Mixing
Elements in ring that satisfy $x^m+n=x^m$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am trying to recall a statement about ring which talks about the existence of an element $x$ that has the property $x^m+n=x^n$ for some positive integers $m$ and $n$, but I cannot remember where I read it before and cannot find any reference talking about this, and don't even remember whether this element or the ring has a specific name or not. Perhaps anyone can give me any reference or maybe explanation about this? I really appreciate the help. Thanks!
abstract-algebra ring-theory
asked Jul 27 at 3:13
bms
243
add a comment |Â
up vote
1
down vote
favorite
I am trying to recall a statement about ring which talks about the existence of an element $x$ that has the property $x^m+n=x^n$ for some positive integers $m$ and $n$, but I cannot remember where I read it before and cannot find any reference talking about this, and don't even remember whether this element or the ring has a specific name or not. Perhaps anyone can give me any reference or maybe explanation about this? I really appreciate the help. Thanks!
abstract-algebra ring-theory
asked Jul 27 at 3:13
bms
243
Every element in a finite ring has this property, proven by pigeonhole principle. In rings which are not finite, nilpotent and idempotent elements satisfy the property but are not the only examples.
– JMoravitz
Jul 27 at 3:27
1
You want to solve $x^n(x^m-1)=0$. $x=0$ and $x=1$ are obvious possibilities. More generally, if your ring has no zerodivisors, you look for roots of the factors, i.e.: $m$-th roots of unity; and nilpotent elements. In general rings, there will be many more.
– Torsten Schoeneberg
Jul 27 at 3:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to recall a statement about ring which talks about the existence of an element $x$ that has the property $x^m+n=x^n$ for some positive integers $m$ and $n$, but I cannot remember where I read it before and cannot find any reference talking about this, and don't even remember whether this element or the ring has a specific name or not. Perhaps anyone can give me any reference or maybe explanation about this? I really appreciate the help. Thanks!
abstract-algebra ring-theory
asked Jul 27 at 3:13
bms
243
I am trying to recall a statement about ring which talks about the existence of an element $x$ that has the property $x^m+n=x^n$ for some positive integers $m$ and $n$, but I cannot remember where I read it before and cannot find any reference talking about this, and don't even remember whether this element or the ring has a specific name or not. Perhaps anyone can give me any reference or maybe explanation about this? I really appreciate the help. Thanks!
abstract-algebra ring-theory
asked Jul 27 at 3:13
bms
243
asked Jul 27 at 3:13
bms
243
asked Jul 27 at 3:13
bms
243
asked Jul 27 at 3:13
bms
243
243
Every element in a finite ring has this property, proven by pigeonhole principle. In rings which are not finite, nilpotent and idempotent elements satisfy the property but are not the only examples.
– JMoravitz
Jul 27 at 3:27
1
You want to solve $x^n(x^m-1)=0$. $x=0$ and $x=1$ are obvious possibilities. More generally, if your ring has no zerodivisors, you look for roots of the factors, i.e.: $m$-th roots of unity; and nilpotent elements. In general rings, there will be many more.
– Torsten Schoeneberg
Jul 27 at 3:30
add a comment |Â
Every element in a finite ring has this property, proven by pigeonhole principle. In rings which are not finite, nilpotent and idempotent elements satisfy the property but are not the only examples.
– JMoravitz
Jul 27 at 3:27
1
You want to solve $x^n(x^m-1)=0$. $x=0$ and $x=1$ are obvious possibilities. More generally, if your ring has no zerodivisors, you look for roots of the factors, i.e.: $m$-th roots of unity; and nilpotent elements. In general rings, there will be many more.
– Torsten Schoeneberg
Jul 27 at 3:30
Every element in a finite ring has this property, proven by pigeonhole principle. In rings which are not finite, nilpotent and idempotent elements satisfy the property but are not the only examples.
– JMoravitz
Jul 27 at 3:27
Every element in a finite ring has this property, proven by pigeonhole principle. In rings which are not finite, nilpotent and idempotent elements satisfy the property but are not the only examples.
– JMoravitz
Jul 27 at 3:27
1
1
You want to solve $x^n(x^m-1)=0$. $x=0$ and $x=1$ are obvious possibilities. More generally, if your ring has no zerodivisors, you look for roots of the factors, i.e.: $m$-th roots of unity; and nilpotent elements. In general rings, there will be many more.
– Torsten Schoeneberg
Jul 27 at 3:30
You want to solve $x^n(x^m-1)=0$. $x=0$ and $x=1$ are obvious possibilities. More generally, if your ring has no zerodivisors, you look for roots of the factors, i.e.: $m$-th roots of unity; and nilpotent elements. In general rings, there will be many more.
– Torsten Schoeneberg
Jul 27 at 3:30
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864013%2felements-in-ring-that-satisfy-xmn-xm%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Every element in a finite ring has this property, proven by pigeonhole principle. In rings which are not finite, nilpotent and idempotent elements satisfy the property but are not the only examples.
– JMoravitz
Jul 27 at 3:27
1
You want to solve $x^n(x^m-1)=0$. $x=0$ and $x=1$ are obvious possibilities. More generally, if your ring has no zerodivisors, you look for roots of the factors, i.e.: $m$-th roots of unity; and nilpotent elements. In general rings, there will be many more.
– Torsten Schoeneberg
Jul 27 at 3:30