Ideal quotients - when does $I:h^2 = I:h$ hold? A professor gave me the following problem: prove the fact that $I : h^2 = I:h$, where $I \subset k[x_1,\dots,x_n]$ is a zero-dimensional ideal, and $h$ has the property $I + (h) = I + (h^2)$.
Now I don't know much about ideal quotients, but for general $I$ and $h$ the equality $I:h^2 = I:h$ seems wrong. Consider for example $I = (x^2)$ and $h=x$. Then $I:h^2 = (x^2):(x^2)=k[x]$, but $I:h = (x^2) : (x) = (x)$.
In which cases does $I:h = I:h^2$  hold? Do I miss something here, are the properties above enough? Or might this be a mistake on my professor's side?
 A: I got it. According to Wikipedia the following identity holds for ideals $I, J \subset R$:
$$(I:J) = Ann_R\left(\frac{I+J}{I}\right)$$
Because in my case we have $I + (h) = I + (h^2)$, the same is true if I mod out $I$, so they will have the same annihilator as well.
A: The problem with your counterexample is that $I + (h) = (x^2) + (x) = (x)$, but $I + (h^2) = (x^2) + (x^2) = (x^2)$.
What you are asked to show seems to be: assume that $I \subset k[x_1,\dots,x_n]$ is a zero-dimensional ideal, and $h$ has the property $I + (h) = I + (h^2)$. Then $I : h = I : h^2$.
Since $I : h \subseteq I : h^2$, the problem is really to show that given any polynomial $p \in I : h^2$, $p \in I : h$. That is, if $ph^2 \in I$, then $ph \in I$.
So we assume $ph^2 \in I$. Since $h \in I + (h) = I + (h^2)$, $h = i + qh^2$ for some polynomial $q$ and some $i \in I$.
So now multiply $ph^2 \in I$ by $q$ to get $pqh^2 \in I$, and replace $qh^2 = h - i$ to get $p(h-i) \in I$. Finally, $ph - ip \in I$ and $ip \in I$ so $ph \in I$. This completes the proof.

Note: In the professor's problem, it is stated that $I$ is a "zero-dimensional" ideal. I.e. (reminder mainly for myself) that $k[x_1, \ldots, x_n] / I$ has finite dimension as a vector space over $k$.
If you have stated the problem correctly, then it is true even though we did not use the fact it is a zero-dimensional ideal.
An equivalent condition for an ideal $I$ of $k[x_1, \ldots, x_n]$ to be zero-dimensional is that, for each $x_i$, $I$ contains some nonzero element of $k[x_i]$. (See Lemma 64 here.)
