I am trying to solve the following problem from Artin's algebra textbook, but I am confused about what the question is asking for. I'm not asking y'all for a solution (yet), I'm asking for help interpreting the problem
Let $R=\mathbb{Z}[\sqrt{-5}]$, and let $V$ be the module presented by the matrix $A=\begin{bmatrix} 2 \\ 1+δ \end{bmatrix}$ where $δ=\sqrt{-5}$. Prove that the residue of $A$ in $R/P$ has rank $1$ for every prime ideal $P$ of $R$, but that $V$ is not a free module.
I am confused by this, "Prove that the residue of $A$ in $R/P$ has rank $1$ for every prime ideal $P$ of $R$". What is the "residue" of $A$ in $R/P$ (in layman's terms)?
My theory is that the "residue" of $A$ in $R/P$ is simply the matrix $\begin{bmatrix} 2+P \\ 1+δ+P \end{bmatrix}$, a $2\times1$ matrix with entries in $R/P$. Now, I'm also a little confused regarding what it means to prove that this "has rank 1". My best guess is that we're being asked to show that the (R/P)-module $(R/P)^2/\begin{bmatrix} 2+P \\ 1+δ+P \end{bmatrix}(R/P)$ is isomorphic to $(R/P)^1$. Do you guys agree?
Your guesses are almost correct. The "residue" of the matrix in this case is the matrix obtained by taking the mod $P$ residues of all its entries. The matrix $\begin{bmatrix} 2+P \\ 1+\delta+P \end{bmatrix}$ is then a matrix with entries in a field (the fraction field of $R/P$), and "rank" has exactly its usual meaning (the dimension of the span of the columns). When $P\neq 0$ (so that $R/P$ is its own fraction field), this is equivalent to the condition you stated (though be careful: it's not because of the $1$ in $(R/P)^1$, but rather because that $1$ is equal to $2-1$).