Over $\mathbb{R}$, if $Z(p') \subset Z(p)$ when does $p' \vert p$? I'm mainly wondering about the planar case, when $p', p \in \mathbb{R}[x,y]$. For instance the simplest case would be when $Z(p')$ is a line contained in $Z(p)$, does it follow that $p' \vert p$? I know that over an algebraically closed field the answer would be yes by the Nullstellensatz, but I don't know anything about the real case.
EDIT: Based on the comments I will ask something more concrete:

Let $p \in \mathbb{R}[x,y]$ be irreducible and $p' = x$ so that $Z(p')$ is the $y$-axis. 
If $Z(p') \subset Z(p)$ do we necessarily have $p' \vert p$ (i.e. $x \vert p$)?

 A: Write $p(x,y)=p_0(y)+xp_1(x,y)$ with $p_0(y)\in\mathbb R[y]$. Then for all $b\in \mathbb R$, $p_0(b)=p(0,b)=0$. So $p_0\equiv 0$ and $p\in x\mathbb R[x,y]$. 
A more general question is : let $I, I'$ be ideals of $\mathbb R[x_1,\dots, x_n]$ such that $Z(I')\subseteq Z(I)$, under which condition we have $I\subseteq \sqrt{I'}$ ? A trivial sufficient condition is $I'$ satisfies Nullstellensatz, i.e. if the set $\mathcal I(Z(I'))$ of polynomials vanishing at $Z(I')$, is equal to $\sqrt{I'}$. This condition is also necessary if we want the above conclusion to hold for all $I$ (take $I=\mathcal I(Z(I'))$).   

Claim: If $Z(I')\subseteq \mathbb R^n$ is Zariski dense in $Z(I'\mathbb C[x_1,\dots, x_n])\subseteq \mathbb C^n$ (equivalently, the real points of $\mathrm{Spec}(\mathbb R[x_1,\dots, x_n]/I')$ are Zariski dense in the complex points), then $I'$ satisfies Nullstellensatz. 

Example: if $I'=(x_1)$. Then $I'$ satisfies Nullstellensatz. 
Proof of the claim: let $f\in \mathcal I(Z(I'))$. In $\mathbb C^n$, $Z(f)$ contains $Z(I')\subseteq \mathbb R^n$. By hypothesis, this implies that $Z(f)\supseteq Z(I'\mathbb C[x_1, \dots, x_n])$. Hence by Nullstellensatz, $f^m\in I'\mathbb C[x_1, \dots, x_n]$ for some $m\ge 1$:
$$f^m=\sum_{i} g_i z_i, \quad g_i\in I', z_i\in \mathbb C[x_1, \dots, x_n].$$
Taking the real parts in this equality we get $f^m\in I'$.

Proposition (Generalization of the above example). Suppose the complex variety $Z(I'\mathbb C[x_1,\dots, x_n])$ is irreducible, and its smooth locus contains a real point. Then $I'$ satisfies Nullstellensatz.

EDIT Idea of the proof: the scheme $X:=\mathrm{Spec} (\mathbb R[x_1,\dots, x_n]/I')$ is irreducible and smooth at a rational point $P$. By general results, restricting $X$ to a Zariski open neighborhood of $P$ (this doesn't change the property that real points are Zariski dense) there exists an étale morphism $f : X\to \mathbb A^d_{\mathbb R}$ such that $f(P)=0$. By implicit function theorem, there exists a (real) open neighborhood $V$ of $0$ and a real open neighborhood $U$ of $P$ such that $f|_U : U(\mathbb R)\to V(\mathbb R)$ is an isomorphism of real manifolds. If $X(\mathbb R)$ was not Zariski dense, it would be contained in some hypersurface $Z(g)$ (in scheme-theoretical sense, or think of complex points) with $g\ne 0$. The image $f(Z(g))$ has Krull dimension $<d$ hence contained in $Z(h)$ for some $h\in \mathbb R[t_1,\dots, t_d]$. As
$$V(\mathbb R)\subseteq f(X(\mathbb R))\subseteq Z(h)(\mathbb R),$$ 
and $V(\mathbb R)$ contains a set of the form $I_1\times ...\times I_d$ with 
$I_i$ non-empty open intervals, this forces $h$ to be zero. Contradiction. 
EDIT 2 Reference: Bochnak-Coste-Roy: Real algebraic geometry, Chapter 4. 
An ideal $I$ of $\mathbb R[x_1,\dots, x_n]$ is called a real ideal if for any $f_1, \dots, f_s\in \mathbb R[x_1, \dots, x_n]$ such that $f_1^2+\cdots+f_s^2\in I$, we have $f_1, \dots, f_s\in I$. 

Theorem 4.1.4: $I$ satisfies nullstellensatz ($I=\mathcal I(Z(I))$) if and only if $I$ is real. 

Related to Proposition above is  

Theorem 4.5.1 (iii): $f\in \mathbb R[x_1,\dots, x_n]$ generates a real ideal if and only if the hypersurface $Z(f)$ has a smooth real point.  

