AAR5

From Exampleproblems

Show that $1+3\sqrt{-5}\,$ is irreducible but not a prime in $\mathbb{Z}[\sqrt{-5}]\,$ .

The norm of an element in this ring is $N(\alpha + \beta\sqrt{-5})=\alpha^2+5\beta^2\,$ .

If $1+3\sqrt{-5} = \alpha\beta\,$ with $\alpha,\beta\isin\mathbb{Z}\,$ , then $46=1+5\cdot 9=N(1+3\sqrt{-5}) = N(\alpha\beta) = N(\alpha)N(\beta)\,$ .

So either $N(\alpha)=1\,$ and $N(\beta)=46\,$ or $N(\alpha)=2\,$ and $N(\beta)=23\,$ .

If $\alpha = a+b\sqrt{-5}\,$ then $N(\alpha) = a^2 + 5b^2 = 2\,$ which is impossible.

Therefore $N(\alpha) = 1\,$ , $\alpha\,$ is a unit, and $1+3\sqrt{-5}\,$ is irreducible.

$(1+3\sqrt{-5})(1-3\sqrt{-5}) = 46 = 2\cdot 23\,$ , so $1+3\sqrt{-5}\,$ divides 46 but $1+3\sqrt{-5}\,$ does not divide 2 or 23. Therefore $1+3\sqrt{-5}\,$ is not prime.

Main Page : Abstract Algebra : Rings

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