Let $R$ be a graded $k$-algebra and $X = Proj(R)$. Suppose that $I \subset R$ has dimension one (i.e. $\dim(R/I)=1$), then $V(I)$ consists of some closed points $\{\mathfrak{p}_1,\ldots,\mathfrak{p}_r\}$.
I really we would like to understand the formula: $$ \dim_k\big(H^0(X,\frac{\mathcal{O}_X}{I\mathcal{O}_X})\big) = \sum_{\mathfrak{p}_i} \dim_k\big(\frac{\mathcal{O}_{X,\mathfrak{p}_i}}{I\mathcal{O}_{X,\mathfrak{p}_i}} \big)=\sum_{\mathfrak{p}_i}length_{\mathcal{O}_{X,\mathfrak{p}_i}}\big(\frac{\mathcal{O}_{X,\mathfrak{p}_i}}{I\mathcal{O}_{X,\mathfrak{p}_i}}\big)=\sum_{\mathfrak{p}_i}length_{R_{\mathfrak{p}_i}} \big(\frac{R_{\mathfrak{p}_i}}{{IR_{\mathfrak{p}_i}}}\big). $$
I think that my only concern is in the part $$ \dim_k\big(\frac{\mathcal{O}_{X,\mathfrak{p}_i}}{I\mathcal{O}_{X,\mathfrak{p}_i}} \big)=length_{\mathcal{O}_{X,\mathfrak{p}_i}}\big(\frac{\mathcal{O}_{X,\mathfrak{p}_i}}{I\mathcal{O}_{X,\mathfrak{p}_i}}\big). $$
How can one proceed? It is not the case that actually we have? $$ \dim_k\big(\frac{\mathcal{O}_{X,\mathfrak{p}_i}}{I\mathcal{O}_{X,\mathfrak{p}_i}} \big)=[k(\mathfrak{p}_i):k]\cdot length_{\mathcal{O}_{X,\mathfrak{p}_i}}\big(\frac{\mathcal{O}_{X,\mathfrak{p}_i}}{I\mathcal{O}_{X,\mathfrak{p}_i}}\big). $$
In general if we assume that the field $k$ is algebraically closed then we have the equality. But in general how to proceed?
Please, help!