Show that a sequence is a free resolution Let $I \subset R = k[x_1,\dots,x_n]$ be an ideal and $f \in R$ such that $I = \left < f \right >$ ($k$ is a field, so R is commutative ring). How do I show that
(1) $I$ has a free resolution $$0 \to Re_1 \to I \to 0.$$
(2) If $f_1,f_2 \in R$ (assume neither divides the other) and $I = \left <f_1 ,f_2 \right >$, then it has a free resolution of the form $$0 \to Re_1 \to Re_1 \oplus Re_2 \to I \to 0.$$

Definitions: Let $A$ be an $R-$ module, a resolution of $A$ (over $R$) is an exact sequence
$$\to C_n \to \dots C_1 \to C_0 \to A \to 0.$$ If each $C_q$ is a free $R-$ module, then the resolution is free.

So something I am not sure if i take for free is
(1) Because $I$ is finitely generated, we get a surjective homomorphism $Re_1 \to I$ for free where the map sends $e_1 \to f$ in a one-to-one fashion, so we have a sequence  $$0 \to Re_1 \to I \to 0.$$ So $I$ is a free resolution (because $I$ is also finitely generated).
(2) How do I deal with the map $ Re_1 \to Re_1 \oplus Re_2 $? How do I know it is 1-1? Do I fix the second slot to be $0$? For the map, $Re_1 \oplus Re_2 \to I$, it is a surjective homomorphism (for free) because $e_i \to f_i$.
 A: $1)$ is correct. For $2)$ you should send $e_1 \mapsto f_1, e_2 \mapsto f_2$. Then $Re_1 \oplus Re_2 \to I$ is surjective, as you suggested. To conclude you have to show that kernel is generated by $f_2e_1-f_1e_2$, hence is free (any cyclic torsion-free submodule is obviously free).
Note that you still need some assumptions on $f_1,f_2$ for this to hold, namely they have to be co-prime.
If $f_1,f_2$ are not co-prime, the kernel will be generated by $\frac{1}{d}(f_2e_1-f_1e_2)$, where $d$ is the greatest common divisor.
You also have to take care of the case $f_1=0$, but actually this case is included, since the kernel will be generated by $e_1$ then, which coincides with $\frac{1}{d}(f_2e_1-f_1e_2)$ in this case.
Another method is to consider $J=\langle \frac{f_1}{d}, \frac{f_2}{d} \rangle$. $I \cong J$ as $R$-modules, i.e. if $I$ has a finite free resolution, $J$ has a resolution of the same length and with the same ranks of free modules. This shows we can actually assume $f_1, f_2$ to be co-prime without loss of generality.
