 emancipezN

2020-11-02

Let R be a commutative ring with identity and let I be a proper ideal of R. proe that $\frac{R}{I}$ is a commutative ring with identity. sovienesY

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

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