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I have a suggestion for you. Try it when A=k[G]$A=k[G]$ for a finite group G$G$ and E=k[H]$E=k[H]$ for a subgroup H$H$. Then R$R$ should be k[G/H]$k[G/H]$, which of course will only be a coalgebra and not a Hopf algebra if H$H$ is not normal. This example leads me to doubt your claim that R$R$ is a coalgebra in the category of A$A$-algebras, since I don't think R$R$ is an A$A$-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of H$H$. A tensor over E$E$ with N$N$ is just the induced G-representation.

I have a suggestion for you. Try it when A=k[G] for a finite group G and E=k[H] for a subgroup H. Then R should be k[G/H], which of course will only be a coalgebra and not a Hopf algebra if H is not normal. This example leads me to doubt your claim that R is a coalgebra in the category of A-algebras, since I don't think R is an A-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of H. A tensor over E with N is just the induced G-representation.

I have a suggestion for you. Try it when $A=k[G]$ for a finite group $G$ and $E=k[H]$ for a subgroup $H$. Then $R$ should be $k[G/H]$, which of course will only be a coalgebra and not a Hopf algebra if $H$ is not normal. This example leads me to doubt your claim that $R$ is a coalgebra in the category of $A$-algebras, since I don't think $R$ is an $A$-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of $H$. A tensor over $E$ with $N$ is just the induced G-representation.

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Greg Kuperberg
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I have a suggestion for you. Try it when A=k[G] for a finite group G and E=k[H] for a subgroup H. Then R should be k[G/H], which of course will only be a coalgebra and not a Hopf algebra if H is not normal. This example leads me to doubt your claim that R is a coalgebra in the category of A-algebras, since I don't think R is an A-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of H. A tensor over E with N is just the induced G-representation.

                     Mark

I have a suggestion for you. Try it when A=k[G] for a finite group G and E=k[H] for a subgroup H. Then R should be k[G/H], which of course will only be a coalgebra and not a Hopf algebra if H is not normal. This example leads me to doubt your claim that R is a coalgebra in the category of A-algebras, since I don't think R is an A-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of H. A tensor over E with N is just the induced G-representation.

                     Mark

I have a suggestion for you. Try it when A=k[G] for a finite group G and E=k[H] for a subgroup H. Then R should be k[G/H], which of course will only be a coalgebra and not a Hopf algebra if H is not normal. This example leads me to doubt your claim that R is a coalgebra in the category of A-algebras, since I don't think R is an A-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of H. A tensor over E with N is just the induced G-representation.

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Mark Hovey
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I have a suggestion for you. Try it when A=k[G] for a finite group G and E=k[H] for a subgroup H. Then R should be k[G/H], which of course will only be a coalgebra and not a Hopf algebra if H is not normal. This example leads me to doubt your claim that R is a coalgebra in the category of A-algebras, since I don't think R is an A-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of H. A tensor over E with N is just the induced G-representation.

                     Mark