Here is what I know.
Let $R = \mathbb Z[\delta],$ where $\delta = \sqrt{-5}.$ If we denote by $I = (2, 1 + \delta),$ define the surjective map $\varphi: R^2 \rightarrow I$ by $(r,s) \mapsto 2r + (1 + \delta)s.$ We have that $$\operatorname{ker}\varphi =
\left(\begin{matrix}-\frac 1 2(1 + \delta)s\\s\end{matrix}\right)=\left(\begin{matrix}-\frac 1 2(1+\delta)\\1\end{matrix}\right)s,$$ hence a generator for $\operatorname{ker}\varphi$ is $\left(\begin{matrix}-\frac 1 2(1+\delta)\\1\end{matrix}\right).$
But below in a solution of exercise 14.5.1 of Algebra (Second Edition) by Artin, the author did not use this generator when the author needed to use a set of generators of the module of relations to be expressed as a linear combination of the set of generators of $I$ to find a presentation matrix as an $R−$module for the ideal $I.$ So, my question is this.
Does it make any difference in the set of generators I should find that we are working in the ring $R = \mathbb Z[\sqrt{-5}]$?
Here is the solution I referenced above.
Exercise 14.5.1. Let $R=\mathbb{Z}[\delta]$ where $\delta=\sqrt{-5}$. Determine a presentation matrix as an $R$-module for the ideal $(2,1+\delta).$
Proof. Let $\varphi:R^2\longrightarrow I$ that sends $(x,y)\rightsquigarrow2x+(1+\delta)y$. Now
$$\ker\varphi=\left\{(x,y)\in R^2\,\middle|\,x=-\frac{1+\delta}{2}y\right\}\leftrightarrow\left\{y\in R^2\,\middle|\,\frac{1+\delta}{2}y\in R\right\}$$
where $\leftrightarrow$ denotes bijection. But $y=2,1-\delta$ are the elements in $R$ with smallest norm satisfying this condition, and moreover the corresponding vectors $(-3,1-\delta),(-1-\delta,2)\in R$ are independent over $R$ since $2r\ne1-\delta$ for any $r\in R$. Thus, $\ker\varphi$ is generated by these two vectors, and by pp. 424-425 we have a presentation matrix
$$A=\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).\qquad\square$$
Also, why do we need the $s$ satisfying the condition of $\operatorname{ker}\varphi$ to be of minimum norm. Could someone clarify this to me please?
EDIT: Also, is the solution in the picture above correct?
Note that this question here helped me a lot in understanding the idea of the presentation matrix.
Best Answer
There is an immediate problem with your generator of $\ker \varphi$: $2$ is not a unit of $R,$ and $2$ does not divide $1 + \sqrt{-5},$ hence the expression $-\frac{1 + \sqrt{-5}} 2$ is not an element of $R.$
Ultimately, the ring $R$ makes all the difference in finding a set of generators for $\ker \varphi$ in the sense that you must pay attention to the structure of $R$ when finding the generators of $\ker \varphi.$ If $R$ were $\mathbb Q(\sqrt{-5}),$ it would make sense to say that $(r, s)$ is in $\ker \varphi$ if and only if $2r + (1 + \sqrt{-5})s = 0$ if and only if $2r = -(1 + \sqrt{-5})s$ if and only if $r = -\frac{1 + \sqrt{-5}} 2$ because the elements $-\frac 1 2$ and $1 + \sqrt{-5}$ belong to the ring $\mathbb Q(\sqrt{-5}).$ Unfortunately, things are not as simple in $\mathbb Z[\sqrt{-5}].$