AAR5

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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 \in {\mathbb  {Z}}[{\sqrt  {-5}}]\,, 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.


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