In many sources I have read I see that for an irreducible affine variety $V \subseteq K^n$ its coordinate ring is defined to be $$K[V] = K[x_1, \ldots, x_n] / I(V),$$ where $I(V) = \{f \in K[x_1, \ldots, x_n] : \forall x \in V, f(x) = 0\}$.
My question is, is this the right definition of the coordinate ring and ideal associated with the variety? I'm confused because $V$ can contain different sets of points depending on the field $K$, and we only get the "full picture" when $K$ is algebraically closed. For example, the variety $V = V(x^2 - 2)$ is empty in $\mathbb Q$, but obviously has points in $\mathbb R$ or $\mathbb C$. So what is $I(V)$? Since $V$ is defined over $\mathbb Q$ (coefficients of the generating polynomials are in $\mathbb Q$), do we have $$I(V) = \{f \in \mathbb Q[x] : \forall x \in V(\mathbb Q), f(x) = 0\} = \mathbb Q[x],$$ so that $$\mathbb Q[V] = \mathbb Q[x] / \mathbb Q[x] \cong \text{trivial ring}?$$ Or do we have $$\begin{align*}I(V) &= \{f \in \mathbb Q[x] : \forall x \in V(\bar{\mathbb Q}), f(x) = 0\} \\ &= \{f \in \mathbb Q[x] : \forall x \in V(\mathbb C), f(x) = 0\} \\ &= (x^2 - 2) \subseteq \mathbb Q[x]\end{align*}$$ so that $$\mathbb Q[V] = \mathbb Q[x] / (x^2 - 2) \cong \mathbb C?$$
Basically, I'm confused as to what points we want to consider to be in $V$ in order to say that $f, g : K[x_1, \ldots, x_n] \to K$ are "equal" on $V$. Do we identify $f \equiv g \in K[V]$ if $f(x) = g(x)$ for all $x \in V(K)$, or is it for all $x \in V(\bar K)$? Is there a standard definition for this?