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There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x +1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

This is with $R=k[x]$ and no evil canonical sheaf. It will be even more interesting if the canonical sheaf is non-trivial.

To answer your question (with my left-right correction for the second part) you need to follow David Speyer's answer, forgetting all about the evil antipode (it just does not exist in any useful for you form). $D$ is an $R$-$R$-bimodule. Now $D\otimes_k D$ has 4 stuctures of $R$-module. I call $R$ and $R^\prime$ to distinguish them. The comultiplication is now a map $$ \Delta : D \rightarrow \,_RD_{R^\prime} \otimes_R \,_RD_{R^\prime} $$ taking image in the equalizer of the two $R^\prime$-module structures. This is all you need this Christmas to define your actions.

There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x +1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

The stuff above is plain wrong. There is an action. See below.

To answer your question (with my left-right correction for the second part) you need to follow David Speyer's answer, forgetting all about the evil antipode (it just does not exist in any useful for you form). $D$ is an $R$-$R$-bimodule. Now $D\otimes_k D$ has 4 stuctures of $R$-module. I call $R$ and $R^\prime$ to distinguish them. The comultiplication is now a map $$ \Delta : D \rightarrow \,_RD_{R^\prime} \otimes_R \,_RD_{R^\prime} $$ taking image in the equalizer of the two $R^\prime$-module structures. This is all you need this Christmas to define your actions.

There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x -1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

This is with $R=k[x]$ and no evil canonical sheaf. It will be even more interesting if the canonical sheaf is non-trivial.

To answer your question (with my left-right correction for the second part) you need to follow David Speyer's answer, forgetting all about the evil antipode (it just does not exist in any useful for you form). $D$ is an $R$-$R$-bimodule. Now $D\otimes_k D$ has 4 stuctures of $R$-module. I call $R$ and $R^\prime$ to distinguish them. The comultiplication is now a map $$ \Delta : D \rightarrow \,_RD_{R^\prime} \otimes_R \,_RD_{R^\prime} $$ taking image in the equalizer of the two $R^\prime$-module structures. This is all you need this Christmas to define your actions.

There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x +1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

This is with $R=k[x]$ and no evil canonical sheaf. It will be even more interesting if the canonical sheaf is non-trivial.

To answer your question (with my left-right correction for the second part) you need to follow David Speyer's answer, forgetting all about the evil antipode (it just does not exist in any useful for you form). $D$ is an $R$-$R$-bimodule. Now $D\otimes_k D$ has 4 stuctures of $R$-module. I call $R$ and $R^\prime$ to distinguish them. The comultiplication is now a map $$ \Delta : D \rightarrow \,_RD_{R^\prime} \otimes_R \,_RD_{R^\prime} $$ taking image in the equalizer of the two $R^\prime$-module structures. This is all you need this Christmas to define your actions.

There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x -1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

This is with $R=k[x]$ and no evil canonical sheaf. It will be even more interesting if the canonical sheaf is non-trivial.

There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x -1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

This is with $R=k[x]$ and no evil canonical sheaf. It will be even more interesting if the canonical sheaf is non-trivial.

To answer your question (with my left-right correction for the second part) you need to follow David Speyer's answer, forgetting all about the evil antipode (it just does not exist in any useful for you form). $D$ is an $R$-$R$-bimodule. Now $D\otimes_k D$ has 4 stuctures of $R$-module. I call $R$ and $R^\prime$ to distinguish them. The comultiplication is now a map $$ \Delta : D \rightarrow \,_RD_{R^\prime} \otimes_R \,_RD_{R^\prime} $$ taking image in the equalizer of the two $R^\prime$-module structures. This is all you need this Christmas to define your actions.