Koszul complex as a graded free resolution I'm trying to compute the Hilbert function of a complete intersection by using the Koszul complex, but I think I'm approaching it incorrectly. If we let $R=k[x,y,z]$ and
$$A= R/(f_1,f_2)$$
the Koszul complex of $(f_1,f_2)$ gives us the following exact sequence
$$ 0 \to R \stackrel{d_2}\to R^2 \stackrel{d_1}\to R \to A \to 0.$$
But $d_2(r) = (f_1r, f_2r)$ doesn't appear to be a homogenous map, and I cant think of a way to shift the grading on $R$ or $R^2$ in such a way that would make it work. I think the grading on $R^2$ is given by $R^2_g = R_a \bigoplus R_b$ where $a+b =g$. Perhaps I am wrong on that, or my misunderstanding lies elsewhere. Any thoughts would be appreciated.
 A: As noted in the comments, one ought to assume that $f_1$ and $f_2$ are homogeneous elements. The real issue, though, is that your Koszul complex definition needs some improvement. Let's start for the case of one graded element: the Koszul complex in this case is $$0\to R(-d)\stackrel{\cdot f}{\to} R\to 0$$ where $\deg f= d$. The Koszul complex associated to $(f_1,\cdots,f_n)$ is then the total complex of the tensor product of the Koszul complexes associated to each $f_i$, which in your case looks like $$0\to R(-d_1-d_2)\to R(-d_1)\oplus R(-d_2) \to R\to 0$$ where the first map sends $r\mapsto (f_2r,f_1r)$ and the second map sends $(r,s)\mapsto f_1r-f_2s$ (up to perhaps a small sign mistake/difference in conventions).
Now let's show that this works just fine with the gradings. First, recall that the shift operator is defined so that $R(s)_d=R_{d+s}$; next, recall that if $M$ and $N$ are graded $R$-modules, then the grading on $M\oplus N$ is such that $(M\oplus N)_d = M_d\oplus N_d$.
To check that our first map is really a graded map, suppose $r\in R_d = R(-d_1-d_2)_{d+d_1+d_2}$. As $f_ir\in R_{d+d_i} = R(-d_j)_{d+d_i+d_j}$, we have that our first map is graded exactly as we want. The second map is similar: if $(r,s)\in [R(-d_1)\oplus R(-d_2)]_d$, then $r\in R_{d-d_1}$ and $s\in R_{d-d_2}$, so $f_1r\in R_d$ and $f_2s\in R_d$, exactly as it should be.
