Here is the question:
Determine a presentation matrix $A$ as an $R-$module for the ideal $ I = (2, 1 + \sqrt{-5})$ where $R = \mathbb Z[\sqrt{-5}].$
Here is a solution of the question:
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
$$A=\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).\qquad\square$$
My questions are:
Here is what I know:
If $I = (2, 1 + \delta )$ where $\delta = \sqrt{-5}.$ Define $\varphi: R^2 \rightarrow I$ by $(r,s) \mapsto 2r + (1+ \delta)s.$ Then $\operatorname{ker}\varphi =
\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)s\\s\end{matrix}\right)=\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right)s.$ And hence a generator for $\operatorname{ker}\varphi$ is $\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right).$ But above in a solution of excercise 14.5.1 in Artin book "Algebra"(second edition) 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 $I$, he did not use that generator.
1- Does $\mathbb Z[\sqrt{-5}]$ make any difference in the set of generators I should find?
2- Also, why we need the $s$ satisfying the condition of $\operatorname{ker}\varphi$ to be of minimum norm, could someone clarify this to me please?
3-Also, is the solution in the picture above correct?
Note that this question here Determining the presentation matrix for a module helped me a lot in understanding the idea of the presentation matrix.
Note there is no need to answer all questions below as they were already answered in the comments
1- why in the second braces $y \in R^2$?
2- How do the author get that that $y =2, 1 – \delta $? And how did we know that they have the smallest norm? and what is the importance of having the smallest the norm in the solution?
3- How did the author get the corresponding vectors $(-3, 1- \delta)$ and $(-1-\delta , 2 )$? how these 2 dimensional vectors belong to $R$? Also, I did not uderstand the argument the author used for showing that they are linearly independent, and why we have to show that they are linearly independent?
4- It seems for me until this point the author is trying to find the coefficients of the set of relations because the columns of the presentation matrix are built from them, is that correct?
5- Finally, is that solution correct?
Could anyone clarify these points to me please?
Here is also another solution:
The surjective map is considered as shown below;
\begin{align*}
R^2 & \longrightarrow\varphi(2,1+\delta),\\
(x,y) & \longrightarrow2x+(1+\delta)y.
\end{align*}
As $\ker\varphi$ has two generators. Then, there is the relation that $2(-1-\delta)+(1+\delta)2=0$ and also the relation $2(-3)+(1+\delta)(1-\delta)=0$.Therefore, two relations cannot be derived from each other. Then $\ker\varphi$ is find by finding $(x,y)\in\mathbb{R}$ such that $2x+(1 +\delta)y=0$.
It means that
$$\ker\varphi=\left\{y:\ y\in\mathbb{R}\ \text{and}\ \left(\frac{1+\delta}{2}\right)y\ \text{is also in}\ \mathbb{R}\right\}.$$Therefore, the presentation matrix is
$$\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).$$
Best Answer
Observe that the ring $R = \mathbb Z[\sqrt{-5}]$ is equipped with a norm $\upsilon : R \to \mathbb Z$ defined by $\upsilon(a + b \sqrt{-5}) = a^2 + 5b^2,$ hence $\upsilon(R)$ is a subset of the non-negative integers. Consequently, $$V = \left \{\upsilon(y) \,\, \bigg| \,\, y \in R - \{0\} \text{ and } \dfrac{1 + \sqrt{-5}} 2 y \in R \right\}$$ is well-ordered, i.e., there is a minimum element of $V.$ Considering that the norm of an element $a + b \sqrt{-5}$ of $R$ is of the form $a^2 + 5b^2,$ the possible norms $\upsilon(y)$ are $1, 4, 5, 6,$ etc. But the only elements of norm $1$ are $\pm 1 = \pm 1 + 0 \sqrt{-5},$ and it is straightforward to show that $y$ cannot be $\pm 1$ (by the second condition defining $V$). Likewise, the only elements of norm $5$ are $\pm \sqrt{-5} = 0 + \pm \sqrt{-5},$ and by the same rationale as before, one can show that $y$ cannot be $\pm \sqrt{-5}.$ Clearly, we have that $2$ satisfies both conditions of $V,$ hence $2$ is an element of least norm $\upsilon(2) = 4$ in $V.$ Further, one can show that the only elements of $R$ of norm $4$ are $\pm 2.$ Last, the element $1 - \sqrt{-5}$ with norm $\upsilon(1 - \sqrt{-5}) = 6$ satisfies both conditions of $V.$ Once you know that $\ker \varphi$ is generated by two elements, it suffices to show that $2$ and $1 - \sqrt{-5}$ do not divide each other in $R$ to conclude that $\ker \varphi$ is generated by $(-1 - \sqrt{-5}, 2)$ and $(-3, 1 - \sqrt{-5}).$*
Ultimately, in order to determine a presentation of a finitely generated ideal $I = (x_1, \dots, x_n)$ of a ring $R,$ it suffices to find a finite generating set of the kernel of the map $\varphi : R^n \to I$ defined by $\varphi(r_1, \dots, r_n) = r_1 x_1 + \cdots + r_n x_n.$ I don't see how the structure of $R$ changes this goal; however, if your ring $R$ is "nice" (i.e., if $R$ is a field or a Euclidean domain or a principal ideal domain), then this will simplify the problem of finding the generators of $\varphi.$ Unfortunately, the ring $R = \mathbb Z[\sqrt{-5}]$ is not even a unique factorization domain, so the argument is a bit tougher.
*Towards this end, observe that every ideal of a Dedekind domain is either principal or generated by two elements. Once you have shown that $\ker \varphi$ is isomorphic (as an $R$-module) to an ideal of $R,$ it follows that $\ker \varphi$ is generated by two elements (because it contains at least two non-associate elements, as computed in the first paragraph above).