Mixing
Does $mathbb{Z}_n$ contain a subfield?
$begingroup$
Is it true that $mathbb{Z}_n$ doesn't contain a proper subring which is a field?
I have a contradiction that shows that this isn't true. Take the ring $mathbb{Z}_{10}$ and the proper subring $langle 2 rangle = { 0,2,4,6,8 }$ which is clearly a field under the same addition and multiplication modulo 10.
But I've come across a statement that says that for a positive integer $n$, $mathbb{Z}_n$ actually doesn't contain a proper subring which is a field. If it is so, then this should contradict the example that I've given above. How can this be possible? Is there any information that I'm missing?
abstract-algebra ring-theory field-theory
edited Jan 24 at 18:50
Fozoro
1265
asked Jan 24 at 18:02
Minto PMinto P
667
$endgroup$
add a comment |
$begingroup$
Is it true that $mathbb{Z}_n$ doesn't contain a proper subring which is a field?
I have a contradiction that shows that this isn't true. Take the ring $mathbb{Z}_{10}$ and the proper subring $langle 2 rangle = { 0,2,4,6,8 }$ which is clearly a field under the same addition and multiplication modulo 10.
But I've come across a statement that says that for a positive integer $n$, $mathbb{Z}_n$ actually doesn't contain a proper subring which is a field. If it is so, then this should contradict the example that I've given above. How can this be possible? Is there any information that I'm missing?
abstract-algebra ring-theory field-theory
edited Jan 24 at 18:50
Fozoro
1265
asked Jan 24 at 18:02
Minto PMinto P
667
$endgroup$
2
$begingroup$
Multiplicative identity?
$endgroup$
– Thomas Shelby
Jan 24 at 18:06
2
$begingroup$
@ThomasShelby: $6$ is a multiplicative identity.
$endgroup$
– tomasz
Jan 24 at 18:27
add a comment |
$begingroup$
Is it true that $mathbb{Z}_n$ doesn't contain a proper subring which is a field?
I have a contradiction that shows that this isn't true. Take the ring $mathbb{Z}_{10}$ and the proper subring $langle 2 rangle = { 0,2,4,6,8 }$ which is clearly a field under the same addition and multiplication modulo 10.
But I've come across a statement that says that for a positive integer $n$, $mathbb{Z}_n$ actually doesn't contain a proper subring which is a field. If it is so, then this should contradict the example that I've given above. How can this be possible? Is there any information that I'm missing?
abstract-algebra ring-theory field-theory
edited Jan 24 at 18:50
Fozoro
1265
asked Jan 24 at 18:02
Minto PMinto P
667
$endgroup$
Is it true that $mathbb{Z}_n$ doesn't contain a proper subring which is a field?
I have a contradiction that shows that this isn't true. Take the ring $mathbb{Z}_{10}$ and the proper subring $langle 2 rangle = { 0,2,4,6,8 }$ which is clearly a field under the same addition and multiplication modulo 10.
But I've come across a statement that says that for a positive integer $n$, $mathbb{Z}_n$ actually doesn't contain a proper subring which is a field. If it is so, then this should contradict the example that I've given above. How can this be possible? Is there any information that I'm missing?
abstract-algebra ring-theory field-theory
abstract-algebra ring-theory field-theory
edited Jan 24 at 18:50
Fozoro
1265
asked Jan 24 at 18:02
Minto PMinto P
667
edited Jan 24 at 18:50
Fozoro
1265
asked Jan 24 at 18:02
Minto PMinto P
667
edited Jan 24 at 18:50
Fozoro
1265
edited Jan 24 at 18:50
Fozoro
1265
edited Jan 24 at 18:50
Fozoro
1265
1265
asked Jan 24 at 18:02
Minto PMinto P
667
asked Jan 24 at 18:02
Minto PMinto P
667
asked Jan 24 at 18:02
Minto PMinto P
667
667
2
$begingroup$
Multiplicative identity?
$endgroup$
– Thomas Shelby
Jan 24 at 18:06
2
$begingroup$
@ThomasShelby: $6$ is a multiplicative identity.
$endgroup$
– tomasz
Jan 24 at 18:27
add a comment |
2
$begingroup$
Multiplicative identity?
$endgroup$
– Thomas Shelby
Jan 24 at 18:06
2
$begingroup$
@ThomasShelby: $6$ is a multiplicative identity.
$endgroup$
– tomasz
Jan 24 at 18:27
2
2
$begingroup$
Multiplicative identity?
$endgroup$
– Thomas Shelby
Jan 24 at 18:06
$begingroup$
Multiplicative identity?
$endgroup$
– Thomas Shelby
Jan 24 at 18:06
2
2
$begingroup$
@ThomasShelby: $6$ is a multiplicative identity.
$endgroup$
– tomasz
Jan 24 at 18:27
$begingroup$
@ThomasShelby: $6$ is a multiplicative identity.
$endgroup$
– tomasz
Jan 24 at 18:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It depends what you consider a ring (and hence, a subring). Many authors include the identity in the definition of a ring. Under this definition, a subring of a ring must contain the original identity. In this case, ${bf Z}_n$ has no proper subrings, so if it is not a field itself, it has no subrings which are fields.
Otherwise, if your rings are not necessarily unital, a subring may not contain the original identity, in which case your example is indeed correct (with $6$ as the multiplicative identity). An easier example is just ${0,5}$.
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
$endgroup$
add a comment |
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$begingroup$
It depends what you consider a ring (and hence, a subring). Many authors include the identity in the definition of a ring. Under this definition, a subring of a ring must contain the original identity. In this case, ${bf Z}_n$ has no proper subrings, so if it is not a field itself, it has no subrings which are fields.
Otherwise, if your rings are not necessarily unital, a subring may not contain the original identity, in which case your example is indeed correct (with $6$ as the multiplicative identity). An easier example is just ${0,5}$.
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
$endgroup$
add a comment |
$begingroup$
It depends what you consider a ring (and hence, a subring). Many authors include the identity in the definition of a ring. Under this definition, a subring of a ring must contain the original identity. In this case, ${bf Z}_n$ has no proper subrings, so if it is not a field itself, it has no subrings which are fields.
Otherwise, if your rings are not necessarily unital, a subring may not contain the original identity, in which case your example is indeed correct (with $6$ as the multiplicative identity). An easier example is just ${0,5}$.
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
$endgroup$
add a comment |
$begingroup$
It depends what you consider a ring (and hence, a subring). Many authors include the identity in the definition of a ring. Under this definition, a subring of a ring must contain the original identity. In this case, ${bf Z}_n$ has no proper subrings, so if it is not a field itself, it has no subrings which are fields.
Otherwise, if your rings are not necessarily unital, a subring may not contain the original identity, in which case your example is indeed correct (with $6$ as the multiplicative identity). An easier example is just ${0,5}$.
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
$endgroup$
It depends what you consider a ring (and hence, a subring). Many authors include the identity in the definition of a ring. Under this definition, a subring of a ring must contain the original identity. In this case, ${bf Z}_n$ has no proper subrings, so if it is not a field itself, it has no subrings which are fields.
Otherwise, if your rings are not necessarily unital, a subring may not contain the original identity, in which case your example is indeed correct (with $6$ as the multiplicative identity). An easier example is just ${0,5}$.
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
answered Jan 24 at 18:27
tomasztomasz
23.9k23482
23.9k23482
add a comment |
add a comment |
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2
$begingroup$
Multiplicative identity?
$endgroup$
– Thomas Shelby
Jan 24 at 18:06
2
$begingroup$
@ThomasShelby: $6$ is a multiplicative identity.
$endgroup$
– tomasz
Jan 24 at 18:27