When does $V(f)\supseteq V(g)$ imply that $g$ divides $f$? Over algebraically closed fields, if $g$ is an irreducible polynomial (in $n$ variables) and $f$ is such that it vanishes on zero-set of $g$ (i.e., $V(f)\supseteq V(g)$), then $g$ divides $f$. This follows easily from Nullstellensatz, since
\begin{align}
V(f)&\supseteq V(g)\\
f\in IV(f)&\subseteq IV(g)=\sqrt{(g)}=(g),
\end{align}
hence $g|f$.
My question is whether this also holds for fields which are not algebraically closed, given that $V(g)\neq\emptyset$ (this is in general clearly needed, as $V(f)\supseteq\emptyset$ for any $f$). The question is actually equivalent to asking whether
$$g\text{ irreducible},\  V(g)\neq\emptyset \ \ \overset{?}{\Longrightarrow}\  IV(g)=(g).$$
Does this hold over any field? If not, does it at least hold over $\mathbb{R}$? Are there any counterexamples?
 A: In general (= not over algebraically closed fields) it is false.
Consider for example the field of real numbers $\mathbb{R}$ and the classic counterexample polynomial $g = X^2+1$. This polynomial is irreducible, and $V(g) = \emptyset$.
Now, take any polynomial $f \notin (g)$. Then we have $g \not| f$ but trivially $V(g) \subseteq V(f)$.

DISCLAIMER: the following works if we consider polynomials in one variable.
In arbitrary fields, the answer is positive if we add the assumption that $V(g) \neq \emptyset$. 
Indeed, because $g$ is also irreducible, this assumption implies that $g$ must be of degree $1$ (an irreducible polynomial with a zero has degree one). Thus, write 
$$g= aX-b = a(X-b/a)$$
This polynomial has zero $b/a$, and thus in fact $V(g) = \{b/a\}$. The condition $V(g) \subseteq V(f)$ implies that $b/a \in V(f)$, thus $b/a$ is a zero of $f$ as well, and thus $f$ contains a linear factor $u(X-b/a)$ where $u$ is a unit. 
It follows that indeed $g | f$ in an arbitrary field, whenever $\emptyset \neq V(g) \subseteq V(f)$ and $g$ is irreducible.

In more than $1$ variable over finite fields, the claim is false.
Let's work in the simplest field you can imagine: $\mathbb{Z}_2 = \mathbb{F}_2$
Consider the irreducible polynomial $g =X-Y$. Then $V(g) = \{(0,0),(1,1)\}$.
Next, consider the polynomial $f=XY(X-1)(Y-1)$. Clearly $V(f) = \{(0,0),(0,1),(1,0),(1,1)\} = \mathbb{F}_2^2$ and thus $\emptyset \neq V(g) \subseteq V(f)$ with $g \not | f$

Here is a counterexample for $\mathbb{R}$:
Consider $\mathbb{R}$ and $g = X^2 + (Y-1)^2$. Then $g$ is irreducible and $V(g) = \{(0,1)\}$. Put $f := X(Y-1)$. Then $V(g) \subseteq V(f)$ and $g \not| f$.
