What is the Characteristic of a local ring? [on hold] Announcing the arrival of Valued...
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What is the Characteristic of a local ring? [on hold]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Jacobson radical of ringsIs every local ring the localization of some other ring?Difference between PID and principal ideal ring$R$ a conmutative ring, $Ksubset R$ a field. ¿Does the characteristic of $text{char}(R)=text{char}(K)$?Consider the ring homomorphism $ϕ : mathbb{R}[x] → mathbb{R}[sqrt{−3}]$ defined by $ϕ(x) = sqrt{−3}$.The localisation of the ring $mathbb{Z}$ at the prime ideal $(p)$ is PIDCharacteristic of a ring $A$ and residue fieldsExample of a commutative noetherian ring with $1$ which is neither domain nor local and has a principal prime ideal of height $1.$Example of a characteristic zero local ring with a quotient of positive characteristicA regular local ring is a UFD.
$begingroup$
What is the Characteristic of a local ring ?
We define Characteristic of a Commutative ring with $1$ say, $A$ in the following way: Define a ring homomorphism $phi: mathbb{Z} to A$ by $phi(n)=n cdot 1.$ Since $mathbb{Z}$ is a PID, $text{ker}(phi)$ is a principal ideal. If $text{ker}(phi)=mmathbb{Z},$ we define the Characteristic of the ring $A$ to be $m.$ We know that Characteristic of a domain
is either $0$ or a prime $p.$
How do I classify the Characteristic of a local ring $(A,m)$
abstract-algebra ring-theory commutative-algebra
asked 2 days ago
user371231user371231
417511
$endgroup$
put on hold as unclear what you're asking by Arnaud D., max_zorn, Javi, José Carlos Santos, Ramiro yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
What is the Characteristic of a local ring ?
We define Characteristic of a Commutative ring with $1$ say, $A$ in the following way: Define a ring homomorphism $phi: mathbb{Z} to A$ by $phi(n)=n cdot 1.$ Since $mathbb{Z}$ is a PID, $text{ker}(phi)$ is a principal ideal. If $text{ker}(phi)=mmathbb{Z},$ we define the Characteristic of the ring $A$ to be $m.$ We know that Characteristic of a domain
is either $0$ or a prime $p.$
How do I classify the Characteristic of a local ring $(A,m)$
abstract-algebra ring-theory commutative-algebra
asked 2 days ago
user371231user371231
417511
$endgroup$
put on hold as unclear what you're asking by Arnaud D., max_zorn, Javi, José Carlos Santos, Ramiro yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The characteristic of a ring need not be prime: for instance, the characteristic of $mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain.
$endgroup$
– Claudius
2 days ago
2
$begingroup$
I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right?
$endgroup$
– Arnaud D.
2 days ago
$begingroup$
It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring?
$endgroup$
– Claudius
2 days ago
$begingroup$
Like integral domain you can classify the characteristic of a local ring.
$endgroup$
– user371231
2 days ago
$begingroup$
I much as I guess $6$ cannot be characteristic of a local ring
$endgroup$
– user371231
2 days ago
|
show 1 more comment
$begingroup$
What is the Characteristic of a local ring ?
We define Characteristic of a Commutative ring with $1$ say, $A$ in the following way: Define a ring homomorphism $phi: mathbb{Z} to A$ by $phi(n)=n cdot 1.$ Since $mathbb{Z}$ is a PID, $text{ker}(phi)$ is a principal ideal. If $text{ker}(phi)=mmathbb{Z},$ we define the Characteristic of the ring $A$ to be $m.$ We know that Characteristic of a domain
is either $0$ or a prime $p.$
How do I classify the Characteristic of a local ring $(A,m)$
abstract-algebra ring-theory commutative-algebra
asked 2 days ago
user371231user371231
417511
$endgroup$
What is the Characteristic of a local ring ?
