I am trying to prove to myself that the structure sheaf of an irreducible variety is indeed a sheaf, where the structure sheaf for an irreducible variety $V$ is defined as $\mathcal O_V(U)=\{ f \in k(V): f \text{ is regular on } U \}$ for $U$ open.
Sheaf condition A is clear. That is, if $U$ is open and $U = \cup U_i$ is an open cover, and $f \in \mathcal O_V(U)$ is such that $\forall i$ $f\vert_{U_i} = 0 $, then $f=0$.
Sheaf condition B or glueability is not clear: if $U$ is open and $U = \cup U_i$ is an open cover, and we have $f_i \in \mathcal O_V(U_i)$ such that $\forall i,j$ we have $f_i \vert_{U_i \cap U_j} = f_j \vert_{U_i \cap U_j}$ then there exists $f \in \mathcal O_V(U)$ such that $f \vert_{U_i} = f_i$.
If the above is too trivial to respond, I'd also appreciate a link to the proof. Thank you.
Define $f: U \rightarrow k, f(x) = f_i(x)$ if $x \in U_i.$ Since $f_i \vert_{U_i \cap U_j} = f_j \vert_{U_i \cap U_j}, \forall i, j$ this map is well defined and also $f|_{U_i} = f_i$. So we just need to check that whether it is a regular function on $U$ or not. Let $p \in U.$ Then $p \in U_i$ for some $i.$ So, there exists an open nbd $X_p \subseteq U_i$ of $p$ where it can be wriiten as $\frac{g}{h}$ with $h$ non-vanishing on $X_p.$ But the same expression holds for $U$ also. Hence it is regular at $p.$