How can I show by considering the product of $\zeta_7^3 + \zeta_7^2 +1$ with $\zeta_7^3 + \zeta_7 +1$ that $(2, \zeta_7^3 + \zeta_7^2 +1 )$ is a principal ideal of $O_{Q(\zeta_7)}$?
Principal ideal by considering a product
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abstract-algebra
number-theory
algebraic-number-theory
ideals
roots-of-unity
2 Answers
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Let $\zeta=\zeta_7$. Multiply the numbers together. Doing so gives you the following: \begin{align*} (\zeta^3+\zeta^2+1)(\zeta^3+\zeta+1)&=\zeta^6+\zeta^5+\zeta^3+\zeta^4+\zeta^3+\zeta+\zeta^3+\zeta^2+1\\&=(\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1)+2\zeta^3\\ &=2\zeta^3 \end{align*} $\zeta$ is a unit in this ring, so $2\in (\zeta^3+\zeta^2+1)$. In other words, $(2,\zeta^3+\zeta^2+1)=(\zeta^3+\zeta^2+1)$.
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Well looking at the product of the two elements, and letting $\omega=\zeta_7$ I get
$$\omega^6+\omega^5+\omega^4+3\omega^3+\omega^2+\omega+1 = 2\omega^3$$
which is an associate of $2$, hence $(\omega^3+\omega^2+1)\supseteq (2)$ proving principality.