We define Characteristic of a Commutative ring with $1$ say, $A$ in the following way: Define a ring homomorphism $phi: mathbb{Z} to A$ by $phi(n)=n cdot 1.$ Since $mathbb{Z}$ is a PID, $text{ker}(phi)$ is a principal ideal. If $text{ker}(phi)=mmathbb{Z},$ we define the Characteristic of the ring $A$ to be $m.$ We know that Characteristic of a domain
is either $0$ or a prime $p.$
How do I classify the Characteristic of a local ring $(A,m)$
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked 2 days ago
user371231user371231
417511
asked 2 days ago
user371231user371231
417511
edited 2 days ago
user371231
asked 2 days ago
user371231user371231
417511
asked 2 days ago
user371231user371231
417511
asked 2 days ago
user371231user371231
417511
417511
put on hold as unclear what you're asking by Arnaud D., max_zorn, Javi, José Carlos Santos, Ramiro yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Arnaud D., max_zorn, Javi, José Carlos Santos, Ramiro yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The characteristic of a ring need not be prime: for instance, the characteristic of $mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain.
$endgroup$
– Claudius
2 days ago
2
$begingroup$
I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right?
$endgroup$
– Arnaud D.
2 days ago
$begingroup$
It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring?
$endgroup$
– Claudius
2 days ago
$begingroup$
Like integral domain you can classify the characteristic of a local ring.
$endgroup$
– user371231
2 days ago
$begingroup$
I much as I guess $6$ cannot be characteristic of a local ring
$endgroup$
– user371231
2 days ago
|
show 1 more comment
2
$begingroup$
The characteristic of a ring need not be prime: for instance, the characteristic of $mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain.
$endgroup$
– Claudius
2 days ago
2
$begingroup$
I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right?
$endgroup$
– Arnaud D.
2 days ago
$begingroup$
It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring?
$endgroup$
– Claudius
2 days ago
$begingroup$
Like integral domain you can classify the characteristic of a local ring.
$endgroup$
– user371231
2 days ago
$begingroup$
I much as I guess $6$ cannot be characteristic of a local ring
$endgroup$
– user371231
2 days ago
2
2
$begingroup$
The characteristic of a ring need not be prime: for instance, the characteristic of $mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain.
$endgroup$
– Claudius
2 days ago
$begingroup$
The characteristic of a ring need not be prime: for instance, the characteristic of $mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain.
$endgroup$
– Claudius
2 days ago
2
2
$begingroup$
I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right?
$endgroup$
– Arnaud D.
2 days ago
$begingroup$
I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right?
$endgroup$
– Arnaud D.
2 days ago
$begingroup$
It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring?
$endgroup$
– Claudius
2 days ago
$begingroup$
It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring?
$endgroup$
– Claudius
2 days ago
$begingroup$
Like integral domain you can classify the characteristic of a local ring.
$endgroup$
– user371231
2 days ago
$begingroup$
Like integral domain you can classify the characteristic of a local ring.
$endgroup$
– user371231
2 days ago
$begingroup$
I much as I guess $6$ cannot be characteristic of a local ring
$endgroup$
– user371231
2 days ago
$begingroup$
I much as I guess $6$ cannot be characteristic of a local ring
$endgroup$
– user371231
2 days ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.
That they all happen is easy : you may look at fields for characteristic $0$, and $mathbb{Z}/p^nmathbb{Z}$ for powers of primes.
Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, aland b = 1$. Then the ideals $I={xin R, ax = 0}$ and $J={xin R, bx=0}$ are comaximal : indeed $ain J, bin I$ and there are $u,v$ with $au+bv=1$ so $1in I+J$.
Therefore by locality, one of them is $R$ (otherwise they would both be $subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 lor b=1$, so that $n$ is a power of a prime.
edited 2 days ago
anomaly
17.9k42666
answered 2 days ago
MaxMax
16.5k11144
$endgroup$
add a comment |
$begingroup$
If you already know a local ring has only trivial idempotents, then you can reason this way:
Suppose the characteristic of a ring $R$ is finite, say $n$, and is divisible by more than one prime. The ring contains a copy of $mathbb Z/nmathbb Z$. So to show $R$ isn't local, it suffices to show that $mathbb Z/nmathbb Z$ contains a nontrivial idempotent, so that $R$ will also contain a nontrivial idempotent.
By reasoning with the Chinese remainder theorem, you can quickly see that $mathbb Z/nmathbb Z$ is isomorphic to $mathbb Z/p^kmathbb Z$ for primes $p$ dividing $n$ and powers $k$ depending on $p$. Since there is more than one prime dividing $n$ (our assumption) there is at least one nontrivial idempotent splitting $mathbb Z/nmathbb Z$ into two pieces. This is clearly a nontrivial idempotent of $R$ too.
