So I am confused with this argument in 3rd Ed, pg 47, Basic Alg. Geo. 1.
Definition 1: if $X \subseteq \Bbb P^n$ is a quasiprojective variety, $x \in X$, and $f=P/Q$ is a homogenous function of degree $0$ with $Q(x)\not=0$. A function on $X$ that is regular at all points $x \in X$ is a regular function on $X$.
Definition 2: If $A$ is a variety of $\Bbb A^n$ then $f:A \rightarrow k$ is regular if exists poylnomial function $F$ such that $F|A=f$.
So Shafarevich sets out to prove definition 1 for a closed subset $X$ of an affine space coincides with 2.
Proof: By assumption each point $x \in X$ has a nhood $U_x$ with $q_x \not=0 $ on $U_x$, in which $f=p_x/q_x$. So $$q_x f= p_x $$ on $U_x$. We can assume that this holds over whole of $X$ by multiplying a regular function equal to $0$ on $X\setminus U_x$ and nonzero at $x$.
What regular function is referred here? I suppose it is definition 2. By how does such a function exist?
Note that $Z:=X\smallsetminus U_x$ is a closed subset of $X$ (and hence of $\Bbb A^n$). If we take elements $f_1,\dots,f_r$ which generate the ideal $I(Z)\subset k[x_1,\dots,x_n]$ (I'm not sure if this is the notation Shafarevich uses, but $I(Z)$ is the ideal given by polynomials vanishing on all of $Z$, and it is finitely generated because $k[x_1,\dots,x_n]$ is noetherian).
If each $f_i$ vanished at $x$, then $x$ would have to be in $Z$ because $Z$ is the vanishing locus of $(f_1,\dots,f_r)$, but this is not the case. So some $f_i$ does not vanish at $x$ and, furthermore, it must vanish on all of $X\smallsetminus U_x$ because it is an element of $I(Z)$. So $f_i$ is the element you want.