Mixing
Finding Zero Divisors and Units of Rings $mathbb{Z}_n times mathbb{Z}_m$
$begingroup$
How do I exactly find the zero divisors and units of a ring in the form of $mathbb{Z}_n times mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $mathbb{Z}_6 times mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $mathbb{Z}_n times mathbb{Z}_m$. Also, the units of $mathbb{Z}_6 times mathbb{Z}_2$ would be elements that send some element $s$ in $mathbb{Z}_n times mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.
abstract-algebra ring-theory
asked Oct 30 '16 at 16:44
MaxMax
687317
$endgroup$
add a comment |
$begingroup$
How do I exactly find the zero divisors and units of a ring in the form of $mathbb{Z}_n times mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $mathbb{Z}_6 times mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $mathbb{Z}_n times mathbb{Z}_m$. Also, the units of $mathbb{Z}_6 times mathbb{Z}_2$ would be elements that send some element $s$ in $mathbb{Z}_n times mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.
abstract-algebra ring-theory
asked Oct 30 '16 at 16:44
MaxMax
687317
$endgroup$
$begingroup$
You can find the zero divisors and units like here.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 16:52
add a comment |
$begingroup$
How do I exactly find the zero divisors and units of a ring in the form of $mathbb{Z}_n times mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $mathbb{Z}_6 times mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $mathbb{Z}_n times mathbb{Z}_m$. Also, the units of $mathbb{Z}_6 times mathbb{Z}_2$ would be elements that send some element $s$ in $mathbb{Z}_n times mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.
abstract-algebra ring-theory
asked Oct 30 '16 at 16:44
MaxMax
687317
$endgroup$
How do I exactly find the zero divisors and units of a ring in the form of $mathbb{Z}_n times mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $mathbb{Z}_6 times mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $mathbb{Z}_n times mathbb{Z}_m$. Also, the units of $mathbb{Z}_6 times mathbb{Z}_2$ would be elements that send some element $s$ in $mathbb{Z}_n times mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Oct 30 '16 at 16:44
MaxMax
687317
asked Oct 30 '16 at 16:44
MaxMax
687317
asked Oct 30 '16 at 16:44
MaxMax
687317
asked Oct 30 '16 at 16:44
MaxMax
687317
asked Oct 30 '16 at 16:44
MaxMax
687317
687317
$begingroup$
You can find the zero divisors and units like here.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 16:52
add a comment |
$begingroup$
You can find the zero divisors and units like here.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 16:52
$begingroup$
You can find the zero divisors and units like here.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 16:52
$begingroup$
You can find the zero divisors and units like here.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 16:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are a few things to look for.
For any two rings $R,S$, and any two elements $rin R, sin S$, we have $(r,0)(0,s)=(0,0)$ in $Rtimes S$. Also, if $r_1,r_2in R$ are non-zero elements such that $r_1r_2=0$, we have $(r_1,0)(r_2,0)=(0,0)$. Similarly for $s_1,s_2in S$. You can even combine these three: $(r_1,0)(r_2,s)=(0,0)$, for instance.
As for units, if both $rin R$ and $sin S$ are invertible, then clearly $(r,s)$ is. If one of them, say $r$ is not a unit, then there is no $(r',s')in Rtimes S$ such that $(r,s)(r',s')=(1,1)$, because $rr'$ cannot be $1$.
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
Every element of the form $(x,y)$ with either $x$ or $y$ a zero divisor is a zero divisor.
Indeed, suppose $x$ is a zero divisor. Then $ax=0$ for some $ane0$, so $(a,0)(x,y)=(0,0)$. Similarly if $y$ is a zero divisor.
Conversely, if $(x,y)$ is a zero divisor, then either $x$ or $y$ is.
Indeed, if $(a,b)(x,y)=(0,0)$, with $(a,b)ne(0,0)$, then either $ane0$ or $bne0$. If $ane0$, then $(a,0)(x,y)=(0,0)$ and similarly if $bne0$.
An element $(x,y)$ is a unit if and only if both $x$ and $y$ are units.
Since every element of $mathbb{Z}_k$ is either a zero divisor or a unit, you should be able to finish.
