Mixing
Examples of finite commutative rings that are not fields
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Would someone be able to provide examples of finite commutative rings that are not fields? I attempted to create commutative rings using matrices (ie permutation matrices) and integers (ie $mathbbZ_p$), however, either they were not rings or they were fields.
abstract-algebra ring-theory field-theory
asked 20 mins ago
Madhav Nakar
460514
add a comment |Â
up vote
1
down vote
favorite
Would someone be able to provide examples of finite commutative rings that are not fields? I attempted to create commutative rings using matrices (ie permutation matrices) and integers (ie $mathbbZ_p$), however, either they were not rings or they were fields.
abstract-algebra ring-theory field-theory
asked 20 mins ago
Madhav Nakar
460514
See also mathoverflow.net/questions/7133/…
– lhf
16 mins ago
2
Here is the DaRT search result set . At present, it's just several quotients of $mathbb Z$ and then a finite quotient of $F_2[x,y]$.
– rschwieb
14 mins ago
1
$mathbf Z/p^kmathbf Z$, $p$ prime, $k>1$.
– Bernard
3 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Would someone be able to provide examples of finite commutative rings that are not fields? I attempted to create commutative rings using matrices (ie permutation matrices) and integers (ie $mathbbZ_p$), however, either they were not rings or they were fields.
abstract-algebra ring-theory field-theory
asked 20 mins ago
Madhav Nakar
460514
Would someone be able to provide examples of finite commutative rings that are not fields? I attempted to create commutative rings using matrices (ie permutation matrices) and integers (ie $mathbbZ_p$), however, either they were not rings or they were fields.
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
asked 20 mins ago
Madhav Nakar
460514
asked 20 mins ago
Madhav Nakar
460514
asked 20 mins ago
Madhav Nakar
460514
asked 20 mins ago
Madhav Nakar
460514
asked 20 mins ago
Madhav Nakar
460514
460514
See also mathoverflow.net/questions/7133/…
– lhf
16 mins ago
2
Here is the DaRT search result set . At present, it's just several quotients of $mathbb Z$ and then a finite quotient of $F_2[x,y]$.
– rschwieb
14 mins ago
1
$mathbf Z/p^kmathbf Z$, $p$ prime, $k>1$.
– Bernard
3 mins ago
add a comment |Â
See also mathoverflow.net/questions/7133/…
– lhf
16 mins ago
2
Here is the DaRT search result set . At present, it's just several quotients of $mathbb Z$ and then a finite quotient of $F_2[x,y]$.
– rschwieb
14 mins ago
1
$mathbf Z/p^kmathbf Z$, $p$ prime, $k>1$.
– Bernard
3 mins ago
See also mathoverflow.net/questions/7133/…
– lhf
16 mins ago
See also mathoverflow.net/questions/7133/…
– lhf
16 mins ago
2
2
Here is the DaRT search result set . At present, it's just several quotients of $mathbb Z$ and then a finite quotient of $F_2[x,y]$.
– rschwieb
14 mins ago
Here is the DaRT search result set . At present, it's just several quotients of $mathbb Z$ and then a finite quotient of $F_2[x,y]$.
– rschwieb
14 mins ago
1
1
$mathbf Z/p^kmathbf Z$, $p$ prime, $k>1$.
– Bernard
3 mins ago
$mathbf Z/p^kmathbf Z$, $p$ prime, $k>1$.
– Bernard
3 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
$mathbb Z_4$, why not? Beginning students may think that the field of $4$ element is this, but it is not, so this is a good example to talk about.
A finite commutative ring with no zero divisors is a field, so we have to look for zero divisors to get an example that you ask for.
edited 1 min ago
Bernard
113k636104
answered 16 mins ago
GEdgar
59.8k265166
add a comment |Â
up vote
1
down vote
Take $A times B$, where $A$ and $B$ are finite commutative rings, even fields.
answered 17 mins ago
lhf
159k9161376
add a comment |Â
up vote
1
down vote
What about the $mathbbZ_2timesmathbbZ_2$?
answered 19 mins ago
José Carlos Santos
128k17102189
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
add a comment |Â
up vote
0
down vote
Let $X$ be a nonempty finite set and $R$ be a finite ring. Take the set $R^X$ of all mappings $f:Xrightarrow R$. This set becomes a ring with $(f+g)(x) = f(x)+g(x)$ and $(fcdot g)(x) = f(x)cdot g(x)$ for all $xin X$ and $f,gin R^X$.
answered 4 mins ago
Wuestenfux
1,444128
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$mathbb Z_4$, why not? Beginning students may think that the field of $4$ element is this, but it is not, so this is a good example to talk about.
A finite commutative ring with no zero divisors is a field, so we have to look for zero divisors to get an example that you ask for.
edited 1 min ago
Bernard
113k636104
answered 16 mins ago
GEdgar
59.8k265166
add a comment |Â
up vote
4
down vote
accepted
$mathbb Z_4$, why not? Beginning students may think that the field of $4$ element is this, but it is not, so this is a good example to talk about.
