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I want to know the left ideals of a matrix ring. For example, given a matrix ring $R = M_2(\mathbb{Z}/N\mathbb{Z})$. When $N$ is a prime, the left ideals of $R$ is $\begin{pmatrix}a & b \\0 & 0\end{pmatrix} $, where $(a,b) \in \mathbb{P}^1(\mathbb{Z}/N\mathbb{Z})$ the number of left ideals are $N+1$. My question is when $N$ is not a prime, left ideals of $R$ are the same as above?

matthew
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  • Following the duplicate, the left ideals correspond to submodules of $R^2$. That means that this is going to increase as the factorization of $N$ gets larger. When $N=pq$ is squarefree, for example, you'll be able to split the ring into $M_2(\mathbb Z/p\mathbb Z)\times M_2(\mathbb Z/q\mathbb Z)$, you'll have ideals contributed from both rings. When $N$ is a power of a prime, you'll get more submodules from the multiple submodules of $\mathbb Z/p^n\mathbb Z$ that are introduced. – rschwieb Nov 16 '23 at 14:41
  • Your claim that the number of ideals is $N+1$ seems to only be counting the nontrivial ideals of $M^2(\mathbb Z/p\mathbb Z)$, by the way. It's easy even by hand to see that there are at least 5 left ideals of $M^2(\mathbb Z/2\mathbb Z)$, counting the additional trivial left ideals. That's the number of subspaces of $(\mathbb Z/2\mathbb Z)^2$ too, by the way. – rschwieb Nov 16 '23 at 14:52

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