# Minimal free resolution of ideal generated by three homogeneous polynomials

I am trying to solve the following exercise;

Let $R=k[x_0,x_1,x_2]$ and $f_i$ homogeneous polynomials of degree $d_i, 0\leq i \leq 2$. Suppose $f_0,f_1,f_2$ have no common roots in $\mathbb P^2$. Construct a minimal free resolution of $(f_0,f_1,f_2)$.

Initially i thought that Koszul complex will do but since $\{f_0,f_1,f_2\}$ may not form a regular sequence therefore Koszul complex is not necessarily a minimal free resolution. Any other ideas to construct this resolution?

The Koszul complex is the right choice, you can show that your polynomials must form a regular sequence. First of all, let $I:=\langle f_1,f_2,f_3\rangle$, then $\sqrt{I}=R_+=(x_0,x_1,x_2)$ by assumption and hence, $I$ has height $3$. In particular, $I$ contains a regular sequence of length $3$.
Let $k$ be the maximal index such that $f_1,\ldots,f_k$ form a regular sequence, possibly $k=1$ ($R$ is a domain). We also order the polynomials such that $k$ is maximal, in other words $f_i$ is a zero divisor in $R/\langle f_1,\ldots,f_k\rangle$ for all $i>k$. If we had $k<3=\operatorname{ht}(I)$, then there would be some $g\in I$, say $g=g_1f_1+g_2f_2+g_3f_3$ , such that $f_1,\ldots,f_k,g$ is a regular sequence. However, by choice of $k$, the image of $g_1f_1+g_2f_2+g_3f_3$ is a zero divisor in $R/\langle f_1,\ldots,f_k\rangle$, which is a contradiction.
Remark: This argument would work for $n$ in place of $3$.