Characters of $mathbbG_m$ The Next CEO of Stack OverflowAre there always nontrivial primitive elements in a Hopf algebra?Isomorphism of affine group schemes of rank 2Identities that connect antipode with multiplication and comultiplicationCocommutative k-Hopf algebra finite as k-vector space represents a constant group functorDo complete Hopf algebras have an antipode?Primitive elements of finite dimensional Hopf algebrasComultiplication in Convolution algebra: Where is the mistake?A (higher) categorical approach to representation theoryHopf Ideal and Normal Hopf idealgroup-like elements of a Hopf algebra and the group algebra
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Characters of $mathbbG_m$
The Next CEO of Stack OverflowAre there always nontrivial primitive elements in a Hopf algebra?Isomorphism of affine group schemes of rank 2Identities that connect antipode with multiplication and comultiplicationCocommutative k-Hopf algebra finite as k-vector space represents a constant group functorDo complete Hopf algebras have an antipode?Primitive elements of finite dimensional Hopf algebrasComultiplication in Convolution algebra: Where is the mistake?A (higher) categorical approach to representation theoryHopf Ideal and Normal Hopf idealgroup-like elements of a Hopf algebra and the group algebra
$begingroup$
Fix a field $k$. Let $mathbbG_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbbG_m$ is an endomorphism of affine group schemes $mathbbG_m to mathbbG_m$, or, equivalently, the data of a group homomorphisms $chi_R: R^times to R^times$ for each $k-$algebra $R$ such that for each map of $k-$algebras $f: R to S$, we have $f circ chi_R = chi_S circ f$.
Another approach, which uses the Hopf algebra representing $mathbbG_m$, is to look for group-like elements $P in k[mathbbG_m]:=k[X,X^-1]$. These are invertible elements such that $Delta(P)=Potimes P$. Since $Delta$ is a map of $k-$algebras satisfying $Delta(X)=X otimes X$, if we write
$$P = sum_i=-m^n a_iX^i,$$
the above condition translates to
$$Delta(P) = sum_i=-m^n a_iX^iotimes X^i overset!= sum_i,j=-m^n a_ia_jX^iotimes X^j = P otimes P.$$
By comparing coefficients, we see immediately that $P=X^i$ for some $i in mathbbZ$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbbG_m$ are in bijection with $mathbbZ$.
My question is: is this fact true also if $k$ is not a field? In particular: my proof uses the fact that $k$ is a field when comparing coefficients. In a general ring $k$, there can be zerodivisors and idempotents that are not $0$ or $1$.
abstract-algebra representation-theory hopf-algebras group-schemes
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
$endgroup$
add a comment |
$begingroup$
Fix a field $k$. Let $mathbbG_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbbG_m$ is an endomorphism of affine group schemes $mathbbG_m to mathbbG_m$, or, equivalently, the data of a group homomorphisms $chi_R: R^times to R^times$ for each $k-$algebra $R$ such that for each map of $k-$algebras $f: R to S$, we have $f circ chi_R = chi_S circ f$.
Another approach, which uses the Hopf algebra representing $mathbbG_m$, is to look for group-like elements $P in k[mathbbG_m]:=k[X,X^-1]$. These are invertible elements such that $Delta(P)=Potimes P$. Since $Delta$ is a map of $k-$algebras satisfying $Delta(X)=X otimes X$, if we write
$$P = sum_i=-m^n a_iX^i,$$
the above condition translates to
$$Delta(P) = sum_i=-m^n a_iX^iotimes X^i overset!= sum_i,j=-m^n a_ia_jX^iotimes X^j = P otimes P.$$
By comparing coefficients, we see immediately that $P=X^i$ for some $i in mathbbZ$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbbG_m$ are in bijection with $mathbbZ$.
My question is: is this fact true also if $k$ is not a field? In particular: my proof uses the fact that $k$ is a field when comparing coefficients. In a general ring $k$, there can be zerodivisors and idempotents that are not $0$ or $1$.
abstract-algebra representation-theory hopf-algebras group-schemes
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
$endgroup$
$begingroup$
If $k$ is a direct product of two nontrivial rings $A$ and $B$, then you also have grouplike elements $P$ of the form $P = left(X^a, X^bright)$ for two different integers $a$ and $b$, where the $left(X^a, X^bright)$ roughly means "acts as $X^a$ on the first factor and as $X^b$ on the second". (Note that any $k$-algebra splits into an "$A$-part" and a "$B$-part", and that $k$-algebra homomorphisms respect this decomposition.) So some conditions on $k$ will likely be unavoidable.
$endgroup$
– darij grinberg
Mar 18 at 13:29
add a comment |
$begingroup$
Fix a field $k$. Let $mathbbG_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbbG_m$ is an endomorphism of affine group schemes $mathbbG_m to mathbbG_m$, or, equivalently, the data of a group homomorphisms $chi_R: R^times to R^times$ for each $k-$algebra $R$ such that for each map of $k-$algebras $f: R to S$, we have $f circ chi_R = chi_S circ f$.
