If $R$ is a Boolean ring with $mid R mid > 2$, determine all the zero divisors of $R$.
If $R$ is a Boolean ring with $mid R mid > 2$, determine all the zero divisors of $R$.
My attempt: Let $a, bin R$, $a neq 0$ and $ab = 0$. How do I prove that $b = 0$?
abstract-algebra ring-theory
add a comment |
If $R$ is a Boolean ring with $mid R mid > 2$, determine all the zero divisors of $R$.
My attempt: Let $a, bin R$, $a neq 0$ and $ab = 0$. How do I prove that $b = 0$?
abstract-algebra ring-theory
Problem given: find zero divisors of this ring $R$. Approach chosen: suppose $b$ is a zero divisor and prove it is zero. If you look at this summary of what happened, I think you'll agree you need to think about what you are doing a little more thoroughly.
– rschwieb
Apr 29 '15 at 17:30
add a comment |
If $R$ is a Boolean ring with $mid R mid > 2$, determine all the zero divisors of $R$.
My attempt: Let $a, bin R$, $a neq 0$ and $ab = 0$. How do I prove that $b = 0$?
abstract-algebra ring-theory
If $R$ is a Boolean ring with $mid R mid > 2$, determine all the zero divisors of $R$.
My attempt: Let $a, bin R$, $a neq 0$ and $ab = 0$. How do I prove that $b = 0$?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Apr 29 '15 at 16:13
In78
424311
424311
Problem given: find zero divisors of this ring $R$. Approach chosen: suppose $b$ is a zero divisor and prove it is zero. If you look at this summary of what happened, I think you'll agree you need to think about what you are doing a little more thoroughly.
– rschwieb
Apr 29 '15 at 17:30
add a comment |
Problem given: find zero divisors of this ring $R$. Approach chosen: suppose $b$ is a zero divisor and prove it is zero. If you look at this summary of what happened, I think you'll agree you need to think about what you are doing a little more thoroughly.
– rschwieb
Apr 29 '15 at 17:30
Problem given: find zero divisors of this ring $R$. Approach chosen: suppose $b$ is a zero divisor and prove it is zero. If you look at this summary of what happened, I think you'll agree you need to think about what you are doing a little more thoroughly.
– rschwieb
Apr 29 '15 at 17:30
Problem given: find zero divisors of this ring $R$. Approach chosen: suppose $b$ is a zero divisor and prove it is zero. If you look at this summary of what happened, I think you'll agree you need to think about what you are doing a little more thoroughly.
– rschwieb
Apr 29 '15 at 17:30
add a comment |
1 Answer
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Since $R$ is a Boolean ring, we have $a^2 = a$ for all $a in R$; since $vert R vert > 2$, there must exist an element $a in R$ with $a ne 0$, $a ne 1$.
Then $1 - a ne 0$, but
$a(1 - a) = a - a^2 = a - a = 0, tag{1}$
which shows that every $a in R setminus {0, 1 }$ is a zero divisor. Since $0$ and $1$ are not zero divisors, the set
$N = R setminus {0, 1 } ne emptyset tag{2}$
is precisely the set of all zero divisors of $R$.
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
add a comment |
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1 Answer
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1 Answer
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active
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votes
Since $R$ is a Boolean ring, we have $a^2 = a$ for all $a in R$; since $vert R vert > 2$, there must exist an element $a in R$ with $a ne 0$, $a ne 1$.
Then $1 - a ne 0$, but
$a(1 - a) = a - a^2 = a - a = 0, tag{1}$
which shows that every $a in R setminus {0, 1 }$ is a zero divisor. Since $0$ and $1$ are not zero divisors, the set
$N = R setminus {0, 1 } ne emptyset tag{2}$
is precisely the set of all zero divisors of $R$.
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
add a comment |
Since $R$ is a Boolean ring, we have $a^2 = a$ for all $a in R$; since $vert R vert > 2$, there must exist an element $a in R$ with $a ne 0$, $a ne 1$.
Then $1 - a ne 0$, but
$a(1 - a) = a - a^2 = a - a = 0, tag{1}$
which shows that every $a in R setminus {0, 1 }$ is a zero divisor. Since $0$ and $1$ are not zero divisors, the set
$N = R setminus {0, 1 } ne emptyset tag{2}$
is precisely the set of all zero divisors of $R$.
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
add a comment |
Since $R$ is a Boolean ring, we have $a^2 = a$ for all $a in R$; since $vert R vert > 2$, there must exist an element $a in R$ with $a ne 0$, $a ne 1$.
Then $1 - a ne 0$, but
$a(1 - a) = a - a^2 = a - a = 0, tag{1}$
which shows that every $a in R setminus {0, 1 }$ is a zero divisor. Since $0$ and $1$ are not zero divisors, the set
$N = R setminus {0, 1 } ne emptyset tag{2}$
is precisely the set of all zero divisors of $R$.
Since $R$ is a Boolean ring, we have $a^2 = a$ for all $a in R$; since $vert R vert > 2$, there must exist an element $a in R$ with $a ne 0$, $a ne 1$.
Then $1 - a ne 0$, but
$a(1 - a) = a - a^2 = a - a = 0, tag{1}$
which shows that every $a in R setminus {0, 1 }$ is a zero divisor. Since $0$ and $1$ are not zero divisors, the set
$N = R setminus {0, 1 } ne emptyset tag{2}$
is precisely the set of all zero divisors of $R$.
answered Apr 29 '15 at 16:26
Robert Lewis
43.2k22863
43.2k22863
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
add a comment |
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
Does a boolean ring always contain an identity? @Robert Lewis
– In78
May 1 '15 at 17:21
add a comment |
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Problem given: find zero divisors of this ring $R$. Approach chosen: suppose $b$ is a zero divisor and prove it is zero. If you look at this summary of what happened, I think you'll agree you need to think about what you are doing a little more thoroughly.
– rschwieb
Apr 29 '15 at 17:30