TITLE: What ideal is this?
QUESTION [5 upvotes]: Let $k$ be a field and $R = k[X]$ all polys over $k$ in $X$.  Choose $p \in R$ and define $I_p = \{ f \in R : f\circ p(X) \in I \}$, where $I$ is some ideal in $R$. 
Then $I_p$ is an additive subgroup as $f, g \in I_p \implies (f - g)(p) = f(p) - g(p) \in I$.  And it absorbs $R$: let $h \in R, \ f \in I_p$, then $(hf)\circ p = (h\circ p) (f\circ p) \in I \implies hf \in I_p$.   What is the name of this ideal?
Let's see if it generalizes to $R = k[x_1, \dots, x_n]$.  

Let $I \subset R$ be an ideal and $p = \{p_1, \dots, p_n\}$ a collection of polynomials in $R$. Then define $I_p = \{ f \in R : f(p_1(x), \dots, p_n(x)) \in I \}$. 

Let $f, g \in I_p$.  Then $(f - g)(p_1, \dots, p_n) = \dots  \in I \implies f - g \in I_p \implies I_p$ is an additive subgroup of $R$.  Let's see if it absorbs $R$.  Let $h \in R$.  Then $(hf)(p_1(x), \dots, p_n(x)) = h(p_1(x),\dots, p_n(x))f(p_1(x),\dots, p_n(x))$, the right factor being in $I$ so the whole thing being in $I \implies hf \in I_p$.  So it generalizes nicely to multivariate polynomials.
Some examples.
Ex 1.  Let $I = (g), \ p(x) = x$, then clearly $I_p = I$ since $I_p = \{ f \in R : f(x) \in I\} = I$.  This shows that the ideal is not always trivial.

REPLY [4 votes]: The ideal $I$ determines an affine scheme $X \subset \def\A{\mathbb{A}}\A^n$, and the $n$ polynomials determine a map $p \colon \A^n \to \A^n$.  The condition $f\circ (p_1, \dots, p_n) \in I$ is equivalent to $f \circ p|_X = 0$.  Thus, $I_p$ is the ideal of the closure of the image $p(X)$ (it has to be the closure, since an ideal defines a closed subscheme, and a polynomial vanishing on a set necessarily vanishes on its closure).  Or, more precisely, it's the ideal defining the scheme-theoretic image of $X$ under $p$; you may want to do a search for the various keywords you might not know.