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1

$begingroup$

Fix a field $k$. Let $mathbb{G}_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbb{G}_m$ is an endomorphism of affine group schemes $mathbb{G}_m to mathbb{G}_m$, 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 $mathbb{G}_m$, is to look for group-like elements $P in k[mathbb{G}_m]:=k[X,X^{-1}]$. 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 $i in mathbb{Z}$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbb{G}_m$ are in bijection with $mathbb{Z}$.

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

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asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

1

$begingroup$

Fix a field $k$. Let $mathbb{G}_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbb{G}_m$ is an endomorphism of affine group schemes $mathbb{G}_m to mathbb{G}_m$, 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 $mathbb{G}_m$, is to look for group-like elements $P in k[mathbb{G}_m]:=k[X,X^{-1}]$. 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 $i in mathbb{Z}$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbb{G}_m$ are in bijection with $mathbb{Z}$.

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

share|cite|improve this question

asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $mathbb{G}_m$ be the multiplicative affine group scheme over $k$. A $k-$character $chi$ of $mathbb{G}_m$ is an endomorphism of affine group schemes $mathbb{G}_m to mathbb{G}_m$, 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 $mathbb{G}_m$, is to look for group-like elements $P in k[mathbb{G}_m]:=k[X,X^{-1}]$. 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 $i in mathbb{Z}$, which tells us that $chi$ is exponentiation to some integer power. Hence, the characters of $mathbb{G}_m$ are in bijection with $mathbb{Z}$.

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

share|cite|improve this question

asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422

$endgroup$

share|cite|improve this question

asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422

share|cite|improve this question

asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422

asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422

asked Mar 18 at 12:54

57Jimmy57Jimmy

3,467422