Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Specifically I came across the following:

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


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

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

share|cite|improve this question
Newtonian mechanics or Lagrangian? Honestly I find your title uninspired. – user26857 Mar 18 '13 at 0:18
up vote 5 down vote accepted

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 $H_1\oplus\cdots\oplus H_n$.

share|cite|improve this answer
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
Thanks. That's what I needed. – TheBirdistheWord Mar 17 '13 at 23:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.