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
$endgroup$
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
add a comment |
$begingroup$
In a finite commutative ring, a nonzero element is either a unit or a zero divisor. So it is enough to look for, say, the units. Here we just have
$$
U(mathbb{Z}/ntimes mathbb{Z}/m)cong U(mathbb{Z}/n)times U(mathbb{Z}/m).
$$
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
$endgroup$
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are a few things to look for.
For any two rings $R,S$, and any two elements $rin R, sin S$, we have $(r,0)(0,s)=(0,0)$ in $Rtimes S$. Also, if $r_1,r_2in R$ are non-zero elements such that $r_1r_2=0$, we have $(r_1,0)(r_2,0)=(0,0)$. Similarly for $s_1,s_2in S$. You can even combine these three: $(r_1,0)(r_2,s)=(0,0)$, for instance.
As for units, if both $rin R$ and $sin S$ are invertible, then clearly $(r,s)$ is. If one of them, say $r$ is not a unit, then there is no $(r',s')in Rtimes S$ such that $(r,s)(r',s')=(1,1)$, because $rr'$ cannot be $1$.
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
There are a few things to look for.
For any two rings $R,S$, and any two elements $rin R, sin S$, we have $(r,0)(0,s)=(0,0)$ in $Rtimes S$. Also, if $r_1,r_2in R$ are non-zero elements such that $r_1r_2=0$, we have $(r_1,0)(r_2,0)=(0,0)$. Similarly for $s_1,s_2in S$. You can even combine these three: $(r_1,0)(r_2,s)=(0,0)$, for instance.
As for units, if both $rin R$ and $sin S$ are invertible, then clearly $(r,s)$ is. If one of them, say $r$ is not a unit, then there is no $(r',s')in Rtimes S$ such that $(r,s)(r',s')=(1,1)$, because $rr'$ cannot be $1$.
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
There are a few things to look for.
For any two rings $R,S$, and any two elements $rin R, sin S$, we have $(r,0)(0,s)=(0,0)$ in $Rtimes S$. Also, if $r_1,r_2in R$ are non-zero elements such that $r_1r_2=0$, we have $(r_1,0)(r_2,0)=(0,0)$. Similarly for $s_1,s_2in S$. You can even combine these three: $(r_1,0)(r_2,s)=(0,0)$, for instance.
As for units, if both $rin R$ and $sin S$ are invertible, then clearly $(r,s)$ is. If one of them, say $r$ is not a unit, then there is no $(r',s')in Rtimes S$ such that $(r,s)(r',s')=(1,1)$, because $rr'$ cannot be $1$.
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
$endgroup$
There are a few things to look for.
For any two rings $R,S$, and any two elements $rin R, sin S$, we have $(r,0)(0,s)=(0,0)$ in $Rtimes S$. Also, if $r_1,r_2in R$ are non-zero elements such that $r_1r_2=0$, we have $(r_1,0)(r_2,0)=(0,0)$. Similarly for $s_1,s_2in S$. You can even combine these three: $(r_1,0)(r_2,s)=(0,0)$, for instance.
As for units, if both $rin R$ and $sin S$ are invertible, then clearly $(r,s)$ is. If one of them, say $r$ is not a unit, then there is no $(r',s')in Rtimes S$ such that $(r,s)(r',s')=(1,1)$, because $rr'$ cannot be $1$.
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
answered Oct 30 '16 at 16:50
ArthurArthur
116k7116198
116k7116198
add a comment |
add a comment |
$begingroup$
Every element of the form $(x,y)$ with either $x$ or $y$ a zero divisor is a zero divisor.
Indeed, suppose $x$ is a zero divisor. Then $ax=0$ for some $ane0$, so $(a,0)(x,y)=(0,0)$. Similarly if $y$ is a zero divisor.
Conversely, if $(x,y)$ is a zero divisor, then either $x$ or $y$ is.
Indeed, if $(a,b)(x,y)=(0,0)$, with $(a,b)ne(0,0)$, then either $ane0$ or $bne0$. If $ane0$, then $(a,0)(x,y)=(0,0)$ and similarly if $bne0$.
An element $(x,y)$ is a unit if and only if both $x$ and $y$ are units.
Since every element of $mathbb{Z}_k$ is either a zero divisor or a unit, you should be able to finish.
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
$endgroup$
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
add a comment |
$begingroup$
Every element of the form $(x,y)$ with either $x$ or $y$ a zero divisor is a zero divisor.
