Question

1. Question 13: If I is an ideal in a ring R, then $\mathbf{(}\mathbf{I}\mathbf{\times }\mathbf{I}\mathbf{)}$is an ideal in by$\mathbf{(}\mathbf{R}\mathbf{\times }\mathbf{R}\mathbf{)}$Exercise 8 of Section 6.1. Prove that $\left(R\times R\right)\mathbf{/}\left(I\times I\right)$is isomorphic to$\mathbf{(}\mathbf{R}\mathbf{/}\mathbf{l}\mathbf{)}\times \mathbf{(}\mathbf{R}\mathbf{/}\mathbf{I}\mathbf{)}$ . [Hint: Show that the function $f:\mathbf{R}\times \mathbf{R}\to \mathbf{(}\mathbf{R}\mathbf{/}\mathbf{l}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{R}\mathbf{/}\mathbf{I}\mathbf{)}$given by is a surjective homomorphism of rings with the kernel $\mathbf{(}\mathbf{I}\mathbf{\times }\mathbf{I}\mathbf{)}$.]

Short answer

Answer:

It is proved that$\left(R\times R\right)/\left(I\times I\right)$ is isomorphic to $R/I\times R/I$.

Step-by-step solution

 Step 1: First isomorphism Theorem

Consider that$f:R\to S$is a surjective homomorphism of rings with the kernel. Then, thequotient ring would be R/K isomorphic to S.

Show that R×R/I×I is isomorphic to" width="9" height="19" role="math" style="max-width: none; vertical-align: -4px;" localid="1659434917684">R/I x R/I 

Establish .$Iff:R\times R\to R/I\times R/I\left(a,b\right):=\left(a+I,b+I\right)$

This function would be a ring homomorphism.

$$\begin{gathered} f\left(a_1+a_2,b_1+b_2\right)=\left(a_1+a_2+I,b_1+b_2+I\right) \\ =\left(a_1+I,b_1+I\right)+\left(a_2+I,b_2+I\right) \\ =f\left(a_1,b_1\right)+f\left(a_2,b_2\right) \\ f\left(a_1{a}_2,b_1{b}_2\right)=\left(a_1{a}_2+I,b_1{b}_2+I\right) \\ =\left(a_1+I,b_1+I\right)\left(a_2+I,b_2+I\right) \\ =f\left(a_1,b_1\right)f\left(a_2,b_2\right) \end{gathered}$$

This is surjective because for every , $\left(a+I,a+I\right)\in R/I\times R/I$and$f\left(a,b\right)=\left(a+I,b+I\right)$ .

There is$\left(a,b\right)\in \ker f$ , such that if$\left(a+I,b+I\right)=\left(I,I\right)$ in $R/I\times R/I$. This is equivalent to ,$a\in I$ and $b\in I$. As a result,$\ker f=I\times I$ .

It concludes that$R\times R/I\times I\cong R/I\times R/I$ .