Minimal set of generators of an ideal

Question

Let $R$ be a commutative local ring with maximal ideal $\mathfrak{m}$. Let $I$ be an ideal and $x\in R$ such that $x$ is not a zero divisor on $R/I$. Then a minimal set of generators for $I$ is sent to a minimal set of generators for $(I +(x))/(x)$ in $R/(x)$.

My idea was to try to prove that $I/\mathfrak{m}I\cong (I + (x))/(\mathfrak{m}I +(x))$, but my attempts have been unsuccessful.

Answer 1

Lemma: For any ideal $I$ and $x$ in $R$, one has $I \cap (x) = x(I:_R x)$. In addition if $x$ is a nonzerodivior on $R/I$, $I:x = I$, so $I \cap (x) = xI$.

Now, apply the 3rd and 2nd isomorphism theorems to have $$ \frac{I+(x)/(x)}{m(I+(x)) +(x) / (x)} \cong \frac{I + (x)}{m(I+(x)) + (x)} \cong \frac{I+(x)}{mI + (x)} = \frac{I}{ mI + I \cap (x)}. $$ By the lemma above, $$ I/ (mI + I \cap (x)) \cong I / (mI + xI) = I/mI. $$