Do taking a general hyperplane section and taking a colon ideal commute? Let $I$ be an ideal and $f$ be an element of $R = \mathbb{C}[x_1,\ldots,x_n]$, where $\mathbb{C}$ is an algebraically closed field of characteristic $0$. Does 
$$
(I+H):f = (I:f)+H
$$
hold for a general hyperplane in $\operatorname{Spec}(R)$, i.e., for $H = \langle \text{general linear function} \rangle$?
 A: This is true for a general affine-linear form. First we assume $(I,f)\neq R$. The point is that $(I,f)$ has finitely many associated primes, so a general $H$ would be a  non-zerodivisor on $R/(I,f)$. 
For any ideal $I$ and any element $f$, there is an exact sequence:
$$0\to R/(I:f) \to R/I \to R/(I,f) \to 0$$
(The first map is simply multiplication by $f$. Now tensor the exact sequence with $R/(H)$, keep in mind that $H$ is nzd on $R/(I,f)$, so $Tor_1^R(R/(I,f), R/H)=0$, we get an exact sequence:
$$0\to R/(I:f, H) \to R/(I,H) \to R/(I,f,H) \to 0$$
On the other hand, the first exact sequence with $I$ replaced by $(I,H)$ says that the leftmost term should be isomorphic to $R/(I,H):f$, so we are done. 
Note that the exact sequence also takes care of the case $(I,f)=R$, since it implies in that case that $(I:f)=I$. Since $(I+H):f=(I+H)$ then, the equality is clear. 
A: A silly counter-example. Take $n:=2$, $I:=Rx_1$, $f:=x_2$. Let $l\notin Rx_1$ be a general linear form in $x_1,x_2$ and $H:=Rl$. Then $I+H=Rx_1+Rl=Rx_1+Rx_2$ and $(I:f)=(Rx_1:x_2)=Rx_1$. Hence, $(I+H):f=(Rx_1+Rx_2):x_2=R$, whereas $(I:f)+H=Rx_1+Rl=Rx_1+Rx_2$.
That was really silly counter-example, as promised. Thanks to Allen Knutson and to Karl Schwede. Actually, I should only observe that the problem could be reformulated: Given an affine $\Bbb C$-algebra $\Bbb C[L]$ generated by a finite-dimensional $L$ and $f\in L$, for a generic $l\in L$, we have $Af\cap A(l+1)=Af(l+1)$.
