Let $X$ be a smooth quasiprojective scheme defined over $\mathbb{C}$, and let $\Omega^{\bullet}_X$ denote its cotangent complex, explicitly, we have:
$\Omega^{\bullet}_X:=\mathcal{O}_X\longrightarrow \Omega^1_X\longrightarrow...\longrightarrow \Omega^{\dim X}_X$.
We define the naive filtration, $\Omega^{\bullet,\ge p}_X$, as the subcomplex, given explicitly by:
$\Omega^{\bullet,\ge p}_X := 0\longrightarrow ...\longrightarrow 0\longrightarrow \Omega_X^p\longrightarrow...\longrightarrow \Omega_X^{\dim X}.$
Now, the Hodge filtration is defined by:
$\mathcal{F^p}H^i_{\text{dR}}(X) := \text{Im}(H^i(X, \Omega_X^{\bullet, \ge p})\longrightarrow H^i(X, \Omega_X^{\bullet})).$
My issue is that even though I clearly see that $\Omega_X^{\bullet, \ge p}$ is a subcomplex of $\Omega_X^{\bullet}$, I don't see why $\mathcal{F}^pH^i_{\text{dR}}(X)$ forms a vector subspace of $H^i_{\text{dR}}(X)$ and isn't the whole thing.
For example, in the case where $X$ is affine, consider $\mathcal{F}^1H^1_{\text{dR}}(X)$. In this case I believe the de Rham complex is acyclic so de Rham cohomology is simply the hypercohomology of said complex, in particular, it seems that $H^1_{\text{dR}}(X, \Omega_X^{\bullet, \ge 1}) = H^0(X, \Omega^1_X)$, however, $H^1_{dR}(X, \Omega_X^{\bullet})$ is isomorphic to a quotient of $H^0(X, \Omega^1_X)$ by the span of exact 1-forms.
It seems my reasoning implies that the first Hodge filtration of the $H^1_{\text{dR}}$ of an affine curve is trivial. Is this correct, or am I missing something?