🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

We start by showing $\varphi$ is a crone homomorphism, note that we already know that $R$ and $\mathbb{Z}$ are crones, now note the following: $${\displaystyle \varphi \left(\right(\begin{array}{cc}a& b\\ 0& a\end{array}\left)\right(\begin{array}{cc}c& d\\ 0& c\end{array}\left)\right)=\varphi \left(\right(\begin{array}{cc}ac& ac+bd\\ 0& ac\end{array}\left)\right)=ac=\varphi \left(\right(\begin{array}{cc}a& b\\ 0& a\end{array}\left)\right)\varphi \left(\right(\begin{array}{cc}c& d\\ 0& c\end{array}\left)\right)}$$ and that $${\displaystyle \varphi \left(\right(\begin{array}{cc}a& b\\ 0& a\end{array})+(\begin{array}{cc}c& d\\ 0& c\end{array}\left)\right)=\varphi \left(\right(\begin{array}{cc}a+c& b+d\\ 0& a+c\end{array}\left)\right)=a+c=\varphi \left(\right(\begin{array}{cc}a& b\\ 0& a\end{array}\left)\right)+\varphi \left(\right(\begin{array}{cc}c& d\\ 0& c\end{array}\left)\right)}$$

We recognize that the $2\times 2$ identity is the identity in this ring as well, therefore we confirm that $${\displaystyle \varphi \left(\right(\begin{array}{cc}1& 0\\ 0& 1\end{array}\left)\right)=1}$$ Thus $\varphi$ is indeed a crone homomorphism

From the definition of $\varphi$ we can conclude that $\varphi \left(\right(\begin{array}{cc}a& b\\ 0& a\end{array}\left)\right)=0$ iff $a=0$, therefore the kernel is all matrices of this form $${\displaystyle ker\left(\varphi \right)=\left\{\right(\begin{array}{cc}0& b\\ 0& 0\end{array}):b\in \mathbb{Z}\}}$$

Recall the definition of $R/ker\left(\varphi \right)$, it is $$\begin{gathered} {\displaystyle \{M+ker\left(\varphi \right):M\in R\}\&=\{\{M+N:N\in ker\left(\varphi \right)\}:M\in R\} \\ \&=\{\left\{\right(\begin{array}{cc}a& b\\ 0& a\end{array})+(\begin{array}{cc}0& c\\ 0& 0\end{array}):c\in \mathbb{Z}\}:a,b\in \mathbb{Z}\} \\ \&=\{\left\{\right(\begin{array}{cc}a& b+c\\ 0& a\end{array}):c\in \mathbb{Z}\}:a,b\in \mathbb{Z}\} \\ \&=\{\left\{\right(\begin{array}{cc}a& d\\ 0& a\end{array}):d\in \mathbb{Z}\}:a\in \mathbb{Z}\}} \end{gathered}$$ Let $tl\left(M\right)$ be defined as the top left coefficent in $M$, then for two matrices $M,N$ it should be clear that if $tl\left(N\right)=tl\left(M\right)$, then they generate the same coset, also that if $tl\left(M\right)\ne tl\left(N\right)$, then they generate different cosets. Also note that $tl\left(MN\right)=tl\left(M\right)·tl\left(N\right)$ and that $tl(M+N)=tl\left(M\right)+tl\left(N\right)$. Finally given any coset, since we know that $tl\left(X\right)$ agrees for every $X$ in the coset, we can justify the construction of $\overline{tl}(N+ker\left(\varphi \right))=tl\left(N\right)$

We claim that $R/ker\left(\varphi \right)$ is an crone isomorphism to $\mathbb{Z}$ using $\overline{tl}$. We've actually already verified the that it respects addition and multiplication because of our discussion in the previous paragraph and also just note that if $I$ is the the identity then we know it's top left element is $1\in \mathbb{Z}$ so that $\overline{tl}\left(I\right)=tl\left(I\right)=1$ .

At this point we know that it's a homomorphism. So we just have to show it's a bijection to have an isomorphism so suppose that $M+ker\left(\varphi \right)\ne N+ker\left(\varphi \right)$ by the contrapositive of what we noted earlier since they generate different cosets, then $tl\left(N\right)\ne tl\left(M\right)$, therefoer $\overline{tl}(N+ker\left(\varphi \right))\ne \overline{tl}(M+ker\left(\varphi \right))$ , which shows that $tl$ is injective, it is clearly also surjective because since $R$ ranges over all matrices with integer coefficients, then if we pick a particular integer $k\in \mathbb{Z}$ then there is always a matrix with that specific integer in it's top left component, say that matrix is $X$, then $\overline{tl}\left(X\right)=tl\left(X\right)=k$ as needed, so $\overline{tl}$ is an isomorphism.

Now we verify that $ker\left(\varphi \right)$ is a prime ideal, so suppose that $MN\in ker\left(\varphi \right)$, this means that $tl\left(MN\right)=0$, but as noted earlier $tl\left(MN\right)=tl\left(M\right)tl\left(N\right)$, but we know that $\mathbb{Z}$ so in this context we have $tl\left(M\right)tl\left(N\right)=0$, but $\mathbb{Z}$ is a domain therefore we must have wlog $tl\left(M\right)=0$, but that implies that $M\in ker\left(\varphi \right)$ which shows that $ker\left(\varphi \right)$ is a prime ideal as neeeded.