Mixing
The group ring of a ring.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.
An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?
abstract-algebra group-theory ring-theory group-rings
asked 3 hours ago
palio
4,02832460
add a comment |Â
up vote
2
down vote
favorite
Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.
An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?
abstract-algebra group-theory ring-theory group-rings
asked 3 hours ago
palio
4,02832460
4
Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
3 hours ago
1
I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago
@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago
@palio No, I did not.
– Lord Shark the Unknown
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.
An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?
abstract-algebra group-theory ring-theory group-rings
asked 3 hours ago
palio
4,02832460
Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.
An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?
abstract-algebra group-theory ring-theory group-rings
abstract-algebra group-theory ring-theory group-rings
asked 3 hours ago
palio
4,02832460
asked 3 hours ago
palio
4,02832460
asked 3 hours ago
palio
4,02832460
asked 3 hours ago
palio
4,02832460
asked 3 hours ago
palio
4,02832460
4,02832460
4
Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
3 hours ago
1
I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago
@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago
@palio No, I did not.
– Lord Shark the Unknown
1 hour ago
add a comment |Â
4
Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
3 hours ago
1
I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago
@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago
@palio No, I did not.
– Lord Shark the Unknown
1 hour ago
4
4
Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
3 hours ago
Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
3 hours ago
1
1
I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago
I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago
@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago
@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago
@palio No, I did not.
– Lord Shark the Unknown
1 hour ago
@palio No, I did not.
– Lord Shark the Unknown
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.
answered 2 hours ago
David Hill
8,3821618
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
1
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
1
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
1
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
add a comment |Â
up vote
4
down vote
It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.
Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.
One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)
answered 2 hours ago
rschwieb
102k1299235
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.
answered 2 hours ago
David Hill
8,3821618
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
1
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
1
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
1
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.
answered 2 hours ago
David Hill
8,3821618
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
1
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
1
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
1
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.
answered 2 hours ago
David Hill
8,3821618
It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.
answered 2 hours ago
David Hill
8,3821618
answered 2 hours ago
David Hill
8,3821618
answered 2 hours ago
David Hill
8,3821618
answered 2 hours ago
David Hill
8,3821618
8,3821618
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
1
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
1
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
1
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
add a comment |Â
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
1
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
1
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
1
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
– palio
2 hours ago
1
1
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
$r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
– David Hill
2 hours ago
1
1
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
– David Hill
2 hours ago
1
1
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
– palio
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
yes, that is correct.
– David Hill
2 hours ago
add a comment |Â
up vote
4
down vote
It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.
Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.
One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)
answered 2 hours ago
rschwieb
102k1299235
add a comment |Â
up vote
4
down vote
It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.
Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.
One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)
answered 2 hours ago
rschwieb
102k1299235
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.
Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.
One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)
answered 2 hours ago
rschwieb
102k1299235
It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.
Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.
One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)
answered 2 hours ago
rschwieb
102k1299235
answered 2 hours ago
rschwieb
102k1299235
answered 2 hours ago
rschwieb
102k1299235
answered 2 hours ago
rschwieb
102k1299235
102k1299235
add a comment |Â
add a comment |Â
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4
Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
3 hours ago
1
I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago
@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago
@palio No, I did not.
– Lord Shark the Unknown
1 hour ago