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 $\overline{a}\in\langle{}a\rangle{}^n$.


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

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