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Theorem (CRT) : Let $a_1,\cdots ,a_r$ be ideals in a ring $R$, such that $a_i + a_j = R$, for all $i \ne j$. Let $\pi_i: R \mapsto R/a_i$ denote the surjection. Then the map $$\eta : R \mapsto R/a_1 \times \cdots R/a_r$$

given by $\eta(x) = (\pi_1(x_1),\cdots \pi_r(x_r))$ is onto.

Proof : A typical element in $R/a_1 \times \cdots R/a_r$ is of the form $(\pi_1(x_1),\cdots \pi_r(x_r)) = \pi_1(x_1)e_1 + \cdots +\pi_r(x_r)e_r$, where $e_i = (0,\cdots,1,\cdots,0)$ one is at $i th$ postion.

Here I have a doubt should I write $e_i = (0,\cdots,1,\cdots,0)$ or $e_i = ([0],\cdots,[1],\cdots,[0])$, where [0] is a remainder class (class of all element that are zero when mod with $a_1$).

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  • $\begingroup$ One is correct and one is an acceptable abuse of notation. $\endgroup$
    – Randall
    Commented Aug 16, 2017 at 13:21
  • $\begingroup$ This is not so much a <s>doubt</s> question of proof as it is a question of notation. $\endgroup$
    – rschwieb
    Commented Aug 16, 2017 at 13:35

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$(0,\cdots,1,\cdots,0)$ is an acceptable abuse of notation, but $([0],\cdots,[1],\cdots,[0])$ does not improve on it much, if the goal is to have very descriptive notation.

I would recommend $e_i=(0+a_1,\cdots,1+a_i,\cdots,0+a_n)$, faithfully using the coset notation in each position, if the goal is more descriptive notation.

Using a lowercase roman letter for ideals also looks a little hard to read (to me at least) and it might be worth switching to capital or fraktur letters.

And I hope I don't sound like I'm blowing the importance of notation out of proportion here. All of these are perfectly acceptable depending on your audience and how you present your work.

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  • $\begingroup$ Thanks . Is it depends on whether $R/a_1 \times \cdots R/a_r$ is a product of rings or not ? $\endgroup$
    – user275490
    Commented Aug 16, 2017 at 13:35
  • $\begingroup$ @Haha No, I would do the same thing to express an element of a product of quotients of any Abelian group. For nonabelian groups, I would use whatever coset notation was appropriate (it could be multiplicative, for example.) $\endgroup$
    – rschwieb
    Commented Aug 16, 2017 at 13:36

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