non-flatness of $k[\epsilon] \to k$

Question

Let $k$ be a field and $k[\epsilon] = k[x]/(x^2) \to k, x \mapsto 0$ the morphism, which is an example of a non-flat morphism. Although I already know a direct way to show why it's not flat, at wikipedia site I found another argument for it's non-flatness I not understand. The argument says that the morphism above is not flat because

$$ k \otimes_{k[\epsilon]}^L k $$

is an infinite complex with explicit resolution

$$ ... \to k[\epsilon] \to k[\epsilon] \to k[\epsilon] \to k$$

where the maps are the multiplication maps with $\epsilon$.
My question is why the infinity of this complex implies that $k[\epsilon] \to k$ is not flat? Or how exactly this argument works?

Answer 1

If $k$ is flat over $k[\epsilon],$ then $(k\otimes_{k[\epsilon]}k)[0]$ would be quasi-isomorphic to $k\otimes^L_{k[\epsilon]}k,$ because we could use $k[0]$ as a flat resolution of $k$ to compute the derived tensor product. This would give \begin{align*} k\otimes^L_{k[\epsilon]}k&\simeq k\otimes_{k[\epsilon]}(k[0])\\ &\simeq (k\otimes_{k[\epsilon]}k)[0]. \end{align*}

However, $k\otimes^L_{k[\epsilon]}k\simeq\bigoplus_{i\geq0} k[i],$ showing that $k\otimes^L_{k[\epsilon]}k$ has nonzero cohomology in infinitely many degrees, unlike $(k\otimes_{k[\epsilon]}k)[0]$.