In $\mathbb{Z}[\sqrt{5}]$, we defnie the norma $N(a+b\sqrt{5})=|a^2-5b^2|$. I'm trying show that if $4\leq N(x)<16$, then $x$ is irreducible.

We know that an element is a unity is equivalent to tha its norm is $1$. I have try and I think is enougt prove that for all $x$, $N(x)\notin \{2,3\}$. Because if there exists an $x$ such that $N(x)\in \{2,3\}$, then $x$ is not a unity and $N(x^2)\in \{4,9\}$. So $4\leq N(x^2)<16$ and $x^2$ is not irreducible. Some have a suggestion for me? Thanks.

  • $\begingroup$ It is false that being a unit is equivalent to having norm $1$, because your function $N$ can take negative values. Specifically, $N(2+\sqrt{5}) = 4-5 =-1$, and $2+\sqrt{5}$ is a unit, since $(2+\sqrt{5})(-2+\sqrt{5}) = -4+5 = 1$. I cannot really understand the rest of your argument. $\endgroup$ Commented May 20 at 19:53
  • $\begingroup$ I already corrected the definition of the norm. What I know is enought for show the result is that $N(x)\notin \{2,3\}$ for all $x$. Do you know if that is thrue? $\endgroup$ Commented May 20 at 20:13

1 Answer 1


Throughout this argument, we will use the fact that the norm is multiplicative, that is, $N(x)N(y)=N(xy)$ (I implore you to check this).

A unit is an element $x$ with $N(x)=1$.

An element $x$ is irreducible if every factorisation $x=ab$ forces $a$ or $b$ to be a unit.

As you mentioned, there are no elements in $\mathbb{Z}[\sqrt5]$ of norm $2$ or $3$. This follows from the fact that we cannot solve $x^2=2$ or $3$ (mod$5$) (check this too!).

Any element of prime norm is irreducible, by the multiplicativity of the norm, so that does $N(x)=5,7,11,13$.

If $N(x)=4,6,8,10,14$, then the factorisation of $x$ must include an element of norm $2$, which is impossible, so these are irreducible.

If $N(x)=9,15$, then the factorisation of $x$ must include an element of norm $3$, which is again impossible.

Finally, if $N(x)=12$, the factorisation of $x$ must include an element of norm $2$ or $3$, which is impossible.

  • $\begingroup$ What about in general? How one can find all irreducible elements of that ring? $\endgroup$
    – nozalp10
    Commented May 20 at 21:42
  • $\begingroup$ @nozalp10 With difficulty. This is a particularly tricky ring to work with for several reasons. It is not a UFD and it is not a ring of integers (the ring of integers of $\mathbb{Q}(\sqrt 5)$ is much larger). Also this ring has infinitely many units. $\endgroup$
    – ljfirth
    Commented May 21 at 6:28

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