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We have $\alpha=(1+i)^7$ and $\beta=16$ in the ring $R=\mathbb Z[i]$. How can we prove that $\beta R\subseteq\alpha R$?

My try: $\beta R=\{a+bi:a,b\in 16\mathbb Z\}$ and $\alpha =(1+i)^7\in\mathbb Z$. I am clueless.

2 Answers 2

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Hint: $\dfrac{1+i}{\sqrt2}$ is an $8$-th root of unity.

So, $(1+i)^8=16$ and then $\beta=16=\alpha(1+i)\in \alpha R$.

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    $(1+i)^8=16$ but then how should i proceed? – 2017-02-20
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    can i say $\alpha=(1+i)^7=\beta\frac{(1-i)}{2}\in \beta R?$ i.e can i conclude $\beta R= \alpha R?$ – 2017-02-20
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    @MatheMagic, no because $\frac{(1-i)}{2}\notin R$. And you want to prove the reverse inclusion. – 2017-02-20
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    oh..that means $\beta R\subseteq \alpha R$.Right? – 2017-02-20
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You just have to prove that $16\in\alpha R$, that is, $16=(1+i)^7(a+ib)$ for some $a,b\in\mathbb{Z}$.

Consider $2=(1+i)(1-i)=-i(1+i)^2$, so $$ 16=2^4=(-i(1+i)^2)^4=(1+i)^8 $$

Note. The only prime in $\mathbb{Z}[i]$ that divides $16$ is (up to associates), $1+i$. Therefore $16=u(1+i)^k$, for some positive integer $k$ and an invertible $u$.