A finite commutative ring with no zero divisors is a field, so we have to look for zero divisors to get an example that you ask for.
edited 1 min ago
Bernard
113k636104
answered 16 mins ago
GEdgar
59.8k265166
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$mathbb Z_4$, why not? Beginning students may think that the field of $4$ element is this, but it is not, so this is a good example to talk about.
A finite commutative ring with no zero divisors is a field, so we have to look for zero divisors to get an example that you ask for.
edited 1 min ago
Bernard
113k636104
answered 16 mins ago
GEdgar
59.8k265166
$mathbb Z_4$, why not? Beginning students may think that the field of $4$ element is this, but it is not, so this is a good example to talk about.
A finite commutative ring with no zero divisors is a field, so we have to look for zero divisors to get an example that you ask for.
edited 1 min ago
Bernard
113k636104
answered 16 mins ago
GEdgar
59.8k265166
edited 1 min ago
Bernard
113k636104
edited 1 min ago
Bernard
113k636104
edited 1 min ago
Bernard
113k636104
113k636104
answered 16 mins ago
GEdgar
59.8k265166
answered 16 mins ago
GEdgar
59.8k265166
answered 16 mins ago
GEdgar
59.8k265166
59.8k265166
add a comment |Â
add a comment |Â
up vote
1
down vote
Take $A times B$, where $A$ and $B$ are finite commutative rings, even fields.
answered 17 mins ago
lhf
159k9161376
add a comment |Â
up vote
1
down vote
Take $A times B$, where $A$ and $B$ are finite commutative rings, even fields.
answered 17 mins ago
lhf
159k9161376
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take $A times B$, where $A$ and $B$ are finite commutative rings, even fields.
answered 17 mins ago
lhf
159k9161376
Take $A times B$, where $A$ and $B$ are finite commutative rings, even fields.
answered 17 mins ago
lhf
159k9161376
answered 17 mins ago
lhf
159k9161376
answered 17 mins ago
lhf
159k9161376
answered 17 mins ago
lhf
159k9161376
159k9161376
add a comment |Â
add a comment |Â
up vote
1
down vote
What about the $mathbbZ_2timesmathbbZ_2$?
answered 19 mins ago
José Carlos Santos
128k17102189
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
add a comment |Â
up vote
1
down vote
What about the $mathbbZ_2timesmathbbZ_2$?
answered 19 mins ago
José Carlos Santos
128k17102189
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What about the $mathbbZ_2timesmathbbZ_2$?
answered 19 mins ago
José Carlos Santos
128k17102189
What about the $mathbbZ_2timesmathbbZ_2$?
answered 19 mins ago
José Carlos Santos
128k17102189
edited 16 mins ago
answered 19 mins ago
José Carlos Santos
128k17102189
answered 19 mins ago
José Carlos Santos
128k17102189
answered 19 mins ago
José Carlos Santos
128k17102189
128k17102189
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
add a comment |Â
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
Would that be a commutative ring? Multiplication is not commutative on matrices.
– Madhav Nakar
17 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
You are right. I've edited my answer.
– José Carlos Santos
16 mins ago
add a comment |Â
up vote
0
down vote
Let $X$ be a nonempty finite set and $R$ be a finite ring. Take the set $R^X$ of all mappings $f:Xrightarrow R$. This set becomes a ring with $(f+g)(x) = f(x)+g(x)$ and $(fcdot g)(x) = f(x)cdot g(x)$ for all $xin X$ and $f,gin R^X$.
answered 4 mins ago
Wuestenfux
1,444128
add a comment |Â
up vote
0
down vote
Let $X$ be a nonempty finite set and $R$ be a finite ring. Take the set $R^X$ of all mappings $f:Xrightarrow R$. This set becomes a ring with $(f+g)(x) = f(x)+g(x)$ and $(fcdot g)(x) = f(x)cdot g(x)$ for all $xin X$ and $f,gin R^X$.
answered 4 mins ago
Wuestenfux
1,444128
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $X$ be a nonempty finite set and $R$ be a finite ring. Take the set $R^X$ of all mappings $f:Xrightarrow R$. This set becomes a ring with $(f+g)(x) = f(x)+g(x)$ and $(fcdot g)(x) = f(x)cdot g(x)$ for all $xin X$ and $f,gin R^X$.
answered 4 mins ago
Wuestenfux
1,444128
Let $X$ be a nonempty finite set and $R$ be a finite ring. Take the set $R^X$ of all mappings $f:Xrightarrow R$. This set becomes a ring with $(f+g)(x) = f(x)+g(x)$ and $(fcdot g)(x) = f(x)cdot g(x)$ for all $xin X$ and $f,gin R^X$.
answered 4 mins ago
Wuestenfux
1,444128
answered 4 mins ago
Wuestenfux
1,444128
answered 4 mins ago
Wuestenfux
1,444128
answered 4 mins ago
Wuestenfux
1,444128
1,444128
add a comment |Â
add a comment |Â
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See also mathoverflow.net/questions/7133/…
– lhf
16 mins ago
2
Here is the DaRT search result set . At present, it's just several quotients of $mathbb Z$ and then a finite quotient of $F_2[x,y]$.
– rschwieb
14 mins ago
1
$mathbf Z/p^kmathbf Z$, $p$ prime, $k>1$.
– Bernard
3 mins ago