Why quotient ring $R/R$ is zero ring $\{ 0\}$? There is already similar question.
Factor rings $R/R$ and $R/0$
Of course, I read it. However I still don't know, how and why the quotient ring $R/R$ is zero ring. So, I ask your help to check my thinking logic.

In my opinion, it seems to be $R/R=R$.
The definition of the quotient ring is $R/P=\{r+P|r \in R\}$.
So, $R/R=\{r+R|r \in R\}$. Does $r+R$ make all of element of $R$?
And if $R$ is a commutative ring with identity, is that something different result?

I think I have a big problem with that logic, but I'm blind to find it. 
I hope your brighter sight.
Thank you in advance.
 A: The elements of the quotient ring $R/P$ are equivalence classes of elements in $R$. That is, two different elements $r_1, r_2 \in R$ produce the same class in $R/P$ if $r_1 - r_2 \in P$. This is equivalent to the definition you wrote above. 
Now $R/0 = R$ because $r_1, r_2 \in R$ are in the same class in $R/0$ only if $r_1 - r_2 = 0$, which means that $r_1 = r_2$. As a result, every element is in its own class, and no two different elements become equal. Thus $R/0$ looks exactly the same as $R$.
However, for any two elements $r_1, r_2 \in R$ we know that $r_1 - r_2 \in R$. Thus in the quotient ring $R/R$, every element is in the same class, so there is only one class! This means we have a ring with only one element -- the zero ring.
A: See the first thing you should understand is the following: 
$$ R/P:= \text{Set of equivalence classes where the relation is defined as $r\sim s\iff (r-s)\in P$}   $$
Now, what will be $R/R$? It is the set of equivalence classes of the above relation. Now we claim that there is only one class and that class will be $[0]$ class. The intuitive answer is given in the problem that you linked. But if you wanted to be more precise just show the following:
$R/R=\{[0]\}$. 
Let $[r]=r+R\in R/R. \ [r]= r+R=0+(r+R)=[0] \implies R/R\subseteqq \{[0]\}.$ The other inclusion is trivial. 
A: Here is an example for the ring $\mathbb{Z}/ 3\mathbb{Z}$. The elements of this quotient ring are the cosets $ n +  3\mathbb{Z}$ for $n \in \mathbb{Z}$. What do these look like?
$$\begin{aligned}
0 +  3\mathbb{Z} &= \{\ldots, -6, -3, 0, 3, 6, \ldots\} \\
1 +  3\mathbb{Z} &= \{\ldots, -5, -2, 1, 4, 7, \ldots\} \\
2 +  3\mathbb{Z} &= \{\ldots, -4, -1, 2, 5, 8, \ldots\}
\end{aligned}$$
So this ring has at least three distinct elements. Are there any more? We can check that
$$\begin{aligned}
3 +  3\mathbb{Z} &= \{\ldots, -3, 0, 3, 6, 9, \ldots\} \\
&= \{\ldots, -6, -3, 0, 3, 6, \ldots\} \\
&= 0 + 3\mathbb{Z}
\end{aligned}$$
Demonstrating that the cosets $0 + 3\mathbb{Z}$ and $3+\mathbb{Z}$ are equal, and in fact any coset is one of the three listed above. 
Now, what happens for $\mathbb{Z}/\mathbb{Z}$? It's easy to see that for any $n \in \mathbb{Z}$, the coset $n + \mathbb{Z} = \mathbb{Z}$, and so the quotient has only one element (call it 0) with the rules that $ 0+0=0 $ and $0 \times 0 = 0$. So the quotient is the zero ring. 
A: Let $N \leqslant M$ be modules.  $M/N = 0$ iff $M = N$.  $\Leftarrow$ easy, see comments.  $\Rightarrow$ Let $x \in M \notin N$.  But $x + N = 0 + N$ iff $x - 0 = x \in N$ by definition submodule $N$.
Now apply the the above to the $R$-module $R$ (a ring).  The ideals are precisely the $R$-submodules.
Thus we've shown this for two structures at once.
