Question

If ${\mathbf{l}}_{\mathbf{1}}\mathbf{,}{\mathbf{l}}_{\mathbf{2}}\mathbf{,}{\mathbf{l}}_{\mathbf{3}}$are ideals in a ring R with identity such that${\mathbf{l}}_{\mathbf{1}}\mathbf{+}{\mathbf{l}}_{\mathbf{3}}\mathbf{=}\mathbf{R}$and ${\mathit{l}}_{\mathbf{2}}\mathbf{+}{\mathit{l}}_{\mathbf{3}}\mathbf{=}\mathit{R}$, prove that $\left(l_1\cap l_2\right)\mathbf{+}{\mathbf{l}}_{\mathbf{3}}\mathbf{=}\mathbf{R}$. [Hint: If$\mathbf{r}\mathbf{\in }\mathbf{R}$, then $\mathbf{r}\mathbf{=}{\mathbf{i}}_{\mathbf{1}}\mathbf{+}{\mathbf{i}}_{\mathbf{3}}$and for some ${\mathit{i}}_{\mathbf{1}}\mathbf{\in }{\mathit{l}}_{\mathbf{1}}\mathbf{,}\mathbf{}{\mathit{t}}_{\mathbf{2}}\mathbf{\in }{\mathit{l}}_{\mathbf{2}}$, and ,${\mathbf{1}}_{\mathbf{R}}\mathbf{=}{\mathbf{t}}_{\mathbf{2}}\mathbf{+}{\mathbf{t}}_{\mathbf{3}}$. Then $\mathit{r}\mathbf{=}\left(i_1+i_3\right)\left(t_1+t_3\right)$; multiply this out to show that r is in $\left(l_1\cap l_2\right)$. Exercise 2 may be helpful.]

Short answer

It is proved that $\left(l_1\cap l_2\right)+l_3=R$.

Step-by-step solution

Describe the Chinese Remainder Theorem

Consider the relatively prime positive integers ${\mathit{m}}_{\mathbf{1}}\mathbf{,}{\mathit{m}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{m}}_{\mathbf{r}}$and the integers and $a_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{a}}_{\mathbf{r}}$the following system of the congruence equations.

$$\begin{gathered} \mathit{x}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{(}\mathit{m}\mathit{o}\mathit{d}\mathbf{}{\mathit{m}}_{\mathbf{1}}\mathbf{)} \\ \mathit{x}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{(}\mathit{m}\mathit{o}\mathit{d}\mathbf{}{\mathit{m}}_{\mathbf{2}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.} \\ \mathit{x}\mathbf{=}{\mathit{a}}_{\mathbf{r}}\mathbf{(}\mathit{m}\mathit{o}\mathit{d}\mathbf{}{\mathit{m}}_{\mathbf{r}}\mathbf{)}\mathbf{} \end{gathered}$$

This system has a solution. Also, if one of the solutions is t, then an integer z is also a solution of this system if and only if the congruence equation $\mathit{z}\mathbf{=}\mathit{t}\mathbf{}\mathbf{(}{\mathit{m}}_{\mathbf{1}}\mathbf{,}{\mathit{m}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{m}}_{\mathbf{r}}\mathbf{)}$holds true.

Prove that (l1∩l2)+l3=R

Since$l_1,l_2,l_3\subseteq R$and since R is closed under addition and intersection, then

$\left(l_1\cap l_2\right)+l_3\subseteq R......\left(1\right)$

Let $r\in R$

Since localid="1659183059853" $R=l_1+l_2$, there exist $r_1\in l_3$such that $r=r_1+r_3$.

Since$R=l_2+l_3$, there exist localid="1659197416182" $i_2\in l_2$andsuch that localid="1659183172085" $l_R=i_2+i_3$.

$$\begin{gathered} r=\left(r_1+r_3\right).1_R \\ =\left(r_1+r_3\right)\left(i_2+i_3\right) \\ =r_1{i}_2+r_1{i}_3+r_3{i}_2+r_3{i}_3 \end{gathered}$$

Here, localid="1659183550869" $r_1{i}_2\in l_1\cap l_{2}and{r}_1{i}_3+r_3{i}_2+r_3{i}_3\in l_3$and .

Therefore, localid="1659197497959" $r=r_1{i}_3+r_3{i}_2+r_3{i}_3\in \left(l_1\cap l_2\right)+l_3$This holds for every $r\in R$. Thus, localid="1659183783570" $R\subseteq \left(l_1\cap l_2\right)+l_3$

From equations (1) and (2),

$\left(l_1\cap l_2\right)+l_3=R$

Therefore, it is proved that $\left(l_1\cap l_2\right)+l_3=R$.