Find two elements that don't have a gcd in a subring of Gaussian integers

Question

Find two elements in the domain $R := \{ x + 2y \sqrt {-1} \mid x,y \in \mathbb{Z} \}$ that do not have a gcd.

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I have no idea how to start. But I know if we consider $R^\prime = \{ x + y \sqrt {-1} \mid x,y \in \mathbb{Z} \}$ then every two elements have a gcd. So there must be something wrong with the 2 here.

Any help is appreciated.

Answer 1

So there must be something wrong with the 2 here.

Right. Use that $2-2i$ and $2$ are irreducible in $\Bbb Z[\sqrt{-4}]$ but have common factor $2$ in $\Bbb Z[i]$.

So $(2+2i)(2-2i) = 8 = 2^3$ and both $2-2i$ and $2$ are irreducible (consider the multiplicative norm $N(a+b\sqrt{-4}) = a^2+4b^2$ for a proof). Hence $4-4i=2(2-2i)$ and $8$ do not have a greatest common divisor.

Answer 2

Hint $\ $ If $\rm\: a,b,c\: $ fail Euclid's Lemma (below), i.e. if $\rm\ (a,b) = 1,\, $ $\rm\ a\mid bc\ $ but $\rm\ a\nmid c\:$, then one immediately deduces that the gcd $\rm\ \color{#c00}{(ac,bc)}\ $ fails to exist. Apply that to $\ 2\mid (2i)(2i).$

Euclid's Lemma $\rm\quad a\ |\ bc\ $ and $\rm\ (a,b)=1\ \Rightarrow\ a\ |\ c\quad$ if $\rm\ \color{#c00}{(ac,bc)}\ $ exists.

Proof $\ \ $ If $\rm\ (ac,bc)\ $ exists then $\rm\ a\ |\ ac,bc\ \Rightarrow\ a\ |\ (ac,bc) = (a,b)\:c = c\ $ by the following

Lemma $\rm\ \ (a,b)\ =\ (ac,bc)/c\quad$ if $\rm\ (ac,bc)\ $ exists. $\ $ [GCD Distributive Law]

Proof $\rm\quad d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$