What algebraic structure does the set of endomorphisms of a ring have?
$begingroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
Jan 15 at 20:56
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
Jan 15 at 21:56
add a comment |
$begingroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
$endgroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
abstract-algebra ring-theory modules terminology ring-homomorphism
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
asked Jan 15 at 20:54
Keshav SrinivasanKeshav Srinivasan
2,08011443
2,08011443
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
Jan 15 at 20:56
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
Jan 15 at 21:56
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
Jan 15 at 20:56
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
Jan 15 at 21:56
1
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
Jan 15 at 20:56
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
Jan 15 at 20:56
1
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
Jan 15 at 21:56
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
Jan 15 at 21:56
add a comment |
1 Answer
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$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
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$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
$endgroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
answered Jan 15 at 20:58
Eric WofseyEric Wofsey
183k13209337
183k13209337
add a comment |
add a comment |
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1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
Jan 15 at 20:56
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
Jan 15 at 21:56