So by contrapositive, we have shown that a local ring $R$ must have either characteristic $0$, or else it has finite characteristic that is a power of a prime.
Since we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible.
answered 2 days ago
rschwiebrschwieb
108k12105253
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.
That they all happen is easy : you may look at fields for characteristic $0$, and $mathbb{Z}/p^nmathbb{Z}$ for powers of primes.
Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, aland b = 1$. Then the ideals $I={xin R, ax = 0}$ and $J={xin R, bx=0}$ are comaximal : indeed $ain J, bin I$ and there are $u,v$ with $au+bv=1$ so $1in I+J$.
Therefore by locality, one of them is $R$ (otherwise they would both be $subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 lor b=1$, so that $n$ is a power of a prime.
edited 2 days ago
anomaly
17.9k42666
answered 2 days ago
MaxMax
16.5k11144
$endgroup$
add a comment |
$begingroup$
The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.
That they all happen is easy : you may look at fields for characteristic $0$, and $mathbb{Z}/p^nmathbb{Z}$ for powers of primes.
Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, aland b = 1$. Then the ideals $I={xin R, ax = 0}$ and $J={xin R, bx=0}$ are comaximal : indeed $ain J, bin I$ and there are $u,v$ with $au+bv=1$ so $1in I+J$.
Therefore by locality, one of them is $R$ (otherwise they would both be $subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 lor b=1$, so that $n$ is a power of a prime.
edited 2 days ago
anomaly
17.9k42666
answered 2 days ago
MaxMax
16.5k11144
$endgroup$
add a comment |
$begingroup$
The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.
That they all happen is easy : you may look at fields for characteristic $0$, and $mathbb{Z}/p^nmathbb{Z}$ for powers of primes.
Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, aland b = 1$. Then the ideals $I={xin R, ax = 0}$ and $J={xin R, bx=0}$ are comaximal : indeed $ain J, bin I$ and there are $u,v$ with $au+bv=1$ so $1in I+J$.
Therefore by locality, one of them is $R$ (otherwise they would both be $subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 lor b=1$, so that $n$ is a power of a prime.
edited 2 days ago
anomaly
17.9k42666
answered 2 days ago
MaxMax
16.5k11144
$endgroup$
The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.
That they all happen is easy : you may look at fields for characteristic $0$, and $mathbb{Z}/p^nmathbb{Z}$ for powers of primes.
Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, aland b = 1$. Then the ideals $I={xin R, ax = 0}$ and $J={xin R, bx=0}$ are comaximal : indeed $ain J, bin I$ and there are $u,v$ with $au+bv=1$ so $1in I+J$.
Therefore by locality, one of them is $R$ (otherwise they would both be $subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 lor b=1$, so that $n$ is a power of a prime.
edited 2 days ago
anomaly
17.9k42666
answered 2 days ago
MaxMax
16.5k11144
edited 2 days ago
anomaly
17.9k42666
edited 2 days ago
anomaly
17.9k42666
edited 2 days ago
anomaly
17.9k42666
17.9k42666
answered 2 days ago
MaxMax
16.5k11144
answered 2 days ago
MaxMax
16.5k11144
answered 2 days ago
MaxMax
16.5k11144
16.5k11144
add a comment |
add a comment |
$begingroup$
If you already know a local ring has only trivial idempotents, then you can reason this way:
Suppose the characteristic of a ring $R$ is finite, say $n$, and is divisible by more than one prime. The ring contains a copy of $mathbb Z/nmathbb Z$. So to show $R$ isn't local, it suffices to show that $mathbb Z/nmathbb Z$ contains a nontrivial idempotent, so that $R$ will also contain a nontrivial idempotent.
By reasoning with the Chinese remainder theorem, you can quickly see that $mathbb Z/nmathbb Z$ is isomorphic to $mathbb Z/p^kmathbb Z$ for primes $p$ dividing $n$ and powers $k$ depending on $p$. Since there is more than one prime dividing $n$ (our assumption) there is at least one nontrivial idempotent splitting $mathbb Z/nmathbb Z$ into two pieces. This is clearly a nontrivial idempotent of $R$ too.
So by contrapositive, we have shown that a local ring $R$ must have either characteristic $0$, or else it has finite characteristic that is a power of a prime.
