indeed this characterizes reduced rings). Then note $xR + Ann(x)$ is never contained in a minimal prime of a reduced ring, as $PR_P = 0$. Since $R$ is assumed to be $0$-dimensional, moreover $xR + Ann(x)$ is not contained in any prime ideal, and must be the entire ring.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
            $endgroup$
            – ray
            Dec 31 '18 at 9:28










          • $begingroup$
            I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
            $endgroup$
            – ray
            Dec 31 '18 at 12:14
















          1












          1








          1





          $begingroup$

          (2) <-> (3) is pretty much trivial. The key observations are that $x^2R$ being radical implies that $x$ is a Von Neumann Regular (VNR) element, and that $x$ being a VNR element implies that $x^n$ divides $x$.



          So if you're good with (3) -> (1), we might as well just show (1) -> (3) directly.



          For good measure we'll do it two ways.



          First Way: (This is the way I've seen most often). We start by considering the module $xR / x^2R$. If we can show that this is the zero module, then we'll have $xR = x^2R$ — end of proof. Being zero is a local property, so in fact we only need to show that $(xR / x^2R)_mathfrak{m} = 0$ for each maximal ideal of $R$. Since localization distributes over quotients, we're actually looking at $xR_mathfrak{m}/x^2R_mathfrak{m}$. By assumption that $R$ is $0$-dimensional and reduced, $R_mathfrak{m}$ is a field. Hence the image of $x$ in $R_mathfrak{m}$ is either $0$ or a unit. In either case, certainly $xR_mathfrak{m}/x^2R_mathfrak{m} = 0$.



          Second Way: Note that $x in R$ is a VNR element iff $xR$ is a direct summand of $R$. Indeed, if $xR$ is a direct summand of $R$, then $xR + J = R$ with $xR cap J = 0$, so we can write $xa + j = 1$ for some $a in R, j in J$. Then $xj = 0$ and $x^2a = x$ follows. Conversely, if $x^2a = x$, then $axR = xR$ and $ax$ is an idempotent. When $e in R$ is an idempotent, one always has $eR oplus (1-e)R = R$.



          With this in mind, our approach will be to show that $xR oplus Ann(x) = R$ for any element $x$ of a reduced $0$-dimensional ring. First check that $xR cap Ann(x) = 0$ because $R$ is reduced (indeed this characterizes reduced rings). Then note $xR + Ann(x)$ is never contained in a minimal prime of a reduced ring, as $PR_P = 0$. Since $R$ is assumed to be $0$-dimensional, moreover $xR + Ann(x)$ is not contained in any prime ideal, and must be the entire ring.






          share|cite|improve this answer









          $endgroup$



          (2) <-> (3) is pretty much trivial. The key observations are that $x^2R$ being radical implies that $x$ is a Von Neumann Regular (VNR) element, and that $x$ being a VNR element implies that $x^n$ divides $x$.



          So if you're good with (3) -> (1), we might as well just show (1) -> (3) directly.



          For good measure we'll do it two ways.



          First Way: (This is the way I've seen most often). We start by considering the module $xR / x^2R$. If we can show that this is the zero module, then we'll have $xR = x^2R$ — end of proof. Being zero is a local property, so in fact we only need to show that $(xR / x^2R)_mathfrak{m} = 0$ for each maximal ideal of $R$. Since localization distributes over quotients, we're actually looking at $xR_mathfrak{m}/x^2R_mathfrak{m}$. By assumption that $R$ is $0$-dimensional and reduced, $R_mathfrak{m}$ is a field. Hence the image of $x$ in $R_mathfrak{m}$ is either $0$ or a unit. In either case, certainly $xR_mathfrak{m}/x^2R_mathfrak{m} = 0$.



          Second Way: Note that $x in R$ is a VNR element iff $xR$ is a direct summand of $R$. Indeed, if $xR$ is a direct summand of $R$, then $xR + J = R$ with $xR cap J = 0$, so we can write $xa + j = 1$ for some $a in R, j in J$. Then $xj = 0$ and $x^2a = x$ follows. Conversely, if $x^2a = x$, then $axR = xR$ and $ax$ is an idempotent. When $e in R$ is an idempotent, one always has $eR oplus (1-e)R = R$.



          With this in mind, our approach will be to show that $xR oplus Ann(x) = R$ for any element $x$ of a reduced $0$-dimensional ring. First check that $xR cap Ann(x) = 0$ because $R$ is reduced (indeed this characterizes reduced rings). Then note $xR + Ann(x)$ is never contained in a minimal prime of a reduced ring, as $PR_P = 0$. Since $R$ is assumed to be $0$-dimensional, moreover $xR + Ann(x)$ is not contained in any prime ideal, and must be the entire ring.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 1:01









          Badam BaplanBadam Baplan

          4,611722




          4,611722












          • $begingroup$
            Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
            $endgroup$
            – ray
            Dec 31 '18 at 9:28










          • $begingroup$
            I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
            $endgroup$
            – ray
            Dec 31 '18 at 12:14




















          • $begingroup$
            Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
            $endgroup$
            – ray
            Dec 31 '18 at 9:28










          • $begingroup$
            I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
            $endgroup$
            – ray
            Dec 31 '18 at 12:14


















          $begingroup$
          Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
          $endgroup$
          – ray
          Dec 31 '18 at 9:28




          $begingroup$
          Could one prove the assertion that $xR+Ann(x)$ is not contained in a minimal prime without using localisations? (I am not yet familiar with this power tool in algebra; I am an analyst).
          $endgroup$
          – ray
          Dec 31 '18 at 9:28












          $begingroup$
          I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
          $endgroup$
          – ray
          Dec 31 '18 at 12:14






          $begingroup$
          I think one can proceed as follows: use that the Krull dimension of $R$ is $0$ if and only if for all $ain R$ there is $yin R$ and $nin N$ such that $a^n(1+ay)=0$. Since $aRcap Ann(a)=0$ we get that $n=1$.
          $endgroup$
          – ray
          Dec 31 '18 at 12:14




















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