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I am having trouble this problem in my textbook. It ask us to determine whether the following elements are irreducible in $\mathbb{Z} [ \sqrt{-3} ] = \{a+b\sqrt{-3} : a,b \in \mathbb{Z}\}.$

$\sqrt{-3}, 1, 2, 1 + \sqrt{-3}, 5$

I am having trouble even figuring out what the units are in this ring.

Rebekah
  • 997

1 Answers1

5

Hint: the “norm function” $N\colon a+b\sqrt{-3}\mapsto a^2+3b^2$ is multiplicative, that is, for $x,y\in\mathbb{Z}[\sqrt{-3}]$, $$ N(xy)=N(x)N(y) $$ Note also that $N(x)\ge0$.

Suppose $\sqrt{-3}=xy$; then $N(\sqrt{-3})=N(xy)$ so $3=N(x)N(y)$. This is an equality in the positive integers $\mathbb{Z}$, so you conclude that $N(x)=1$ or $N(y)=1$.

Similarly, if $x$ is invertible, then $1=N(1)=N(xx^{-1})=N(x)N(x^{-1})$, so $N(x)=1$.

Can you go on?

egreg
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