I'm reading a book "Complex Geometry" by Daniel Huybrechts. In this book he says that a simply connected dga satisfying some conditions must be minimal. (p.147, Remark 3.A.13) I tried to prove this statement but eventually failed. It seems that there is a counter example.


The statement in the book is as follows:



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> If $ (M=\bigoplus \_{i\ge0} M^i ,d) $ is a simply connected dga over a field $ k $ satisfying
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> 1) $ M^0 =k $
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> 2) $ M^+ := \bigoplus \_{i>0} M^i $ is free, i.e. there exist homogeneous elements $ x\_1 , x\_2 , ... $ of positive degree $d\_1 , d\_2 , ...$, such that $M^+ =S^+ <x\_1 , x\_2 , ...>$ (the positive part of the (super)symmetric algebra of the vector space generated by $x\_1 , x\_2 , ...$.
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> and such that $d(M)\subset M^+ \cdot M^+$, then $M$ is minimal. In fact, 1) and 2) suffice to deduce $M^1 =0$ as in the proof of the previous lemma.
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> 


And the counterexample I found is as follows:



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> Let $M=S^\*<x,y,z>$ with $x, y$ and $z$ homogeneous elements of degree $1$.
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> Set $dx=yz, dy=zx$ and $dz=xy$.
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> Using Leibniz rule and $x^2 = y^2 =z^2 = 0$ by graded commutativity, we can show that $d$ satisfies $d^2 = 0$.
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> e.g. $d(dx)=d(yz)=(dy)z-ydz=zxz-yxy=-xzz+xyy=0$
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> Moreover, since $d(ax+by+cz)=ayz+bzx+cxy=0$ implies $a=b=c=0$, we also know that $H^1(M,d)=0$, i.e. $(M,d)$ is simply connected.
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> Therefore, $(M,d)$ satisfies all assumptions in the statement but fails to have $M^1 =0$.
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> Moreover, $(M,d)$ is not minimal as $dx, dy, dz \ne0$
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**Question** : Is this indeed a counterexample and the statement is wrong? Or is there something I am confused or missing?


And then what is the importance of this remark? It seems that this is not used in the rest of the book.



A condition for a dga to be minimal
You are correct and the statement, as you cite it, is wrong. The counterexample you describe is well-known, and is usually cited to demonstrate precisely this failure.


