Do these matrix rings have non-zero elements that are neither units nor zero divisors?
Multi tool use
$begingroup$
Let $R$ be a commutative ring (with $1$) and $R^{n times n}$ be the ring of $n times n$ matrices with entries in $R$.
In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question:
Is every non-zero element of $R^{n times n}$ a zero divisor or a unit as well?
We know that if $A in R^{n times n}$, then $AC=CA=mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix.
This means that if $mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n times n}$ (since $A^{-1}=(mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n times n}$, then $mathrm{det}(A)$ is a unit.
I would like to know if one can show $0 not= A in R^{n times n}$ is a zero divisor if $mathrm{det}(A)$ is zero or a zero divisor.
Things to consider:
1) This is true when $R=mathbb{F}$ a field. Since over a field (no zero divisors) and if $mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x|0|cdots|0]$ gives us a right zero divisor $AB=0$.
2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example:
$$A=begin{bmatrix} 1 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 end{bmatrix} qquad mathrm{implies} qquad mathrm{classical;adjoint} = 0 $$
(All $2 times 2$ sub-determinants are zero.)
3) This is true when $R$ is finite (since $R^{n times n}$ would be finite as well).
4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$
and construct the diagonal matrix $D = mathrm{diag}(r,1,dots,1)$ (this is non-zero, not a zero divisor, and is not a unit).
Edit: Not totally unrelated...
https://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor
Edit: One more thing to consider...
5) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations):
$$ A = begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix} qquad Longrightarrow qquad mathrm{classical;adjoint} = C = begin{bmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} end{bmatrix} $$
Thus if $mathrm{det}(A)b=0$ for some $b not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC not=0$ and so $A(Cb)=mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor.
linear-algebra abstract-algebra matrices ring-theory
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring (with $1$) and $R^{n times n}$ be the ring of $n times n$ matrices with entries in $R$.
In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question:
Is every non-zero element of $R^{n times n}$ a zero divisor or a unit as well?
We know that if $A in R^{n times n}$, then $AC=CA=mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix.
This means that if $mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n times n}$ (since $A^{-1}=(mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n times n}$, then $mathrm{det}(A)$ is a unit.
I would like to know if one can show $0 not= A in R^{n times n}$ is a zero divisor if $mathrm{det}(A)$ is zero or a zero divisor.
Things to consider:
1) This is true when $R=mathbb{F}$ a field. Since over a field (no zero divisors) and if $mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x|0|cdots|0]$ gives us a right zero divisor $AB=0$.
2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example:
$$A=begin{bmatrix} 1 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 end{bmatrix} qquad mathrm{implies} qquad mathrm{classical;adjoint} = 0 $$
(All $2 times 2$ sub-determinants are zero.)
3) This is true when $R$ is finite (since $R^{n times n}$ would be finite as well).
4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$
and construct the diagonal matrix $D = mathrm{diag}(r,1,dots,1)$ (this is non-zero, not a zero divisor, and is not a unit).
Edit: Not totally unrelated...
https://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor
Edit: One more thing to consider...
5) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations):
$$ A = begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix} qquad Longrightarrow qquad mathrm{classical;adjoint} = C = begin{bmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} end{bmatrix} $$
Thus if $mathrm{det}(A)b=0$ for some $b not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC not=0$ and so $A(Cb)=mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor.
linear-algebra abstract-algebra matrices ring-theory
$endgroup$
$begingroup$
I am going to try reposting this on Mathoverflow.
$endgroup$
– Bill Cook
Oct 11 '11 at 13:36
1
$begingroup$
Thanks for catching the typo.
$endgroup$
– Bill Cook
Dec 4 '18 at 17:59
add a comment |
$begingroup$
Let $R$ be a commutative ring (with $1$) and $R^{n times n}$ be the ring of $n times n$ matrices with entries in $R$.
In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question:
Is every non-zero element of $R^{n times n}$ a zero divisor or a unit as well?
We know that if $A in R^{n times n}$, then $AC=CA=mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix.
This means that if $mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n times n}$ (since $A^{-1}=(mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n times n}$, then $mathrm{det}(A)$ is a unit.
I would like to know if one can show $0 not= A in R^{n times n}$ is a zero divisor if $mathrm{det}(A)$ is zero or a zero divisor.
Things to consider:
1) This is true when $R=mathbb{F}$ a field. Since over a field (no zero divisors) and if $mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x|0|cdots|0]$ gives us a right zero divisor $AB=0$.
2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example:
$$A=begin{bmatrix} 1 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 end{bmatrix} qquad mathrm{implies} qquad mathrm{classical;adjoint} = 0 $$
(All $2 times 2$ sub-determinants are zero.)
3) This is true when $R$ is finite (since $R^{n times n}$ would be finite as well).
4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$
and construct the diagonal matrix $D = mathrm{diag}(r,1,dots,1)$ (this is non-zero, not a zero divisor, and is not a unit).
Edit: Not totally unrelated...
https://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor
Edit: One more thing to consider...
5) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations):
$$ A = begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix} qquad Longrightarrow qquad mathrm{classical;adjoint} = C = begin{bmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} end{bmatrix} $$
Thus if $mathrm{det}(A)b=0$ for some $b not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC not=0$ and so $A(Cb)=mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor.
linear-algebra abstract-algebra matrices ring-theory
$endgroup$
Let $R$ be a commutative ring (with $1$) and $R^{n times n}$ be the ring of $n times n$ matrices with entries in $R$.
In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question:
Is every non-zero element of $R^{n times n}$ a zero divisor or a unit as well?
We know that if $A in R^{n times n}$, then $AC=CA=mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix.