Since we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible.
answered 2 days ago
rschwiebrschwieb
108k12105253
$endgroup$
add a comment |
$begingroup$
If you already know a local ring has only trivial idempotents, then you can reason this way:
Suppose the characteristic of a ring $R$ is finite, say $n$, and is divisible by more than one prime. The ring contains a copy of $mathbb Z/nmathbb Z$. So to show $R$ isn't local, it suffices to show that $mathbb Z/nmathbb Z$ contains a nontrivial idempotent, so that $R$ will also contain a nontrivial idempotent.
By reasoning with the Chinese remainder theorem, you can quickly see that $mathbb Z/nmathbb Z$ is isomorphic to $mathbb Z/p^kmathbb Z$ for primes $p$ dividing $n$ and powers $k$ depending on $p$. Since there is more than one prime dividing $n$ (our assumption) there is at least one nontrivial idempotent splitting $mathbb Z/nmathbb Z$ into two pieces. This is clearly a nontrivial idempotent of $R$ too.
So by contrapositive, we have shown that a local ring $R$ must have either characteristic $0$, or else it has finite characteristic that is a power of a prime.
Since we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible.
answered 2 days ago
rschwiebrschwieb
108k12105253
$endgroup$
add a comment |
$begingroup$
If you already know a local ring has only trivial idempotents, then you can reason this way:
Suppose the characteristic of a ring $R$ is finite, say $n$, and is divisible by more than one prime. The ring contains a copy of $mathbb Z/nmathbb Z$. So to show $R$ isn't local, it suffices to show that $mathbb Z/nmathbb Z$ contains a nontrivial idempotent, so that $R$ will also contain a nontrivial idempotent.
By reasoning with the Chinese remainder theorem, you can quickly see that $mathbb Z/nmathbb Z$ is isomorphic to $mathbb Z/p^kmathbb Z$ for primes $p$ dividing $n$ and powers $k$ depending on $p$. Since there is more than one prime dividing $n$ (our assumption) there is at least one nontrivial idempotent splitting $mathbb Z/nmathbb Z$ into two pieces. This is clearly a nontrivial idempotent of $R$ too.
So by contrapositive, we have shown that a local ring $R$ must have either characteristic $0$, or else it has finite characteristic that is a power of a prime.
Since we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible.
answered 2 days ago
rschwiebrschwieb
108k12105253
$endgroup$
If you already know a local ring has only trivial idempotents, then you can reason this way:
Suppose the characteristic of a ring $R$ is finite, say $n$, and is divisible by more than one prime. The ring contains a copy of $mathbb Z/nmathbb Z$. So to show $R$ isn't local, it suffices to show that $mathbb Z/nmathbb Z$ contains a nontrivial idempotent, so that $R$ will also contain a nontrivial idempotent.
By reasoning with the Chinese remainder theorem, you can quickly see that $mathbb Z/nmathbb Z$ is isomorphic to $mathbb Z/p^kmathbb Z$ for primes $p$ dividing $n$ and powers $k$ depending on $p$. Since there is more than one prime dividing $n$ (our assumption) there is at least one nontrivial idempotent splitting $mathbb Z/nmathbb Z$ into two pieces. This is clearly a nontrivial idempotent of $R$ too.
So by contrapositive, we have shown that a local ring $R$ must have either characteristic $0$, or else it has finite characteristic that is a power of a prime.
Since we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible.
answered 2 days ago
rschwiebrschwieb
108k12105253
answered 2 days ago
rschwiebrschwieb
108k12105253
answered 2 days ago
rschwiebrschwieb
108k12105253
answered 2 days ago
rschwiebrschwieb
108k12105253
108k12105253
add a comment |
add a comment |
2
$begingroup$
The characteristic of a ring need not be prime: for instance, the characteristic of $mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain.
$endgroup$
– Claudius
2 days ago
2
$begingroup$
I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right?
$endgroup$
– Arnaud D.
2 days ago
$begingroup$
It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring?
$endgroup$
– Claudius
2 days ago
$begingroup$
Like integral domain you can classify the characteristic of a local ring.
$endgroup$
– user371231
2 days ago
$begingroup$
I much as I guess $6$ cannot be characteristic of a local ring
$endgroup$
– user371231
2 days ago