# Mechanics of a direct sum mod a direct sum

I would appreciate help in understanding the mechanics of how a direct sum mod a direct sum works.

Specifically I came across the following:

$\mathcal{O}_K$ = $\mathbb{Z}e_1 \oplus\mathbb{Z}e_2 \oplus \dots \oplus \mathbb{Z}e_n$

and

$m\mathcal{O}_K$ = $m\mathbb{Z}e_1 \oplus m\mathbb{Z}e_2 \oplus \dots \oplus m \mathbb{Z}e_n$

Hence $\mathcal{O}_K/m \mathcal{O}_K$ = ($\mathbb{Z}/m \mathbb{Z})\bar{e}_1 \oplus(\mathbb{Z}/m \mathbb{Z})\bar{e}_2 \oplus \dots \oplus (\mathbb{Z}/m \mathbb{Z})\bar{e}_n$

While this looks quite reasonable, I would appreciate help understanding what steps are taken.

And what do the $\bar{e}_i$'s mean (i.e. the bar over the $e$). Although it was not explicitly stated, I presume the $e_i$'s are the standard basis. So what's happens to elements of the basis here.

Thanks very much

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Newtonian mechanics or Lagrangian? Honestly I find your title uninspired. – user26857 Mar 18 '13 at 0:18

The $e_i$'s are probably formal generators, or you could think of them as the standard basis of a vector space. Then $\bar e_i$ means the image of $e_i$ under the quotient map $\mathcal O_k \to \mathcal O_k/m \mathcal O_k$.
This is a special case of the following fact: if $G_1,\ldots, G_n$ are groups and $H_1 \lhd G_1,\ldots, H_n \lhd G_n$ are normal subgroups then $$(G_1 \oplus \cdots \oplus G_n) / (H_1 \oplus \cdots\oplus H_n) \simeq (G_1/H_1) \oplus \cdots \oplus (G_n/H_n).$$ You can see this from the first isomorphism theorem by show that the map $$G_1 \oplus\cdots\oplus G_n \to (G_1/H_1) \oplus \cdots \oplus (G_n/H_n) \\ (g_1,\ldots,g_n) \mapsto (g_1H_1, \ldots, g_nH_n).$$ is surjective and has kernel $H_1\oplus\cdots\oplus H_n$.
Thanks, Eric. Maybe you would please elaborate a bit as to what happens as to, e.g., what is $G_1/(H_2 \oplus \cdots\oplus H_n)$. Also, I realize what you are saying about $\bar{e}_i$, but why would $e_i$ change as you mod out by the direct sum of the $H$'s in your example. Regards, Andrew – TheBirdistheWord Mar 17 '13 at 22:10
@Andrew I'm not sure I understand your first question... $G_1/(H_2 \oplus \cdots \oplus H_n)$ does not make sense. The $e_i$'s are really formal: $\mathcal O_k$ is just $\mathbb Z \oplus \cdots \oplus \mathbb Z$. Then $e_j$ is the element $(0,\ldots,0,1,0,\ldots,0)$ in $\mathcal O_k$ and $\bar e_j$ denotes the coset $e_j m\mathcal O_k$. – Eric O. Korman Mar 17 '13 at 22:17
I'm sure I'm missing something, perhaps how a direct sum mod a direct sum is defined. What I was driving at is the presumption that each $G_i$ is being modded out by $H_1 \oplus \cdots\oplus H_n$. So you clearly show what happens with the $H_1$ on $G_1$, but what is the effect of the other $H_i$'s on the $G_1$? Thanks, again. – TheBirdistheWord Mar 17 '13 at 22:25
@Andrew We consider $H_1 \oplus \cdots \oplus H_n$ as the subgroup of $G_1 \oplus \cdots \oplus G_n$ consisting of all elements $(h_1,\ldots, h_n)$ with $h_j \in H_j < G_j$. The quotient is defined just as the quotient of any group by a subgroup. – Eric O. Korman Mar 17 '13 at 22:58