Let $V(f)$ be a real irreducible codimension $1$ hypersurface of $\mathbb R^n$. Suppose further $f$ is a real polynomial in $n$ variable s.t. $f$ is irreducible over $\mathbb C$. Now consider $I(V(f))\subset\mathbb R[x_1,\dots, x_n]$. Is $I(V(f))=(f)$?
Note that $f$ vanishes on some real hypersurface but $f$ does not vanish on open ball of $C^n$ along $V(f)\subset\mathbb C^n$. So I cannot really say $g\in I(V(f))\implies$ $g$ vanishes along complex variety $V(f)$ (i.e. accumulation point argument through analyticity does not work here). It seems that I could run the argument for $n=2$ if the curve has some sort of parametrization.
If $f$ is irreducible and $V(f)$ has a smooth real point $y\in\Bbb R^n$, then in a neighborhood of $y$, $V(f)(\Bbb R)$ is a smooth real hypersurface which is Zariski-dense in the complex hypersurface $V(f)(\Bbb C)$. Therefore if $g$ vanishes on $V(f)(\Bbb R)$, it must also vanish on $V(f)(\Bbb C)$, and thus $f|g$ in $\Bbb C[x_1,\cdots,x_n]$, but as both polynomials are real, it is true that $f|g$ in $\Bbb R[x_1,\cdots,x_n]$.
Your condition on $V(f)(\Bbb R)$ being real-codimension-one guarantees you a smooth point, and you're done.