Why is the quotient ring $\frac{\mathbb{R}}{\mathbb{R}} = \left \{ 0 \right \}$?

Question

As the question says, why is the quotient ring $\frac{\mathbb{R}}{\mathbb{R}} = \left \{ 0 \right \}$ ? Shouldn't it be $\mathbb{R}$? We have that $\frac{\mathbb{R}}{\mathbb{R}} = \left \{ x + \mathbb{R} \mid x \in \mathbb{R} \right \}$ and the sum of two real numbers is just another real number. And in a more general case, is $\frac{A}{A} = \left \{ 0 \right \}$ or $\frac{A}{A} = A $?

Answer 1

Remember that $x + \mathbb{R}$ is a set, not a number. As $x + r$ is a real number for all $r \in \mathbb{R}$ (as you've correctly said), the set $x + \mathbb{R}$ is just the set of all real numbers, $\mathbb{R}$ for every $x \in \mathbb{R}$. So $\mathbb{R}/\mathbb{R} = \{\mathbb{R}\} \cong \{0\}$.

In the more general case, for a group $A$, the quotient group $A/A = \{x + A \mid x \in A\}$. But as $x + a \in A$ for every $x, a \in A$, again $x + A = A$ itself. So $A/A = \{A\} = \{0\}$.

Answer 2

The quotient $A/B$ means "the elements of $A$ partitioned into equivalence classes where two elements $x,y \in A$ are equivalent if their difference is in $B$: $x \sim y \iff x-y \in B$." Since every two real numbers in $A = \mathbb{R}$ differ by some element of $B = \mathbb{R}$, there is only one equivalence class. Pick a representative of this very large equivalence class. I like "$0$".

Since in addition, the quotient of a ring with one of its ideals, itself for instance, is a ring, we would like to pick the representative so that the result is obviously a ring. Since every ring contains $0$ ...

Answer 3

Another way to see this is to look at it with the perspective of the isomorphism theorem. In general, if you have a ring homomorphism $\phi:R_1 \rightarrow R_2$, then $Im(\phi) \cong R_1/\ker(\phi)$.

Now, consider the ring homomorphism $\phi: \mathbb{R} \rightarrow \{0\}$ defined as follows:

$$\phi(x) \mapsto 0$$

You can check that this is indeed a ring homomorphism (usually referred to as the 'trivial' homomorphism). Now, $\ker(\phi) = \mathbb{R}$ and $Im(\phi) = \{0\}$. Therefore, by the isomorphism theorem, we have:

$$\mathbb{R}/\mathbb{R} \cong \{0\}$$