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What is the prime factorization of $7$ in $\Bbb{Z}[\sqrt{2}]$? Also, why are the elements in the factorization prime elements?

Xam
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Hostile Amigo
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1 Answers1

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$$7 = (3 - \sqrt{2})(3 + \sqrt{2})$$ These elements are prime in $\Bbb Z[\sqrt{2}]$ because their norms ($N(3 - \sqrt{2}) = N(3 + \sqrt{2}) = 7$) are prime in $\Bbb Z$.

To find this factorization, it's helpful to know that $\Bbb Z[\sqrt{2}]$ is a PID and the proposition in my answer here which gives the factorization of the ideal $(7)\subseteq\Bbb Z[\sqrt{2}]$ into prime ideals. From there, one finds a generator of each prime ideal lying over $(7)$, which gives a factorization of the element $7$ up to a unit. In this case, $x^2 - 2\equiv x^2 - 9\equiv (x + 3)(x - 3)\pmod{7}$, so $(7) = (7,\sqrt{2} + 3)(7,\sqrt{2} - 3)$ in $\Bbb Z[\sqrt{2}]$, and one notes that $7$ is a multiple of $\sqrt{2} + 3$ and $\sqrt{2} - 3$, so that $(7) = (\sqrt{2} + 3)(\sqrt{2} - 3)$.

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Stahl
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  • Elements whose norm is prime are only irreducible. Fortunately our ring is Euclidean, so irreducible are also prime. – user26857 Jan 15 '17 at 11:44
  • @user26857 : Would you have a counter-example? If $x$ belongs to the ring of integers $O_K$ of a number field $K$ and the norm of $x$ w.r.t to $K/\Bbb Q$ is a prime number, then $x$ is always prime in $O_K$. – Watson Jan 15 '17 at 14:25
  • Yeah, the same happens in all rings $\mathbb Z[\sqrt d]$. What I wanted to say is that this fact needs to be proved, and such proofs are not entirely trivial. – user26857 Jan 15 '17 at 15:08
  • @user26857 Yes, I'm aware that I'm implicitly using the fact that $\Bbb Z[\sqrt{2}]$ is a PID in that argument; that's one reason it says it in the spoiler. I wanted to leave something for the OP to justify :) – Stahl Jan 15 '17 at 18:04
  • @Stahl: even if the ring of integers $O_K$ of $K$ is not a UFD, an element $x \in O_K$ of prime norm is a prime element of $O_K$, since $O_K/(x)$ has prime cardinality, or using the decomposition of $(x)$ into primes. – Watson Jan 15 '17 at 20:14
  • @Watson that's a nice argument, all I had in mind was multiplicativity of the norm. – Stahl Jan 15 '17 at 20:17