3

I have the following ideal in $\mathbb{Z}[\sqrt{-5}]$, $I = (2 - \sqrt{-5})$. I want to know how to explicitly represent $\mathbb{Z}[\sqrt{-5}]/I$ and determine if this is a field.

I have tried $$\mathbb{Z}[\sqrt{-5}]/I \cong \mathbb{Z}[x]/(2-x,x^2+5)$$ but I am stuck.

Shaun
  • 44,997
user100000001
  • 379

2 Answers2

3

We have: $$ \begin{aligned} \mathbb{Z}[\sqrt{-5}]/I &= (\mathbb{Z}[X]/(X^2+5))\ /\ (2-X) \\ &= \mathbb{Z}[X]/(X^2+5,\ 2-X) \\ &= (\mathbb{Z}[X]/(2-X))\ /\ (X^2+5) \text{ and via } X\to 2 \\ &= \mathbb{Z}\ /\ (2^2+5) \\ &= \mathbb{Z}\ /\ (9) \end{aligned} $$ which is not a field.

dan_fulea
  • 32,856
  • In fact, working modulo $I$ in $\Bbb Z[\sqrt{-5}]$ we have $3\cdot 3=9=(2-\sqrt{-5})(2+\sqrt{-5})$. It remains to show $3\ne 0$ modulo $I$. Well, if $3\in I$, then we would have a relation of the shape $3=(a+b\sqrt{-5})(2-\sqrt{-5})$, taking norms then would give $9=(a^2+5b^2)\cdot 9$, so $a^2+5b^2=1$, so $b=0$ and $a=\pm1$, contradiction. – dan_fulea Aug 11 '19 at 14:29
1

Is this what you had in mind? $$(2-x,x^2+5)=(2-x,2x-x^2,x^2+5)=(2-x,5+2x)=(2-x,9)$$

In $\Bbb Z[x]/I$, $x$ acts like $2$ and $9$ as $0$.

cansomeonehelpmeout
  • 12,782
  • 3
  • 22
  • 49