1

The similarly looking ring $\mathbb{Z}\Bigl[\dfrac{1 + \sqrt{-7}}{2}\Bigr]$ actually is Euclidean (see here).

Now I want to show that $\mathbb{Z}\bigl[1 + \sqrt{-7}\bigr]$ is not Euclidean.

Proof: First we note that $\mathbb{Z}\bigl[1 + \sqrt{-7}\bigr] = \mathbb{Z}\bigl[\sqrt{-7}\bigr]$. Assume that $1 + \sqrt{-7}$ is reducible, that means there exist $k, l, m, n \in \mathbb{Z}$ so that $$\bigl(k + l \sqrt{-7}\bigr)\bigl(m + n \sqrt{-7}\bigr) = k m - 7 l n + (k n + l m)\sqrt{-7} = 1 + \sqrt{-7}.$$ Do the equations $$k m - 7 l n = 1 \\ k n + l m = 1$$ have a solution?

In the case $k m \leq 0 \land l n \geq 0$ there is obviously no solution. The case $k m \geq 0 \land l n \leq 0$ only has the solutions $k = m = \pm 1$ and $(l = 0 \land n = k) \lor (l = 1 \land n = k)$, which is just a unit times an associated element of $1 + \sqrt{-7}$.

In the other two cases (that is $k m$ and $l n$ have the same sign) $k n$ and $l m$ must have the same sign, so the only two solution of the second equation would be $k n = 1 \land l m = 0$ or $k n = 0 \land l m = 1$, which again gives us just a unit times an associated element. So $1 + \sqrt{-7}$ is irreducible.

Since $\bigl(1 + \sqrt{-7}\bigr)\bigl(1 - \sqrt{-7}\bigr) = 8 = 2^3$ the ring $\mathbb{Z}\bigl[1+\sqrt{-7}\bigr]$ is not even an unique factorization domain, so it can't be Euclidean. $\square$

Everything ok?

The irreducibility proof of $1 + \sqrt{-7}$ was pretty cumbersome... perhaps using the norm is faster?

Thank you!

Community
  • 1
Ystar
  • 2,966
  • 4
  • 22
  • 39

2 Answers2

4

This can be shown without having to deal directly with any kind of factorization. A Euclidean domain is a UFD, and every UFD is integrally closed, which means something like the rational roots theorem holds: an integral domain $R$ with fraction field $K$ is called integrally closed if any root in $K$ of a monic polynomial in $R[x]$ lies in $R$.

The ring $\mathbf Z[\sqrt{-7}]$ is not integrally closed because $x^2 - x + 2$ is monic with coefficients in $\mathbf Z[\sqrt{-7}]$ but it has a root in the fraction field $\mathbf Q(\sqrt{-7})$ that is not in $\mathbf Z[\sqrt{-7}]$. The same argument shows $\mathbf Z[\sqrt{d}]$ is not integrally closed if $d$ is any integer that is not a perfect square and $d \equiv 1 \bmod 4$: the polynomial $x^2 - x + (1-d)/4$ is monic with coefficients in $\mathbf Z[\sqrt{d}]$ and has root $(1+\sqrt{d})/2$, which is in the fraction field $\mathbf Q(\sqrt{d})$ but is not in $\mathbf Z[\sqrt{d}]$. Therefore $\mathbf Z[\sqrt{d}]$ for such $d$ is not Euclidean or a UFD.

KCd
  • 55,662
0

No,you are wrong to prove this just show any one property of Euclidean domain is violated.Here $2|(1+\sqrt7i)(1-\sqrt 7i)$ but $2$ does not divide either $(1+\sqrt7i)$ or $(1-\sqrt 7i)$ hence $2$ is not prime but $2$ is irreducible as

$2=(a+b\sqrt7i)(c+d\sqrt 7i)$

Taking conjugate and multiplying

$\implies 4=(a^2+b^2)(c^2+d^2)$

either $(a^2+b^2)=1or (c^2+d^2)=1$

i.e either $(a+b\sqrt7i)$ is a unit or $(c+d\sqrt 7i)$ is so .Thus $2$ is irreducible but not prime which cant happen in a ED

Learnmore
  • 31,675
  • 10
  • 110
  • 250