ring contains a subring isomorphic to $Z$ implies it is integral domain?
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If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?
I know that it must have characteristic 0.However I cannot proceed further.
ring-theory
asked Aug 28 at 14:34
Anwi
338112
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up vote
0
down vote
favorite
If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?
I know that it must have characteristic 0.However I cannot proceed further.
ring-theory
asked Aug 28 at 14:34
Anwi
338112
This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?
I know that it must have characteristic 0.However I cannot proceed further.
ring-theory
asked Aug 28 at 14:34
Anwi
338112
If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?
I know that it must have characteristic 0.However I cannot proceed further.
ring-theory
asked Aug 28 at 14:34
Anwi
338112
asked Aug 28 at 14:34
Anwi
338112
asked Aug 28 at 14:34
Anwi
338112
asked Aug 28 at 14:34
Anwi
338112
338112
This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36
add a comment |Â
This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36
This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36
This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36
add a comment |Â
2 Answers
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oldest
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0
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In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
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up vote
1
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No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
$$
a + bepsilonqquad a,binmathbb Z,
$$
with obvious addition and multiplication given by
$$
(a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
$$
Then $epsilon^2 = 0$, even though $epsilon neq 0$.
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
add a comment |Â
up vote
0
down vote
accepted
In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
answered Aug 28 at 14:37
José Carlos Santos
120k16101182
120k16101182
add a comment |Â
add a comment |Â
up vote
1
down vote
No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
$$
a + bepsilonqquad a,binmathbb Z,
$$
with obvious addition and multiplication given by
$$
(a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
$$
Then $epsilon^2 = 0$, even though $epsilon neq 0$.
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
add a comment |Â
up vote
1
down vote
No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
$$
a + bepsilonqquad a,binmathbb Z,
$$
with obvious addition and multiplication given by
$$
(a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
$$
Then $epsilon^2 = 0$, even though $epsilon neq 0$.
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
$$
a + bepsilonqquad a,binmathbb Z,
$$
with obvious addition and multiplication given by
$$
(a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
$$
Then $epsilon^2 = 0$, even though $epsilon neq 0$.
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
$$
a + bepsilonqquad a,binmathbb Z,
$$
with obvious addition and multiplication given by
$$
(a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
$$
Then $epsilon^2 = 0$, even though $epsilon neq 0$.
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
answered Aug 28 at 14:36
Mees de Vries
14.1k12348
14.1k12348
add a comment |Â
add a comment |Â
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This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36