Indeed, suppose $x$ is a zero divisor. Then $ax=0$ for some $ane0$, so $(a,0)(x,y)=(0,0)$. Similarly if $y$ is a zero divisor.
Conversely, if $(x,y)$ is a zero divisor, then either $x$ or $y$ is.
Indeed, if $(a,b)(x,y)=(0,0)$, with $(a,b)ne(0,0)$, then either $ane0$ or $bne0$. If $ane0$, then $(a,0)(x,y)=(0,0)$ and similarly if $bne0$.
An element $(x,y)$ is a unit if and only if both $x$ and $y$ are units.
Since every element of $mathbb{Z}_k$ is either a zero divisor or a unit, you should be able to finish.
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
$endgroup$
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
add a comment |
$begingroup$
Every element of the form $(x,y)$ with either $x$ or $y$ a zero divisor is a zero divisor.
Indeed, suppose $x$ is a zero divisor. Then $ax=0$ for some $ane0$, so $(a,0)(x,y)=(0,0)$. Similarly if $y$ is a zero divisor.
Conversely, if $(x,y)$ is a zero divisor, then either $x$ or $y$ is.
Indeed, if $(a,b)(x,y)=(0,0)$, with $(a,b)ne(0,0)$, then either $ane0$ or $bne0$. If $ane0$, then $(a,0)(x,y)=(0,0)$ and similarly if $bne0$.
An element $(x,y)$ is a unit if and only if both $x$ and $y$ are units.
Since every element of $mathbb{Z}_k$ is either a zero divisor or a unit, you should be able to finish.
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
$endgroup$
Every element of the form $(x,y)$ with either $x$ or $y$ a zero divisor is a zero divisor.
Indeed, suppose $x$ is a zero divisor. Then $ax=0$ for some $ane0$, so $(a,0)(x,y)=(0,0)$. Similarly if $y$ is a zero divisor.
Conversely, if $(x,y)$ is a zero divisor, then either $x$ or $y$ is.
Indeed, if $(a,b)(x,y)=(0,0)$, with $(a,b)ne(0,0)$, then either $ane0$ or $bne0$. If $ane0$, then $(a,0)(x,y)=(0,0)$ and similarly if $bne0$.
An element $(x,y)$ is a unit if and only if both $x$ and $y$ are units.
Since every element of $mathbb{Z}_k$ is either a zero divisor or a unit, you should be able to finish.
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
answered Oct 30 '16 at 16:56
egregegreg
182k1486204
182k1486204
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
add a comment |
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
$begingroup$
So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
$endgroup$
– Max
Oct 30 '16 at 19:22
add a comment |
$begingroup$
In a finite commutative ring, a nonzero element is either a unit or a zero divisor. So it is enough to look for, say, the units. Here we just have
$$
U(mathbb{Z}/ntimes mathbb{Z}/m)cong U(mathbb{Z}/n)times U(mathbb{Z}/m).
$$
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
$endgroup$
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
add a comment |
$begingroup$
In a finite commutative ring, a nonzero element is either a unit or a zero divisor. So it is enough to look for, say, the units. Here we just have
$$
U(mathbb{Z}/ntimes mathbb{Z}/m)cong U(mathbb{Z}/n)times U(mathbb{Z}/m).
$$
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
$endgroup$
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
add a comment |
$begingroup$
In a finite commutative ring, a nonzero element is either a unit or a zero divisor. So it is enough to look for, say, the units. Here we just have
$$
U(mathbb{Z}/ntimes mathbb{Z}/m)cong U(mathbb{Z}/n)times U(mathbb{Z}/m).
$$
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
$endgroup$
In a finite commutative ring, a nonzero element is either a unit or a zero divisor. So it is enough to look for, say, the units. Here we just have
$$
U(mathbb{Z}/ntimes mathbb{Z}/m)cong U(mathbb{Z}/n)times U(mathbb{Z}/m).
$$
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
answered Oct 30 '16 at 16:56
Dietrich BurdeDietrich Burde
79.6k647103
79.6k647103
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
add a comment |
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
$endgroup$
– Max
Oct 30 '16 at 19:24
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
$begingroup$
Right. Try it first for $mathbb{Z}/6$ alone, to see what happens.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 19:50
add a comment |
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$begingroup$
You can find the zero divisors and units like here.
$endgroup$
– Dietrich Burde
Oct 30 '16 at 16:52