Quotient of quasi-isomorphic cdga's

Question

I'm looking for a theorem about quotient of quasi-isomorphic cdga's:

Let $A, B$ be two cdga's (commutative differential $\mathbb Z$-graded algebra) of nonpositive degrees, and $\mathfrak m \subset A, \mathfrak n \subset B$ be maximal ideals closed under differentials. Suppose $\phi: A \to B$ is a quasi-isomorphism of cdga's and $\phi(\mathfrak m) \subset \mathfrak n$. Is the induced map $\phi: A/ \mathfrak m^2 \to B/\mathfrak n^2$ still a quasi-isomorphism?

I guess more assumptions on $A,B, \mathfrak m, \mathfrak n$ might be needed. Any theorems of this type would be helpful.

Answer 1

Let $A$ be the complex $\Bbb Q$ concentrated in degree zero, and let $B$ be the CDGA $(\Bbb Q[t] \to \Bbb Q[t] dt)$ concentrated in (cohomological) degrees $0$ and $1$, with de Rham differential $d f(t) = f'(t) dt$. Then $B$ is a commutative DGA under the wedge product, and the natural inclusion $\Bbb Q \to \Bbb Q[t] \subset B$ becomes a quasi-isomorphism of CDGAs.

The only maximal ideal $\mathfrak m \subset A$ is zero, whereas there are a lot of maximal ideals of $B$ of the form $\mathfrak n = ((f(t)) \to \Bbb Q[t] dt)$ for $(f(t)) \subset \Bbb Q[t]$ a maximal ideal of $\Bbb Q[t]$ generated by a monic irreducible polynomial $f(t)$. The resulting map on quotients is $\Bbb Q \to \Bbb Q[t] / (f(t))$, which is an isomorphism if and only if $f(t)$ is linear; for example, we could take $f(t) = (t^2 + 1)$.

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For a positive result, you could ask about CDGAs concentrated in nonpositive (cohomological) degree such that $\mathfrak m$ and $\mathfrak n$ are closed under the differential; then any maximal ideal automatically contains all elements in nonzero degree.