$X=Spec(A)$. $X=\bigcup\limits_{i=1}^N D(f_i)\Rightarrow (f_1, ..., f_N)=A$

Question

Let $X=Spec(A)$ and note $D(f)\simeq Spec(A_f)$. $X=\bigcup\limits_{i=1}^N D(f_i)\Rightarrow (f_1, ..., f_N)=A$

We used this to proof a special case of $\mathscr{O}$ the sheaf of rings is a sheaf.
I tried to show the implication but I get stuck:
$X=\bigcup\limits_{i=1}^N D(f_i)=\bigcup\limits_{i=1}^N X\backslash V(f_i) = \bigcup\limits_{i=1}^N \{\mathfrak{p}|f_i\not\subset\mathfrak{p}\}$. But I don't know how to do the step to get $A = (f_1, ..., f_N)$.

Answer 1

Note that if $X = \bigcup_i D(f_i)$ then this is equivalent to saying that $\bigcap_i V(f_i) = \emptyset$.

On the left-hand side, we have that

$$ \bigcap_i V(f_i) = \bigcap_i Spec\big(A/(f_i)\big) = Spec\big(A/(f_1, \ldots, f_n)\big) $$

On the right-hand side, however, we have $\emptyset$ which is Spec of the zero ring. What can you conclude from this?