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Finding irreducible factorizations

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Poirot

Banned
Feb 15, 2012
250
Let $Z[\sqrt{-7}] ={{a+b\sqrt{-7}}}$ , where a,b are integers. Find 2 irreducible factorisations for 8. I can find one, namely $8=2^3$ but how to find another. More generally, what is the best way of finding irreducible factorisations in general rings?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Let $Z[\sqrt{-7}] ={{a+b\sqrt{-7}}}$ , where a,b are integers. Find 2 irreducible factorisations for 8. I can find one, namely $8=2^3$ but how to find another. More generally, what is the best way of finding irreducible factorisations in general rings?
$8 = (1 + \sqrt{-7})(1 - \sqrt{-7})$.

In a ring of the form $\mathbb{Z}[\sqrt{-p}]$, to factorise an element $a + b\sqrt{-p}$, you need to factorise the norm $N(a + b\sqrt{-p}) = a^2+pb^2.$ If $x$ is a factor of $a + b\sqrt{-p}$, then $N(x)$ has to be a factor of $N(a + b\sqrt{-p}).$ In this case, $N(8) = 64$, so the natural thing is to look for an element $x + y\sqrt{-7}$ with $N(x + y\sqrt{-7}) = 8$, in other words $x^2+7y^2=8.$ The obvious solution is to take $x=y=1.$