To find a familiar ring to $\mathbb Z^3/\langle(1,-1,-1)\rangle$, I took a surjective homomorphism $f:\mathbb Z^3\rightarrow \mathbb Z^2$ with kernel $\langle(1,-1,-1)\rangle$ and used the First Isomorphism Theorem.

My doubt is if there is a may do decide when such a familiar ring can be found in the more general case $\mathbb Z^n /\langle z_1,...,z_m\rangle$, with $z_1,...z_m\in \mathbb Z^n$. For the field $\mathbb R$, I guess that $\mathbb R^n /\langle a_1,...,a_m\rangle \simeq \mathbb R^{n-m}$, where $a_1,...,a_m$ are linearly independent vectors, and that the same is not true for all ring $A$. Am I right? I've been searching these observations on abstract algebra books, but I couldn't find them.

I'd be grateful with some help. Thank you!


The ring $\mathbb{Z}^n$ has the elements $e_i$ equal to $1$ in the $i$th component and $0$ in the others. The projection $p: \mathbb{Z}^n \to \mathbb{Z}^n/ \langle z_1, ..., z_m \rangle$ is determined by where the $e_i$ map to. The $e_i$s are orthogonal idempotents in $\mathbb{Z}^n$ which means that ${e_i}^2=e_i$ and that $e_i \cdot e_j=0$ if $i \neq j$. Notice that an idempotent $e$ of a ring $R$ induces a direct sum decomposition as follows: $R \cong Re \oplus R(1-e)$. Thus, for $\mathbb{Z}^n$, the idempotents $\{e_i\}$ induce the decomposition $\mathbb{Z}^n= \oplus_i \mathbb{Z}e_i$.

But the point is, that $p(e_i)$ are also idempotents (or $0$) and so will induce a decomposition in the quotient. Thus, $p: \mathbb{Z}^n \to \mathbb{Z}^n/ \langle z_1, ..., z_m \rangle$ will split into a sum of $n$ factors where each factor is $0$, torsion, or $\mathbb{Z}$ based on if $p(e_i)$ is $0$, torsion, or not torsion.


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