Question

Let I and J be ideals in R. Prove that the set $\mathit{K}\mathbf{=}\left\{a+b/a\in I,b\in J\right\}$ is an ideal in R that contains both I and J. K is called the sum of I and J and is denoted $\mathit{I}\mathbf{+}\mathit{J}$.

Short answer

It is proved that K is an idealin R.

Step-by-step solution

Theorem

Let R be a ring,and Iand Sbenon-empty subsetsof the ring;then,I is said to be the ideal if and only if it has the following two properties:

1. If $a,b\in I$, then$a-b\in I$ .
2. If $r\in R$ and $a\in I$, then $ra\in I$, and $ar\in I$.

Show that K contains I and J

It is given that I and S are ideals, then for subrings, I and S must contain .For any $a\in I$, we have

$a=a+0_R\in K$

Hence, $I\subseteq K$.

Similarly, $J\subseteq K$.

Hence, K contains both I and J,and it is non-empty.

Show that K is ideal

Now, assume that $K_1{K}_2\in K$, then there exists $a_1{a}_2\in I$and $b_1{b}_2\in J$, such that

So, we get $K_1=a_1+b_1$, and $k_2=a_2+b_2$.

Now, subtract k₂ from k₁ .

$k_1-k_2=\left(a_1-a_2\right)+\left(b_1-b_2\right)\in K$

As I and S are ideals, $a_1-a_2\in I$and $b_1-b_2\in J$, which implies $a-b\in I\cap S$, so property (i) of the theorem is satisfied.

Next, assume that $r\in R$. Then, $r{K}_1=r{a}_1+r{b}_2\in K$.

As I and J are ideals, we have $r{a}_1\in I$ and $r{b}_1\in I$$r{b}_1\in I$.

Similarly, $k_{1}r\in K$.

So, property (ii) of the theorem is satisfied.

Hence, it is proved K is an ideal in R.