# Quotient of quasi-isomorphic cdga's

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.

• Are your cdgas $\mathbb{N}$-graded, $\mathbb{Z}$-graded...? – Najib Idrissi Nov 8 '18 at 8:24
• @NajibIdrissi Z-graded – Hsuan-Yi Liao Nov 9 '18 at 3:05
• It seems that you have changed the question in a rather drastic way. It would be better to post a new question. – Mike Miller Nov 9 '18 at 3:06

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 $$f(t) = (t^2 + 1)$$.
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.