Associated graded of filtered module-algebra over a Hopf algebra I ran across the following statement in a paper, and it seems fishy to me: 
Lemma: If $A$ is any Hopf algebra, and if $U$ is an $\mathbb{N}_0$-filtered $A$-module algebra, then $U$ and $\mathrm{gr} (U)$ are isomorphic as $A$-modules.
There is no proof of the lemma, it just states that it is a well-known fact.
Such an isomorphism cannot be canonical: consider just the case that $A = k$ is a field and $U$ is any filtered $k$-algebra.  In this situation there are plenty of vector space isomorphisms of $U$ with $\mathrm{gr}(U)$, just by pulling back a basis of each $U_{n} / U_{n-1}$, but these are hardly canonical.
So if the lemma is true, it is saying that there is some way to choose one of these maps so that it is an isomorphism of $A$-modules.
Question
Is the lemma true?  If no, what is a counterexample?  If yes, could you please provide a proof or a reference?
 A: The statement is definitely false.  For example, let $A = \mathbb k[x]$ be the group algebra of $\mathbb Z$.  Let $U$ be the two-dimensional module in which $x$ acts by $\bigl( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \bigr)$.  The span of $\bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\bigr)$ is a submodule on which $x$ acts by $1$, and the quotient is also one-dimensional with $x=1$.  So we give $U$ a filtration $0 \subset \mathbb k \bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\bigr) \subset U$.  Then the associated graded $\operatorname{gr}(U)$ is the two-dimensional $A$-module on which $x$ acts by $\bigl( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$.  This is not isomorphic to $U$ as and $A$-module, since in $U$ the $x$-action is not diagonalizable.
In general, you should only expect an isomorphism $\operatorname{gr}(U) \cong U$ if you have some semisimplicity.  For example, when $A$ is the universal enveloping algebra of a semisimple Lie algebra, and $U$ is locally finite-dimensional (each filtered piece is finite-dimensional), or the group algebra of a finite group in characteristic $0$ (or prime to the order of the group) and $U$ is arbitrary, then you can build a noncanonical $A$-module isomorphism $\operatorname{gr}(U) \cong U$.
You should write to the author of the paper.
