# True or false: question about generated ideal and his powers. [duplicate]

Let $$R$$ be a commutative ring with identity and $$a\in R$$. Let $$n\in\mathbb{N}$$. The principal ideal generated by $$a$$ is $$\langle{}a\rangle{}=aR=\{ar\;|\;r\in R\}.$$

Question. Is true or false that $$\langle{}a\rangle{}^n=\langle{}a^n\rangle{}$$?

My attempt It's true for me: in fact by definition $$\langle{}a^n\rangle{}=\{a^nr\;|\;r\in R\},$$ and

$$\langle{}a\rangle{}^n=\bigg\{\sum_{\text{finite}}(a_{1}\cdots a_{n})\;|\;a_i\in\langle{}a\rangle{}\bigg\}=\bigg\{\sum_{\text{finite}}a^n(r_1\cdots\ r_n)\;|\;r_i\in R\bigg\}.$$ We observe that $$(a^n)\subseteq(a)^n$$, as if $$\tilde{a}\in (a^n)$$ exists $$r\in R$$ such that $$\tilde{a}=a^nr$$, accordingly $$\tilde{a}=a^n(r\cdot \underbrace{1_R\cdot 1_R\cdots 1_R}_{n-1\;\text{times}})\in\langle{}a\rangle{}^n$$.

Vice versa we prove that $$(a^n)\supseteq(a)^n$$. If $$\overline{a}\in\langle{a}\rangle^n$$, then $$\overline{a}=\sum_{\text{finite}}\overline{a}^n(t_1\cdots\ t_n)$$, where $$t_i\in R$$. Therefore $$\overline{a}=a^n[\underbrace{(s_1\cdots \ s_n)\underbrace{+\cdots+}_{\text{finite}}(v_1\cdots v_n)]}_{:=r}$$, then $$\overline{a}=a^nr$$, where $$r\in R$$. Therefore $$\overline{a}\in\langle{}a\rangle{}^n$$.

Thanks!

• See the answer for the reasons to fail in general. – Dietrich Burde Nov 10 '18 at 9:31
• @Jack Is $R$ commutative? – Gone Nov 10 '18 at 17:27
• @Bill DubuqyeYes! Sorry.... now I correct – Jack J. Nov 10 '18 at 17:33