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For $\alpha = (1+ \sqrt{-3})/2 \in \mathbb{C}$ and $R = \{ x +y\alpha \, | \, x,y \in \mathbb{Z} \}$.

How would you verify that R is a subring of $\mathbb{C}$? Everytime I multiply two 'elements' of $R$ to check closure I get the negative complex conjugate of $\alpha$, I think I'm doing something wrong...


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3 Answers 3



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Hint $\rm\ R\: =\: \mathbb Z + \alpha\:\mathbb Z\ $ is closed under multiplication $\rm \iff \alpha^2 \in R\iff \alpha\:$ is a quadratic integer.

Proof $\:$ (sketch) $\:$ If $\rm\:R\:$ is closed under multiplication then $\rm\:\alpha\in R\:\Rightarrow\: \alpha\cdot \alpha\in R.\:$ This implies that$\rm\:\alpha^2 \:=\: m + n\:\alpha\:$ for $\rm\:m,n\in \mathbb Z,\:$ which implies closure under multiplication

$$\rm (a+b\:\alpha)\:(c+d\:\alpha)\ =\ ac+(ad+bc)\:\alpha + bd\:(m+n\:\alpha)\ \in\ \mathbb Z+\alpha\:\mathbb Z $$

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Well, to verify that $R$ is a subring of $\Bbb C$, you'll need to verify that:

  1. $R$ is a subgroup of $\Bbb C$.
  2. $R$ is closed under multiplication.
  3. $R$ contains the multiplicative identity. (This axiom is to be verified if you consider the multiplicative identity as a part of the structure of your ring.)

To verify the first property:

  • Note that $R$ is non-empty as it contains $\Bbb Z$ inside it, when you set $y=0$ in your definition.
  • Let $x_1+\alpha y_1$ and $x_2 + \alpha y_2$ be elements of $R$. Clearly, $x_1-x_2+\alpha(y_1-y_2) \in R$.

To verify the second property:

  • $(x_1+\alpha y_1)(x_2+\alpha y_2)=x_1x_2+\alpha(x_1y_2+y_1x_2)+\alpha^2y_1y_2$. Now, can we write $\alpha^2$ in terms of $\alpha$ and other integers? If we could it would follow that $R$ is closed under multiplication.

A brief thought and acquaintance with complex numbers tell us that $-\alpha$ is a complex cube root of unity. And, it is know that, the cube roots of unity satisfy $1 +(- \alpha) +(-\alpha)^2=0$. This means, $$ \alpha^2=\alpha-1$$ So the product is, $$(x_1+\alpha y_1)(x_2+\alpha y_2)=x_1x_2-y_1y_2+\alpha(x_1y_2+y_1x_2+y_1y_2) \in R$$

To verify property $3$:

  • Note that by setting $x=1$ and $y=0$, you have $1 \cdot (x_1+\alpha y_1)= x_1 +\alpha y_1$. So, $1$ is the neutral element in $R$.


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You seem to be implying that $R$ is not closed under multiplication, but in fact it is: see my hint. – Brian M. Scott Mar 4 '12 at 12:45
@Brian I think I was. I'll change it to set things right and more explicit. – user21436 Mar 4 '12 at 13:15

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