Let $X = K^r$ for $K$ an algebraically closed (thus infinite) field. Equip $X$ with the Zariski topology. Let $\mathscr{F}(X)$ be the sheaf of germs of functions on $X$ with values in $K$, so that $$\mathscr{F}(X)_U = \{f: U \to K\}$$ where $f$ is any function and $U$ is any Zariski open subset of $X$. The maps that go with functor above are the restriction maps, and the fact that the pre sheaf is a sheaf is easily checked.
For $x = (\mu_1, \dots, \mu_r)$ a point in $X$ let $\mathfrak{m} = \langle x_1 - \mu_1, \dots, x_r - \mu_r \rangle$, the maximal ideal of polynomials vanishing on $x$.
Define $$\begin{align} \mathscr{O}_x &= (K[x_1, \dots, x_r])_{\mathfrak{m}}\\[4pt] &= \{R= P/Q | Q \notin \mathfrak{m} \}. \end{align}$$
What I would like to show is that I can embed $\mathscr{O}_x \hookrightarrow \mathscr{F}(X)_x$, the codomain being the stalk over the point $x$. By definition of the sheaf $\mathscr{F}(X)$, that stalk will consist of equivalence classes of functions who agree on some Zariski open neighborhood of the point $x$.
I know there are some results invoking the fact that $K$ is infinite will let me do this, but a lot of elementary yet important properties of the Zariski topology on affine space are escaping me.
I do have the result that $f = 0$ in $k[x_1, \dots, x_r] \iff f(x) = 0$ for all $x \in K^r$. And the corollary to that that $f = g$ in $k[x_1, \dots, x_r] \iff f(x) = g(x)$ for all $x \in K^r$.
My thought is that since $Q \notin \mathfrak{m}$ then the function $Q(x) \neq 0$. Then can I extend this to some Zariski open subset $U$ containing $x$ and say that the function $Q(x) \neq 0$ on $U$? If so I can define the map taking fractions from the localization to the corresponding equivalence class of $P(x)/Q(x)$ as a function on $U$. Then I can argue injectivity by supposing two such images agree on some neighborhood $V$ of $x$, then for all points $y \in V$, $P_1(y)/Q_1(y) = p_2(y)/Q_2(y)$ thus $P_1(y)Q_2(y) - P_2(y)Q_1(y) = 0$ for all points $Y$ in $V$. But, since $V$ is Zariski open, and Zariski open sets are dense then I can say that the polynomial is zero everywhere? Thus $P_1Q_2 = P_2Q_1$ as elements of $K[x_1, \dots, x_r]$? But this is also satisfies the definition of what it means to be equal in the localization?
Please hold me to a rigorous standard, I am reviewing this material to make sure I remember and understand the subtle nuances here. I want to make sure I am not being naive just because I have gotten used to the identifications that work when $K$ is infinite.