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In $\mathbb{Z}[\sqrt2]$, it is true that $8-3\sqrt2$ has these factorizations into irreducibles (we can check that they're irreducible with the multiplicative norm $N(a+b\sqrt2)=|a^2 - 2b^2|$ and getting a prime for each factor):

$$(5+\sqrt2)(2-\sqrt2)$$ $$(11-7\sqrt2)(2+\sqrt2)$$

But these factorizations can't be different because Euclidean Domains are Unique Factorization Domains, so in what sense are they the same? They must differ by a unit or several units, so my question is really where do the units go and how do we find them?

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    $\begingroup$ $2+\sqrt{2}=(2-\sqrt{2})(3+2\sqrt{2}),11-7\sqrt{2}=(5+\sqrt{2})(3-2\sqrt{2})$ $\endgroup$
    – Wojowu
    Commented Oct 7, 2017 at 17:53
  • $\begingroup$ "how do we find them?" - in this MSE-question. $\endgroup$
    – Dietrich Burde
    Commented Oct 7, 2017 at 18:42

2 Answers 2

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The units in $\Bbb Z[\sqrt2]$ are $\pm(1+\sqrt2)^n$ for $n\in\Bbb Z$. Then $$2+\sqrt2=(2-\sqrt2)(1+\sqrt2)^2$$ and $$5+\sqrt2=(11-7\sqrt2)(1+\sqrt2)^2.$$ These factorisations are the same "up to units".

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Note that $2+\sqrt 2$ and $2-\sqrt 2$ obviously have the same norm, so divide one by the other to get the unit factor you need.

Likewise the other two factors have the same norm, so you can likewise divide and get the unit factor by which they differ.

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  • $\begingroup$ Thanks everyone, I see it's actually quite simple. You've answered my question. $\endgroup$
    – user477203
    Commented Oct 7, 2017 at 18:01

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