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What does the notation $x\equiv y$ mod $2\mathbb{Z}[\zeta_3]$ mean?

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2 Answers 2

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It means that there is an element $z$ of the ring $\mathbb{Z}[\zeta_3] = \{a+b\left(\frac{1+\sqrt{-3}}{2}\right) \ | \ a,b \in \mathbb{Z} \}$ such that $x-y = 2z$.

More generally, if $R$ is a ring and $I$ is an ideal of $R$, then $x \equiv y \pmod I$ means that $x-y \in I$.

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Thanks, that helps. –  Jason Smith Feb 8 '11 at 22:52

It means $\rm\ x - y\ =\ 2\ z\ $ for some $\rm\ z \in \mathbb Z[\zeta_3]\:,\ $ i.e. $\rm\ z = m + n\ \zeta_3,\ $ where $\rm\ \zeta_3^2 + \zeta_3 + 1 = 0\:.$

This is a generalization of the notion of congruence in the ring $\rm\ \mathbb Z/n\ $ of integers modulo $\rm\:n\:$. Recall that when solving integer problems it often proves convenient to attempt to decompose the problem into simpler problems in the finite rings/fields $\rm\ \mathbb Z/n\:.\ $ The same technique proves useful for general rings, where one consider "simpler" images $\rm\ R/I \ $ where $\rm\:I\:$ is the congruence class of $0\:,\ $ the set of all elements that are mapped to $0$ in the quotient ring. This has a fundamental structure known as an ideal, of which the above principal ideal $\rm\ n\ R\ =\ \{ n\:r\ :\ r\in R\}\ $ is a prototypical example.

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Isnt $\Z[\zeta_3]$ a degree two extension so that $z=a+b\zeta_3$. You have helped me a lot in the past and Im just trying to understand why you write $z=m+n\zeta_3+k\zeta_3^2$ ? Thanks. –  Jason Smith Feb 8 '11 at 22:56
    
Yes, that typo is now fixed. –  Bill Dubuque Feb 8 '11 at 23:08

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