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What is the Characteristic of a local ring?

1

$begingroup$

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

share|cite|improve this question

edited 10 hours ago

user371231

asked 13 hours ago

user371231user371231

417511

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  13 hours 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.
  13 hours 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
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Like integral domain you can classify the characteristic of a local ring.
  $endgroup$
  – user371231
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I much as I guess $6$ cannot be characteristic of a local ring
  $endgroup$
  – user371231
  13 hours ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

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

share|cite|improve this question

edited 10 hours ago

user371231

asked 13 hours ago

user371231user371231

417511

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  13 hours 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.
  13 hours 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
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Like integral domain you can classify the characteristic of a local ring.
  $endgroup$
  – user371231
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I much as I guess $6$ cannot be characteristic of a local ring
  $endgroup$
  – user371231
  13 hours 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

share|cite|improve this question

edited 10 hours ago

user371231

asked 13 hours ago

user371231user371231

417511

$endgroup$

share|cite|improve this question

edited 10 hours ago

user371231

asked 13 hours ago

user371231user371231

417511

share|cite|improve this question

edited 10 hours ago

user371231

asked 13 hours ago

user371231user371231

417511

asked 13 hours ago

user371231user371231

417511

asked 13 hours ago

user371231user371231

417511

2 Answers
2

active

oldest

votes

6

$begingroup$

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.

share|cite|improve this answer

edited 12 hours ago

anomaly

17.9k42666

answered 12 hours ago

MaxMax

16.5k11144

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

rschwiebrschwieb

108k12105253

$endgroup$

add a comment | 

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6

$begingroup$

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.

share|cite|improve this answer

edited 12 hours ago

anomaly

17.9k42666

answered 12 hours ago

MaxMax

16.5k11144

$endgroup$

add a comment | 

6

$begingroup$

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.

share|cite|improve this answer

edited 12 hours ago

anomaly

17.9k42666

answered 12 hours 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.

share|cite|improve this answer

edited 12 hours ago

anomaly

17.9k42666

answered 12 hours ago

MaxMax

16.5k11144

$endgroup$

share|cite|improve this answer

edited 12 hours ago

anomaly

17.9k42666

answered 12 hours ago

MaxMax

16.5k11144

edited 12 hours ago

anomaly

17.9k42666

edited 12 hours ago

anomaly

17.9k42666

answered 12 hours ago

MaxMax

16.5k11144

answered 12 hours ago

MaxMax

16.5k11144

1

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

rschwiebrschwieb

108k12105253

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 11 hours 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.

share|cite|improve this answer

answered 11 hours ago

rschwiebrschwieb

108k12105253

$endgroup$

share|cite|improve this answer

answered 11 hours ago

rschwiebrschwieb

108k12105253

answered 11 hours ago

rschwiebrschwieb

108k12105253

answered 11 hours ago

rschwiebrschwieb

108k12105253