Definition of Regular function

Question

In Harsthone page 15, the notion of regular function of a quasi affine variety $Y$ is defined as followds.

A function $f:Y \to k$ is regular at a point $P \in Y$ if there is an open neighborhood $U$ with $p \in U \subset Y$ and polynomial $g,h \in A = k [x_1,...,x_n]$, such that $h$ is nowhere zero on $U$ and $f=g/h$ on $U$. We say that $f$ is regular on $Y$ if it is regular at every point of $Y$. We denote by $\mathcal{O}(Y)$ the ring of all regular functions on $Y$.

On the other hand,

in the page 17, THEOREM 3.3 in the book said that

$\mathcal{O}(Y)$ is isomorphic to the affine coordinate ring of $Y$,
$\mathcal{O}(Y)\cong A(Y)=k[x_1,...,x_n]/I(Y)$.

But I cannot understand it. Because any element of $A(Y)$ can be represented by the polynomial and not its ratio. So why is $\mathcal{O}(Y)$ defined by using ratio $g/h$ of polynomials $g,h$?

Answer 1

Let me give you an easy case of the statement: consider $Y=\mathbb{A}^1$ (over an algebraically closed field $k$). Then you know that $A(Y)=k[x]$.

Consider a regular function $f:Y\to k$. Given a point on $Y$ and an open neighborhood $U$ of this point, you know that you can write $f_U=g_U/h_U$, with $h_U$ never vanishing. But then for any $a\in zero(h_U)$, there is an open neighborhood $V$ containing $a$ with $f_V=g_V/h_V$ and $h_V$ not vanishing on $V$.

On $U\cap V$, you have $f_U=f_V$, ie $g_Vh_U=h_Vg_U$. As $h_V(a)\neq 0$, you necessarily have $g_U(a)=0$ with at least same multiplicity as $h_U$. Since this is true for all $a\in zero(h_U)$, we obtain $h_U |g_U$, so $f_U$ is a polynomial.

Now $f$ is locally given by polynomials, so it is globally a polynomial.