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What is the general form of elements in $\displaystyle \mathbb{Z} \left[\frac{1+\sqrt{-3}}{2} \right] $?

I'm getting muddled.

Thanks

  • 0
    Do you know what the definition of $\mathbb{Z}[\alpha]$ is (for, say, $\alpha \in \mathbb{C}$)?2011-05-22
  • 0
    I do. It's the ring of (finite) sums of powers of $ \alpha $, with integer coefficients.2011-05-22
  • 3
    If $\alpha = \frac{1+\sqrt{-3}}{2}$, can you express the square of $\alpha$ as sum of $1$ and $\alpha$ with integer coefficients ?2011-05-22
  • 0
    since $\alpha^2-\alpha+1=0$ you can write all elements as $a+b\alpha, a,b\in\mathbb{Z}$2011-05-22
  • 1
    So combine that with the fact that $\rm\ \alpha^2 \in \alpha\ \mathbb Z + \mathbb Z\ $ to conclude that $\ \mathbb Z[\alpha] \cong \ldots$2011-05-22

1 Answers 1

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In general, if $\alpha$ is an algebraic integer (satisfies a monic polynomial with integer coefficients), and its monic irreducible is of degree $n$, then the elements of $\mathbb{Z}[\alpha]$ can be written uniquely as $$a_0 + a_1\alpha + \cdots + a_{n-1}\alpha^{n-1},\qquad a_i\in\mathbb{Z}.$$ (A simple application of the remainder theorem for polynomials, and the fact that $\mathbb{Z}[\alpha] = {p(\alpha)\mid p(x)\in\mathbb{Z}[x]}$).

So here, the simplest is to note that $\frac{1+\sqrt{-3}}{2}$ satisfies a monic polynomial of degree $2$ over $\mathbb{Z}$.

In this particular case, there is an alternative possible description: every element of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ can be written uniquely as $$\frac{a + b\sqrt{-3}}{2},\qquad a,b\in\mathbb{Z},\qquad a\equiv b\pmod{2}.$$