# Confusion about commutative differential algebra

In the article "Sullivan minimal model" of the Encyclopedia of Math, we read the following:

Let $(A,d)$ be a commutative differential graded algebra such that $H^0(A,d)=\mathbb Q$, $H^1(A,d)=0$ and $\dim H^p(A,d)<\infty$ for each $p$. There exists then a quasi-isomorphism of commutative differential graded algebras $$\phi: (\wedge V,d)\longrightarrow (A,d),$$ where $\wedge V$ denotes the free commutative algebra on the graded vector space of finite type $V$ and $d(V)\subset \wedge^{\geq 2}V$.

1. What is exactly the vector space $V$ here, do we know it explicitly or is it part of the existence statement?

2. Is the ground field for $V$ is the same as the ground field of the algebra $A$?

3. Do the $(A,d)$ and $(\wedge V,d)$ have the same differential $d$ or is it simply a notation but they are different differentials?

• $V$ is part of the existence statement, the ground field is the same, and the $d$'s are different. – Cheerful Parsnip Feb 6 '13 at 2:47

$V$ is part of the existence statement, the ground field is the same, and the $d$'s are different.