Let $\nu\colon X^\nu\to X$ be normalization of a nodal curve, $L$ be a line bundle on $X$, $\nu_*L$ is a coherent sheaf on $X$, but we can take its determinant bundle, by taking alternating tensor product of a free resolution. Suppose $X$ is the nodal cubic, what is the determinant line bundle of $v_*\mathcal{O}_{X^\nu}(P)$? For $P$ lies in one of inverse image of the nodes?
Is this determinant construction of taking better understood as quotient or sub of the $\pi_*\mathcal{O}_{X^\nu}$?