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1) Show that all element of $\mathbb Z[\sqrt{-5}]$ has the form $a+\sqrt{-5}b$.

2) Show that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible on $\mathbb Z[\sqrt{-5}].$

My attempts

1) Isn't it by definition that $$K[a]=\{\lambda_1+\lambda_2a+...\lambda_na^n\mid \lambda_i\in K\}\ ?$$ Since $\sqrt{-5}*\sqrt{-5}=-5\in\mathbb Z$, the result looks obvious.

2) Since $2$ and $3$ are irreducible in $\mathbb Z$ and that $\mathbb Z\subset \mathbb Z[\sqrt{-5}]$, they are obviously irreducible in $\mathbb Z[\sqrt{-5}]$. For $1\pm\sqrt{-5}$ I'm not sure how to proceed. I would say that if there is $a,b,c,d\in\mathbb Z$ such that $$1+\sqrt{-5}=(a+\sqrt{-5}b)(c+d\sqrt{-5})$$

then $$\begin{cases}ac-5bd=1\\ad+bc=1\end{cases}\implies ac-5bd=ad+bc\implies a(c-d)=b(5d+c)$$ but I can't get a contradiction.

Rick
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1 Answers1

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The key to (b) is to use the norm function $$N(a+b\sqrt{-5})=(a+b\sqrt{-5})\overline{(a+b\sqrt{-5})}=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2$$ Turns out this is multiplicative, which is extremely helpful for testing irreducibility.

vadim123
  • 82,796