Reading Lucian Badescu's Algebraic Surfaces I have encountered a proof I can't understand. That's Theorem 3.28 (M. Artin) at pages 41/42:
Theorem 3.28 (M. Artin). Let $(Y, y)$ be a two-dimensional normal singularity, and let $f: X \to Y$ be a desingularization of $(Y, y)$. If $(Y, y)$ is a rational singularity, $Z$ is the fundamental cycle of the fiber $E = f^{- 1}(y)_{red} = E_1 +··· +E_n$, and $m$ is the maximal ideal of the local ring $O_{Y,y}$, then $X \times_Y Spec(O_{Y,y},m^n) = nZ, H^0(O_{nZ}) \cong O_{Y,y}/m^n$, and $\dim_k(m^n /m^{n+1}) = -n · (Z^2) + 1$ for every $n \ge 1$.
About used notations: $(Y, y)$ be a normal singularity of an algebraic surface $Y$ means that the normal algebraic surface (= $k$-scheme) is everywhere smooth, exept in the singular point $y$. In addition we assume $k$ algebraically closed of arbitrary characeristic.
A desingularization of $(Y, y)$ is a morphism $f : X \to Y$ such that $X$ is nonsingular surface in a neighborhood of $f^{- 1} (y)$ and f is an isomorphism between $X \backslash f^{- 1} (y)$ and $Y \backslash \{y\}$.
The fundamental cycle is defined in 3.20 page 36/37.
The proof: We may obviously assume that $Y = Spec(A)$, with $A = O_{Y,y}$, because even though in that case $X$ is no longer a projective surface, the self-intersection of the divisor $Z$ on $X$ still makes sense, as $Z$ is still a projective curve.
In the next step the proof shows $m^n O_X = O_X(-nZ)$ for all $ n \ge 1$. [...] This part I understand.
Then Badscu uses two isomorphisms which I not understand:
(i) $m^n \cong H^0(m^nO_X)$
(ii) $A \cong H^0(O_X)$
Why are they true?
The (ii) I think I can solve: $Y,X$ are normal and therefore $A=H^0(Y)= H^0(Y \backslash \{y\})= H^0(X \backslash f^{- 1} (y))=H^0(X)$ by Stacks. The equality (i) is harder. Consider only case $n=1$. When $H^0(O_X) \otimes_A m = H^0(mO_X)$ is true?
Since $f:X\to Y$ is a proper morphism with connected fiber, we have $$f_*\mathcal{O}_X=\mathcal{O}_Y.$$ Then (2) is obtained just by taking global section, using the assumption that $Y$ being affine: $$H^0(X,\mathcal{O}_X)=H^0(Y,f_*\mathcal{O}_X)=H^0(Y,\mathcal{O}_Y)=A.$$
For (1), by projection formula, $\mathfrak{m}^n\mathcal{O}_Y=\mathfrak{m}^n\otimes f_*\mathcal{O}_X=f_*(f^*\mathfrak{m}^n\otimes\mathcal{O}_X)$. Now take global section and we are done.