Characterizing von Neumann regularity
$begingroup$
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
asked Dec 30 '18 at 15:45
rayray
809
$endgroup$
add a comment |
$begingroup$
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
asked Dec 30 '18 at 15:45
rayray
809
$endgroup$
add a comment |
$begingroup$
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
asked Dec 30 '18 at 15:45
rayray
809
$endgroup$
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
asked Dec 30 '18 at 15:45
rayray
809
asked Dec 30 '18 at 15:45
rayray
809
asked Dec 30 '18 at 15:45
rayray
809
asked Dec 30 '18 at 15:45
rayray
809
asked Dec 30 '18 at 15:45
rayray
809
809
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(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.
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
$endgroup$
$begingroup$
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
add a comment |
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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.
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
(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.
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
(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.
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
$endgroup$
(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.
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
answered Dec 31 '18 at 1:01
Badam BaplanBadam Baplan
4,611722
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
add a comment |
$begingroup$
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
add a comment |
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