Hilbert's 90 Theorem proof clarification (Milne's Étale Cohomology)

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So I'm reading Milne's Étale Cohomology and stuck in trouble to understand the proof of Proposition 4.9 Hilbert's Theorem 90), from §4; page 124:

The canonical maps $$ H^1(X_{Zar}, \mathcal{O}_X^*) \to H^1(X_{et}, \mathbb{G}_m) \to H^1(X_{fl}, \mathbb{G}_m)$$ are isomorphisms. $X_x$ with $x = Zar,et, fl$ denotes the "topology" of $X$. the proof only shows $ H^1(X_{Zar}, \mathcal{O}_X^*) \to H^1(X_{fl})$. according to the proof by Theroem 1.18 it suffice to show that $R^1f_*\mathbb{G}_m=0$. for this, let $U$ be a Zariski open subset of $X$ and we must show that $H^1(U_{fl}, \mathbb{G}_m)$ becomes zero. if we consider a Zariski open affine cover of $U$ by a family $U_i$, we can assume that $U$ is affine, since we can the $U_i$ choose to be affine, i.e. $U=spec(A)$. this reduction is clearly can be done with sheaf axiom in mind.

The step that confuses me that futhermore is told that is can be also be assumed that $A$ is a local ring. why this assumption can be done here?