I can't find the reference for the following fact:
Let $X$ be an affine variety and let $Y$ be its smooth resolution. $H^0(X,\mathcal{O}_x)=H^0(Y,\mathcal{O}_Y)$ if and only if $X$ is normal.
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4$\begingroup$ I doubt you will find a reference for that statement. By the universal property of the normalization, every proper, birational morphism $\pi:Y\to X$ with $Y$ normal (e.g., with $Y$ smooth) factors through the normalization $\nu:\widetilde{X}\to X$. Thus, $H^0(\widetilde{X},\mathcal{O}_{\widetilde{X}})$ is contained in $H^0(Y,\mathcal{O}_Y)$. Since $\nu$ is affine, $H^0(X,\mathcal{O}_X)\to H^0(\widetilde{X},\mathcal{O}_X)$ is isomorphic if and only if $\nu$ is isomorphisc. If $X$ is normal, use Zariski's Main Theorem, p. 280 of Hartshorne's Algebraic geometry. $\endgroup$– Jason StarrCommented Dec 5, 2017 at 14:26
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3$\begingroup$ ...or instead of ZMT you could argue that if $X$ is normal, then $\pi$ is an isomorphism outside a codimension $2$ subset $Z$ and hence $\mathscr O_X\to \pi_*\mathscr O_Y$ is an isom on $X\setminus Z$ and, again since $X$ is normal and hence $S_2$, then $\mathscr O_X\to \pi_*\mathscr O_Y$ is an isom on $X$. $\endgroup$– Sándor KovácsCommented Dec 5, 2017 at 18:57
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