This means that if $mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n times n}$ (since $A^{-1}=(mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n times n}$, then $mathrm{det}(A)$ is a unit.
I would like to know if one can show $0 not= A in R^{n times n}$ is a zero divisor if $mathrm{det}(A)$ is zero or a zero divisor.
Things to consider:
1) This is true when $R=mathbb{F}$ a field. Since over a field (no zero divisors) and if $mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x|0|cdots|0]$ gives us a right zero divisor $AB=0$.
2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example:
$$A=begin{bmatrix} 1 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 end{bmatrix} qquad mathrm{implies} qquad mathrm{classical;adjoint} = 0 $$
(All $2 times 2$ sub-determinants are zero.)
3) This is true when $R$ is finite (since $R^{n times n}$ would be finite as well).
4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$
and construct the diagonal matrix $D = mathrm{diag}(r,1,dots,1)$ (this is non-zero, not a zero divisor, and is not a unit).
Edit: Not totally unrelated...
https://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor
Edit: One more thing to consider...
5) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations):
$$ A = begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix} qquad Longrightarrow qquad mathrm{classical;adjoint} = C = begin{bmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} end{bmatrix} $$
Thus if $mathrm{det}(A)b=0$ for some $b not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC not=0$ and so $A(Cb)=mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor.
linear-algebra abstract-algebra matrices ring-theory
linear-algebra abstract-algebra matrices ring-theory
edited Dec 4 '18 at 17:59
Bill Cook
asked Oct 9 '11 at 20:27
Bill CookBill Cook
23k4869
23k4869
$begingroup$
I am going to try reposting this on Mathoverflow.
$endgroup$
– Bill Cook
Oct 11 '11 at 13:36
1
$begingroup$
Thanks for catching the typo.
$endgroup$
– Bill Cook
Dec 4 '18 at 17:59
add a comment |
$begingroup$
I am going to try reposting this on Mathoverflow.
$endgroup$
– Bill Cook
Oct 11 '11 at 13:36
1
$begingroup$
Thanks for catching the typo.
$endgroup$
– Bill Cook
Dec 4 '18 at 17:59
$begingroup$
I am going to try reposting this on Mathoverflow.
$endgroup$
– Bill Cook
Oct 11 '11 at 13:36
$begingroup$
I am going to try reposting this on Mathoverflow.
$endgroup$
– Bill Cook
Oct 11 '11 at 13:36
1
1
$begingroup$
Thanks for catching the typo.
$endgroup$
– Bill Cook
Dec 4 '18 at 17:59
$begingroup$
Thanks for catching the typo.
$endgroup$
– Bill Cook
Dec 4 '18 at 17:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $det A$ is a zero divisor, and thus $R^{ntimes n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.
$endgroup$
1
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
add a comment |
$begingroup$
For an artinian ring every element is either a unit or a zero divisor (yo can see this in any book about theory of rings). Moreover, if $R$ is an artinian ring, then $M_n(R)$ is also artinian. So, there are elements in $M_n(R)$ that are neither units nor zero divisor if the ring $R$ is not artinian.
$endgroup$
add a comment |
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2 Answers
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oldest
votes
2 Answers
2
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oldest
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votes
$begingroup$
As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $det A$ is a zero divisor, and thus $R^{ntimes n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.
$endgroup$
1
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
add a comment |
$begingroup$
As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $det A$ is a zero divisor, and thus $R^{ntimes n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.
$endgroup$
1
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
add a comment |
$begingroup$
As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $det A$ is a zero divisor, and thus $R^{ntimes n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.
$endgroup$
As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $det A$ is a zero divisor, and thus $R^{ntimes n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Oct 11 '11 at 16:12
jorikijoriki
170k10183343
170k10183343
1
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
add a comment |
1
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
1
1
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
$begingroup$
Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/…
$endgroup$
– Bill Cook
Oct 11 '11 at 17:26
add a comment |
$begingroup$
For an artinian ring every element is either a unit or a zero divisor (yo can see this in any book about theory of rings). Moreover, if $R$ is an artinian ring, then $M_n(R)$ is also artinian. So, there are elements in $M_n(R)$ that are neither units nor zero divisor if the ring $R$ is not artinian.
$endgroup$
add a comment |
$begingroup$
For an artinian ring every element is either a unit or a zero divisor (yo can see this in any book about theory of rings). Moreover, if $R$ is an artinian ring, then $M_n(R)$ is also artinian. So, there are elements in $M_n(R)$ that are neither units nor zero divisor if the ring $R$ is not artinian.
$endgroup$
add a comment |
$begingroup$
For an artinian ring every element is either a unit or a zero divisor (yo can see this in any book about theory of rings). Moreover, if $R$ is an artinian ring, then $M_n(R)$ is also artinian. So, there are elements in $M_n(R)$ that are neither units nor zero divisor if the ring $R$ is not artinian.
$endgroup$
For an artinian ring every element is either a unit or a zero divisor (yo can see this in any book about theory of rings). Moreover, if $R$ is an artinian ring, then $M_n(R)$ is also artinian. So, there are elements in $M_n(R)$ that are neither units nor zero divisor if the ring $R$ is not artinian.
answered Dec 31 '14 at 20:58
zacariaszacarias
1,573919
1,573919
add a comment |
add a comment |
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6HYH1aGdUHM5jWMUVMD,TnhIXI46rdPtTy,iT 3uvL8QNdgc753u,5rtC9TR8aNuYXD663,4rFincuBPVQb0i2 q347y89g,AKzUvL
$begingroup$
I am going to try reposting this on Mathoverflow.
$endgroup$
– Bill Cook
Oct 11 '11 at 13:36
1
$begingroup$
Thanks for catching the typo.
$endgroup$
– Bill Cook
Dec 4 '18 at 17:59