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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?

Pascal's Wager
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1 Answers1

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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 $2-1$).

Eric Wofsey
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  • You seem to say that $R/P$ is a field. If I'm not mistaken, this would imply that $P$ is a maximal ideal. But I thought that $P$ was assumed to be a prime ideal, not necessarily maximal. Are you suggesting that every prime ideal of $R$ is maximal? – Pascal's Wager Apr 17 '18 at 13:57
  • It seems to me that not every prime ideal of $R$ is maximal. $(0)$ is a prime ideal of $R$ (since $R/(0) \cong R$ is an integral domain) but $(0)$ is not a maximal ideal of $R$ (since $R/(0) \cong R$ is not a field) – Pascal's Wager Apr 17 '18 at 14:07
  • Oops, I misread--I thought the prime $0$ was omitted in the statement. – Eric Wofsey Apr 17 '18 at 14:27
  • According to my knowledge, $(2, 1+δ)$ and $(11)$ are both prime ideals of $R$, yet $\begin{bmatrix} 2+(2, 1+δ) \ 1+δ+(2, 1+δ) \end{bmatrix}$ has rank 0 while $\begin{bmatrix} 2+(11) \ 1+δ+(11) \end{bmatrix}$ has rank 1. Thus, it seems to me that Artin is wrong. What are your thoughts, Eric? – Pascal's Wager Apr 20 '18 at 00:36
  • Hmmm...that seems correct. Yeah, as far as I can tell the problem is wrong. Maybe it meant to say "principal prime ideal" instead of just "prime ideal"? – Eric Wofsey Apr 20 '18 at 01:26
  • FYI, this question is similar to another I have on stackexchange, https://math.stackexchange.com/questions/2733318/prove-that-v-is-not-a-free-module. If you'd like to gain a 50 rep bounty (even though you probably don't need it), you might want to take a stab at helping me prove $V$ is free. – Pascal's Wager Apr 20 '18 at 01:55
  • Actually, I think that this problem was intending for $V$ to be isomorphic to the ideal in $R$ generated by $2$ and $1+\delta$ (in which case an appropriately altered version of the problem statement would be correct). But this matrix $A$ does not work for that purpose. – Eric Wofsey Apr 20 '18 at 02:13