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 $H = \langle \text{general linear function} \rangle$?



Do taking a general hyperplane section and taking a colon ideal commute?
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. 


