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.

emancipezN 2020-11-02 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.

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sovienesY
Answered 2020-11-03 Author has 19764 answers

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.

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