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 $3$.
Recall that any inextendible regular sequence is of maximal length (this is very strong and essential to the following argument. See Theorem 17.4 in Eisenbud's book, and the preceeding discussion).
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 $3$.
