Isomorphism of Associated Graded Algebras

Question

Suppose $A$ and $B$ are two filtered associative $k$-algebras with increasing filtrations $F^{\bullet}$ and $G^{\bullet}$, respectively. It is safe to assume that the filtrations are exhaustive. Assume that there is a filtered algebra morphism $f:A\rightarrow B$, that is, f satisfies $f(F^n)\subseteq G^n$. If the induced map $gr(f): gr_{F^{\bullet}}(A)\rightarrow gr_{G^{\bullet}}(B)$ is an isomorphism of graded $k$-algebras, is it true that the original map $f$ is also a $k$-algebra isomorphism. If not, under what conditions would that be true? Thank you for your help and feedback.

Answer 1

Here's another proof, which is basically the same proof as last time, but references the Five lemma.

Lemma. For the map $f: A \to B$, if the induced map $$gr^{\leq p}_F(A) \to gr^{\leq p}_G(B)$$ is an isomorphism then $F^iA \to F^iB$ is an isomorphism for $i \leq p$.

Proof. If $p=0$ the $gr^0_F(A) = F^0A, gr^0_GB = G^0B$ so there is nothing to prove. Now assume $F^{p-1}A \to F^{p-1}B$ is an isomorphism. We have a morphism between exact sequences:

$0 \to F^{p-1}A \to F^pA \to gr_F^p(A) \to 0$

$0 \to G^{p-1}B \to G^pB \to gr_G^p(B) \to 0$

Where all the arrows except the middle we know are isomorphisms. By the five lemma, the middle is an isomorphism, too. $\square$

Now we know $F^iA \to G^iB$ is an isomorphism for all $i$, and hence that $$\bigcup F^iA \to \bigcup G^iB$$ is an isomorphism. So long as the filtrations are exhaustive, we're done.

Answer 2

You just have to assume the filtrations are exhaustive.

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I'll use a bar to denote the associated graded, e.g. $\bar f = gr(f)$, $\bar A^k = F^kA/F^{k-1}A$ and if $y \in G^kB\setminus G^{k-1}B$ then $\bar y \in \bar B^k$. (It would have been better notation to have defined $\pi_{k}: F^kA \to F^kA/F^{k-1}A$ and kept track of $k$.)

Injectivity: Assume $\bar f$ is an isomorphism. Let $x \in A$ and take the minimal $k$ such that $x \in F^kA$. Then $\bar x \in \bar A^k$ is nonzero, so $\bar f(\bar x) \neq 0$. But $\bar f(\bar x) = \overline{f(x)}$ so this means $f(x) \neq 0$.

Note. It's important that $\bigcup F^kA = A$, otherwise we can't find a minimal such $k$ (the set of $k$ with $x \in F^kA$ may be empty).

Surjectivity: Let $y \in G^kB$ for minimal $k$ and $\bar y \in \bar B^k$ the corresponding element. Take a preimage $\bar x \in \bar A^k$ and lift this to $x \in F^kA$. We can't say that $f(x) = y$, but at least $y_1 := y - f(x) \in G^{k-1}B$. Now we repeat for $y_1$ and find $x_1$ such that $y_1 - f(x_1) \in G^{k-2}B$. And so on until $k-i = 0$ and we can find a preimage $x + x_1 + x_2 + \cdots + x_{k-1}$.

Note. It's important that $\bigcup G^kB = B$, else we can't find a minimal $k$ again.

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Added:

In the case of decreasing sequences, I think you can prove injectivity the same way but you need to assume $\bigcap F^kA=0$. For surjectivity you need to assume completeness with respect to the $G$ filtration so that the infinite sum makes sense and probably also the intersection condition.