Question # Let R be a commutative ring with identity and let I be a proper ideal of R. proe that R/I is a commutative ring with identity.

Commutative Algebra
ANSWERED Let R be a commutative ring with identity and let I be a proper ideal of R. proe that $$\displaystyle\frac{{R}}{{I}}$$ is a commutative ring with identity. 2020-11-03

We know that if R is a ring and I is an ideal of R, $$\displaystyle\frac{{R}}{{I}}$$ is the set of all cosets a+I where $$\displaystyle{a}\in{R}$$.
Then, $$\displaystyle\frac{{R}}{{I}}$$ is a ring under the following operations.
For $$\displaystyle{a},{b}\in{R}$$,
Addition: $$\displaystyle{\left({a}+{I}\right)}+{\left({b}+{I}\right)}={\left({a}+{b}\right)}+{I}$$
Multiplication: $$\displaystyle{\left({a}+{I}\right)}{\left({b}+{I}\right)}={a}{b}+{I}$$
The ring R/I is called as the quotient ring of R by I.
We have to prove that if R is a commutative ring with identity and I is a proper ideal of R, then $$\displaystyle\frac{{R}}{{I}}$$ is a commutative ring with identity.
By the definition of the quotient ring, $$\displaystyle\frac{{R}}{{I}}$$ is a ring under the operations addition and multiplication defined above.
So, it remains to prove that $$\displaystyle\frac{{R}}{{I}}$$ is commutative and has identity.
We say that a ring is commutative if the operation multiplication is commutative.
Consider arbitrary elements $$\displaystyle{\left({a}+{I}\right)},{\left({b}+{I}\right)}\in\frac{{R}}{{I}}.$$
Now,
$$\displaystyle{\left({a}+{I}\right)}{\left({b}+{I}\right)}={a}{b}+{I}$$
$$\displaystyle={b}{a}+{I}$$ [$$\displaystyle:$$' R is commutative, ab=ba]
$$\displaystyle={\left({b}+{I}\right)}{\left({a}+{I}\right)}$$
Since the elements are arbitrarily chosen, $$\displaystyle{\left({a}+{I}\right)}{\left({b}+{I}\right)}={\left({b}+{I}\right)}{\left({a}+{I}\right)}$$ is true for every pair of elements $$\displaystyle{\left({a}+{I}\right)},{\left({b}+{I}\right)}\in\frac{{R}}{{I}}.$$
Hence, $$\displaystyle\frac{{R}}{{I}}$$ is commutative.
Let the identity of R be 1.
We claim that 1+I is the identity $$\displaystyle\in\frac{{R}}{{I}}$$.
The above claim can be proved as follows.
Since $$\displaystyle{1}\in{R},{1}+{I}\in\frac{{R}}{{I}}$$.
Consider an arbitrary element $$\displaystyle{\left({a}+{I}\right)}\in\frac{{R}}{{I}}.$$
Now,
$$\displaystyle{\left({1}+{I}\right)}{\left({a}+{I}\right)}={\left({1}\right)}{\left({a}\right)}+{1}$$
$$\displaystyle={a}+{1}$$
$$\displaystyle{\left({a}+{I}\right)}{\left({1}+{I}\right)}={\left({a}\right)}{\left({1}\right)}+{I}$$
$$\displaystyle={a}+{I}$$
Since $$\displaystyle{\left({1}+{I}\right)}{\left({a}+{I}\right)}={a}+{I}={\left({a}+{I}\right)}{\left({1}+{I}\right)}$$ for each (a+I) in $$\displaystyle\frac{{R}}{{I}}$$, 1+I is the identity element $$\displaystyle\in\frac{{R}}{{I}}$$.
Hence, $$\displaystyle\frac{{R}}{{I}}$$ is a commutative ring with identity.