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Is the dual of a pure module also pure?

Suppose $H$ is a Hopf algebra over a field $K$ and $S$ is a ring which has a right $H$-action.

If ${}_HA$ is a pure left $H$-submodule, will $S\otimes_H A$ be a pure $S$-submodule ? If not, then what are the minimum conditions required (on $H$ or $S$) to make it so?

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  • $\begingroup$ I added the Hopf algebras tag because modules alone is too broad. $\endgroup$
    – David Roberts
    Apr 15, 2011 at 1:30
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    $\begingroup$ What does the first sentence ask? "Pure" is applied to submodules, no? Maybe you are asking if the dual to a pure short exact sequence is pure? As for the rest: something can be a pure submodule of a specific module, but I don't know what being a pure submodule in the abstract means. Can you point to definitions? $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 15, 2011 at 1:36
  • $\begingroup$ I don't know much about Hopf algebras in general, but it seems like if $S$ is flat you're all set. Here's a sketch, with $i:A\hookrightarrow H$ the map $i\otimes id_X:A\otimes X \hookrightarrow H\otimes X$ for all $X$: $S\otimes_H A \otimes X \rightarrow S\otimes_H H \otimes X$ is $id_S\otimes i\otimes id_X$, i.e. it's $id_S\otimes j$ where $j$ is an injection. This is injective for all injective $j$ iff $S$ is flat. Correct me if I've misunderstood your question. $\endgroup$
    – David White
    Apr 15, 2011 at 3:13
  • $\begingroup$ David Roberts: Thanks. Mariano: yes, you are right. Pure is applied to submodules. That was my mistake. here I recall the definition of a Pure submodule. Let M, P be modules over a ring R. If i:P\to M is injective then P is a pure submodule of M if, for any R-module X, the natural induced map on tensor products i\otimes id_X:P\otimes X \to M\otimes X is injective. My question is if A is an H-algebra which is a pure submodule of $H^A$, is the dual H-algebra $A^*$ with values in $H$ also a pure submodule of some extension? David: Thanks, but I dont what any condition on $S$. $\endgroup$
    – Neha
    Apr 15, 2011 at 12:53
  • $\begingroup$ @Neha: please do edit the question body with the new details. $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 15, 2011 at 18:35

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