Another approach, which uses the Hopf algebra representing $mathbbG_m$, is to look for group-like elements $P in k[mathbbG_m]:=k[X,X^-1]$. These are invertible elements such that $Delta(P)=Potimes P$. Since $Delta$ is a map of $k-$algebras satisfying $Delta(X)=X otimes X$, if we write
$$P = sum_i=-m^n a_iX^i,$$
the above condition translates to
$$Delta(P) = sum_i=-m^n a_iX^iotimes X^i overset!= sum_i,j=-m^n a_ia_jX^iotimes X^j = P otimes P.$$
By comparing coefficients, we see immediately that $P=X^i$ for some $i in mathbbZ$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbbG_m$ are in bijection with $mathbbZ$.
My question is: is this fact true also if $k$ is not a field? In particular: my proof uses the fact that $k$ is a field when comparing coefficients. In a general ring $k$, there can be zerodivisors and idempotents that are not $0$ or $1$.
abstract-algebra representation-theory hopf-algebras group-schemes
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
$endgroup$
Fix a field $k$. Let $mathbbG_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbbG_m$ is an endomorphism of affine group schemes $mathbbG_m to mathbbG_m$, or, equivalently, the data of a group homomorphisms $chi_R: R^times to R^times$ for each $k-$algebra $R$ such that for each map of $k-$algebras $f: R to S$, we have $f circ chi_R = chi_S circ f$.
Another approach, which uses the Hopf algebra representing $mathbbG_m$, is to look for group-like elements $P in k[mathbbG_m]:=k[X,X^-1]$. These are invertible elements such that $Delta(P)=Potimes P$. Since $Delta$ is a map of $k-$algebras satisfying $Delta(X)=X otimes X$, if we write
$$P = sum_i=-m^n a_iX^i,$$
the above condition translates to
$$Delta(P) = sum_i=-m^n a_iX^iotimes X^i overset!= sum_i,j=-m^n a_ia_jX^iotimes X^j = P otimes P.$$
By comparing coefficients, we see immediately that $P=X^i$ for some $i in mathbbZ$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbbG_m$ are in bijection with $mathbbZ$.
My question is: is this fact true also if $k$ is not a field? In particular: my proof uses the fact that $k$ is a field when comparing coefficients. In a general ring $k$, there can be zerodivisors and idempotents that are not $0$ or $1$.
abstract-algebra representation-theory hopf-algebras group-schemes
abstract-algebra representation-theory hopf-algebras group-schemes
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
asked Mar 18 at 12:54
57Jimmy57Jimmy
3,470422
3,470422
$begingroup$
If $k$ is a direct product of two nontrivial rings $A$ and $B$, then you also have grouplike elements $P$ of the form $P = left(X^a, X^bright)$ for two different integers $a$ and $b$, where the $left(X^a, X^bright)$ roughly means "acts as $X^a$ on the first factor and as $X^b$ on the second". (Note that any $k$-algebra splits into an "$A$-part" and a "$B$-part", and that $k$-algebra homomorphisms respect this decomposition.) So some conditions on $k$ will likely be unavoidable.
$endgroup$
– darij grinberg
Mar 18 at 13:29
add a comment |
$begingroup$
If $k$ is a direct product of two nontrivial rings $A$ and $B$, then you also have grouplike elements $P$ of the form $P = left(X^a, X^bright)$ for two different integers $a$ and $b$, where the $left(X^a, X^bright)$ roughly means "acts as $X^a$ on the first factor and as $X^b$ on the second". (Note that any $k$-algebra splits into an "$A$-part" and a "$B$-part", and that $k$-algebra homomorphisms respect this decomposition.) So some conditions on $k$ will likely be unavoidable.
$endgroup$
– darij grinberg
Mar 18 at 13:29
$begingroup$
If $k$ is a direct product of two nontrivial rings $A$ and $B$, then you also have grouplike elements $P$ of the form $P = left(X^a, X^bright)$ for two different integers $a$ and $b$, where the $left(X^a, X^bright)$ roughly means "acts as $X^a$ on the first factor and as $X^b$ on the second". (Note that any $k$-algebra splits into an "$A$-part" and a "$B$-part", and that $k$-algebra homomorphisms respect this decomposition.) So some conditions on $k$ will likely be unavoidable.
$endgroup$
– darij grinberg
Mar 18 at 13:29
$begingroup$
If $k$ is a direct product of two nontrivial rings $A$ and $B$, then you also have grouplike elements $P$ of the form $P = left(X^a, X^bright)$ for two different integers $a$ and $b$, where the $left(X^a, X^bright)$ roughly means "acts as $X^a$ on the first factor and as $X^b$ on the second". (Note that any $k$-algebra splits into an "$A$-part" and a "$B$-part", and that $k$-algebra homomorphisms respect this decomposition.) So some conditions on $k$ will likely be unavoidable.
$endgroup$
– darij grinberg
Mar 18 at 13:29
add a comment |
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$begingroup$
If $k$ is a direct product of two nontrivial rings $A$ and $B$, then you also have grouplike elements $P$ of the form $P = left(X^a, X^bright)$ for two different integers $a$ and $b$, where the $left(X^a, X^bright)$ roughly means "acts as $X^a$ on the first factor and as $X^b$ on the second". (Note that any $k$-algebra splits into an "$A$-part" and a "$B$-part", and that $k$-algebra homomorphisms respect this decomposition.) So some conditions on $k$ will likely be unavoidable.
$endgroup$
– darij grinberg
Mar 18 at 13:29