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Let $A$ be an arbitrary commutative ring, and let $X=\mathrm{Proj\,}A[T_0,\ldots,T_n]/\langle T_iT_j\rangle$, so I wish to calculate $\Gamma(X,\mathcal{O}_X)$. We assume $0\le i

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Do you really need to use the Čech complex here? I don't know how to do all the computations with the Čech complex, but I see one way of simplifying it: note that if $k=i$ or $k=j$, then $\Gamma(D_+(\bar{T_i}),\mathcal{O}_X)=A[T_0,\ldots,T_n,T_k^{-1}]/\langle T_iT_j\rangle = A[T_0,\ldots, \hat{T_i},\ldots,T_n]$.

This makes some of the terms of the complex simpler, perhaps making it easier to understand. Try with some low-dimensional examples to see the structure (e.g. $n=1$ or $n=2$).

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If you just want to know the dimension of $H^0(X,\mathscr O_X)=\Gamma(X,\mathscr O_X)$, you can use the machinery of long exact sequences. Note that $X=Proj(A[T_0,\ldots,T_n]/(T_iT_j)$ sits inside $\mathbb P^n$ with ideal sheaf $\mathscr I=(T_iT_j)$. Note also that $\mathscr I=\mathscr O_{\mathbb P^n}(-2)$, the isomorphism being $1 \mapsto T_iT_j$.

Therefore we have an exact sequence

$$ 0 \to \mathscr O_{\mathbb P^n}(-2) \to \mathscr O _{\mathbb P^n}\to \mathscr O_X \to 0. $$ Since $H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(-2))=0$, we get an exact sequence $$ 0 \to H^0(\mathbb P^n,\mathscr O _{\mathbb P^n}) \to H^0(X,\mathscr O_X) \to H^1(\mathbb P^n, \mathscr O_{\mathbb P^n}(-2)) $$ The last group is zero if $n\neq 1$, hence in that case, $H^0(X,\mathscr O_X)=A$ (only the constant functions). If $n=1$, $H^1(\mathbb P^1, \mathscr O_{\mathbb P^1}(-2))=H^0(\mathbb P^1,\mathscr O)=1$ (by Serre duality), implying that $h^0(X,\mathscr O_X)=2$.

PS In this answer, I've used Theorem 5.1 in Hartshorne, chapter III, which assumes that $A$ is Noetherian as well.

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    One should note that in the case $n=1$, we have $D_+(T_0) \cap D_+(T_1) = \emptyset$, i.e. $X = D_+(T_0) \coprod D_+(T_1) = \operatorname{Spec} A \coprod \operatorname{Spec} A$. In particular $H^0(X,\mathcal O_X) = A \oplus A$, which confirms your result).2017-02-22
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    Why is it true that $A[T_0,\ldots,T_n,T_i^{-1}]/\langle T_i T_j\rangle=0$?2017-02-22
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    @MonstrousMoonshine I don't think it is. $A[T_0,\dots, T_n, T_i^{-1}]/(T_i T_j) = A[T_0,\dots, T_n, T_i^{-1}]/(T_j)\cong A[T_0,\dots, \hat{T_j},\dots, T_n, T_i^{-1}]$, if I'm not mistaken (hat, as usual, meaning omit that term).2017-02-22
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    @MonstrousMoonshine Oops, I made a mistake.2017-02-23