Question

A ring $\mathit{R}$ is said to satisfy the ascending chain condition (ACC) on ideals if whenever${\mathit{I}}_{\mathbf{1}}\mathbf{\subseteq }{\mathit{I}}_{\mathbf{2}}\mathbf{\subseteq }{\mathit{I}}_{\mathbf{3}}\mathbf{\cdots }$ is a chain of ideals in$\mathit{R}$ (not necessarily principal ideals), then there is an integer $\mathit{n}$ such that ${\mathit{I}}_{\mathbf{j}}\mathbf{=}{\mathit{I}}_{\mathbf{n}}$ for all $\mathit{j}\mathbf{\ge }\mathit{n}$. Prove that if every ideal in a commutative ring$\mathit{R}$ is finitely generated, then$\mathit{R}$ satisfies the ACC

Short answer

If every ideal in a commutative ring $R$ is finitely generated, then$R$ satisfies the ACC.

Step-by-step solution

Theorem 6.3

Let $R$ be a commutative ring with identity and $c_1,c_2,\dots ,c_n\in R$. Then the set $I=\left\{r_1{c}_1+r_2{c}_2+\cdots +r_n{c}_n|r_1,r_2,\dots ,r_n\in R\right\}$ is an Ideal in $R$.

Im=I

Given that $I$ is finitely generated then $I=(c_1,c_2,\dots ,c_n)$. Since, $R$ is commutative by theorem 6.3, $I$ can be written as$I=\left\{r_1{c}_1+r_2{c}_2+\cdots +r_n{c}_n|r_1,r_2,\dots ,r_n\in R\right\}$.

Let $c_1\in I_{j_1},c_2\in I_{j_1},\dots ,c_n\in I_{j_n}$ then for every $m\ge j_n$, $c_1,c_2,\dots ,c_n\in I_m$.

Thus,$I=\left\{r_1{c}_1+r_2{c}_2+\cdots +r_n{c}_n|r_1,r_2,\dots ,r_n\in R\right\}\subseteq I_m$ …… (1)

for every $m\ge j_n$.

Since, $I_m$ is the sub ideal of$I$implies$I_m\subseteq I$ …… (2)

Here, from (1) and (2), $I_m=I$.

R satisfies the ACC.

Let $l\ge m$ then $I_m\subseteq I_l$. Since, $I_l\subseteq I$ implies $I_l\subseteq I_m$.

Thus,$I_l=I_m$ for every $l\ge m$.

Therefore, if every ideal in a commutative ring $R$ is finitely generated, then $R$ satisfies the ACC.