Let $P$ be a graded commutative differential algebra over the field $GF(2)$ (hence also actually commutative), and let $d$ be the differential. Let $x$ be a homogeneous element of $P$, which is also a cycle, and not a boundary. Let $I$ be the ideal in $P$ generated by $x$. Let $Q$ = $P/x$ be the quotient of $P$ by the ideal generated by $x$. Let $HP = H^*(P,d)$ be the cohomology ring of $P$. Let $HI$ be the ideal in $HP$ generated by the cohomology class of $x$. Then under what conditions will $H^*(Q) = H^*(P/I)$ be the quotient $HP/HI$.
The following example of a "pinched" product is what I would like to learn more about.
Let $A = GF_2(z_a, a_1, a_2, \dots)$ with degree $a_i = i$, and degree $z$ = 2. The differential $d$ is given by $d(z) = 0, d(a_{2i+1}) = 0$, and $ d(a_{2i} ) = z* a_{2i-1}$. The cohomology of this algebra is a polynomial ring with generators $\{a_{2i}^2, i \ge 1, z, a_{2i-1}; i\ge 1\}$, subject to the relations $\{a_{2i-1}z = 0; i\ge 1\}$. We want to take the product of three copies of this algebra, and pinch together at the bottom.
Now let $P = A\otimes A\otimes A$, the product of 3 copies of $A$. Let $I$ be the ideal in $P$ generated by the single homogenous element $x = z\otimes1\otimes 1 + 1\otimes z \otimes 1 + 1\otimes 1 \otimes z$, which is also clearly a non-bounding cycle. I would like to say that the cohomology of the quotient $P/I$ is isomorphic to the quotient of the cohomology of $P$ by the ideal generated by the cohomology class of $x$.
Thanks for your help.