Questions on Vakil’s proof of the qcqs lemma

algebraic-geometryexact-sequencelocalizationschemes

There have been a few questions regarding the qcqs – lemma in Vakil's notes (Lemma 7.3.5), e.g.here and here.
I have some questions about details of his proof.
He covers $X$ by finitely many affine charts $U_i = Spec(A_i)$ and covers their intersections $U_{ij} := U_i\cap U_j$ by finitely many affines $U_{ijk} = Spec(A_{ijk})$. So far so good.
From the exact sequence $\mathcal{O}_X(X)\rightarrow \prod_i A_i \rightrightarrows \prod _{i,j} \mathcal{O}_X(U_{ij})$ he gets an exact sequence
$$\mathcal{O}_X(X)\rightarrow \prod_i A_i \rightrightarrows \prod _{i,j,k} A_{i,j,k}$$ (via the injections $\mathcal{O}_X(U_{ij})\rightarrow \prod_k A_{ijk}$ I assume…?).

Then he says that localizing (an exact functor) at $f\in \mathcal{O}_X(X)$ gives
$$(\mathcal{O}_X(X))_f\rightarrow \left(\prod_i A_i\right)_f \rightrightarrows \left(\prod _{i,j,k} A_{i,j,k}\right)_f.$$

First Question: How does for example $(\prod_i A_i)_f$ make any sense, since $f$ is not an element of the ring. If its meant as localizing at $(f_i)$, where $f_i = f|_{U_i}$ (which is the case as it seems), won't the "exact functor" argument fail to work? Then one should prove exactness by hand…

By finiteness of the products, localization commutes and he gets
$$(\mathcal{O}_X(X))_f\rightarrow \prod_i (A_i)_{f_i} \rightrightarrows \prod _{i,j,k} (A_{i,j,k})_{f_{ijk}}.$$
Then he considers $X_f$ and covers it by finitely many affine opens $Spec(A)_{f_i}$ and their intersections $Spec(A)_{f_i}\cap Spec(A)_{f_j}$ are covered by $Spec(A_{ijk})_{f_{ijk}}$. Then there is an exact sequence
$$(\mathcal{O}_X(X_f))\rightarrow \prod_i (A_i)_{f_i} \rightrightarrows \prod _{i,j,k} (A_{i,j,k})_{f_{ijk}}.$$
Then he says its done, but I am not sure why.

Second Question: How does it follow that $(\mathcal{O}_X(X))_f\rightarrow \mathcal{O}_X(X_f)$ is an isomorphism from the two sequences? Isn't it necessary to prove at least that the two sequences commute?

Best Answer

It would be best to include the statement of the lemma in your post. For reference, here it is:

If $X$ is a quasi compact quasi separated scheme and $s\in \Gamma(X,\mathcal{O}_X)$ then the natural map $\Gamma(X,\mathcal{O}_X)_s\rightarrow \Gamma(X_s,\mathcal{O}_X)$ is an isomorphism.

Question 1): For any open subset $U\subset X$, the restriction map $res:\mathcal{O}_X(X)\to \mathcal{O}_X(U)$ gives $\mathcal{O}_X(U)$ the structure of an $\mathcal{O}_X(X)$-algebra, where for $m\in\mathcal{O}_X(U)$ we have $f\cdot m = res(f)\cdot m = f|_U\cdot m$. So localizing at $f$ translates to inverting $f|_U$ in each $\mathcal{O}_X(U)$, and there are no problems here.

Question 2): These two exact sequences exhibit both $(\mathcal{O}_X(X))_f$ and $\mathcal{O}_X(X_f)$ as the equalizer of the diagram $$ \prod_i (A_i)_{f_i} \rightrightarrows \prod _{i,j,k} (A_{i,j,k})_{f_{ijk}}$$

which means that they are unique up to unique isomorphism by the universal property of equalizers.

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