Let $\mathbb{Z}[\sqrt{10}] = \{a + b \sqrt{10} \mid a, b \in \mathbb{Z}\}$ and the norm function $N: \mathbb{Z}[\sqrt{10}] \rightarrow \mathbb{Z}$ defined by $N(a + b \sqrt{10}) = a^2 - 10b^2$.
I tried to show that greatest common divisor (GCD) of $\{2, 4 + \sqrt{10}\} = 1$ in $\mathbb{Z}[\sqrt{10}]$.
Definition: in a ring $R$, $d$ is the GCD of $\{ a, b \}$, with $a, b \in R$ if
i) $d \mid a$ and $d \mid b$
ii) if there's a $c \in R$ such that $c \mid a$ and $c \mid b$, then $c \mid d$.
We have that $1 \mid 2 $ and $1 \mid 4 + \sqrt{10}$ is trivial. Suppose that there is a $u \in \mathbb{Z}[\sqrt{10}]$ such that $u \mid 2$ and $u \mid 4 + \sqrt{10}$.
From a known fact about norms, $u \mid 4 + \sqrt{10}$ implies that $N(u) \mid N(4 + \sqrt{10})$. Thus $N(u) \mid 6$. The only divisors of 6 in $\mathbb{Z}$ are $\pm 1, \pm 2, \pm 3$ and $\pm 6$ (which equals 1 reducing mod 5).
From a previously proven fact, there is no $u \in \mathbb{Z}[\sqrt{10}]$ such that $N(u) = \pm 2, \pm 3$. Hence, $N(u) = \pm 1$ and $u$ is invertible. Therefore, $u u^{-1} = 1$, implying that $u \mid 1$.
Analogously, if $u \mid 2$, then $u \mid 1$
Edit: If $u \mid 2$, then $N(u) \mid N(2)$. The only divisors of 4 in $\mathbb{Z}$ are $\pm 1, \pm 2, \pm 4$. We know there's no $u \in \mathbb{Z}[\sqrt{10}]$ with $N(u) = \pm 2$. Assuming that $N(u) = 4$, we have that $4 \equiv -1 (\mod 5)$, so $N(u) = - 1$, whence $u$ is invertible and $u \mid 1$.
Hence, GCD$\{2, 4 + \sqrt{10} \} = 1$.
Is this reasoning correct?
PS: My goal is to show that the GCD$\{2, 4 + \sqrt{10} \} = 1$ cannot be expressed as a linear combination of $2$ and $4 + \sqrt{10}$.
Your reasoning is correct but can be simplified.
If $u \mid 2$ and $u \mid 4 + \sqrt{10}$, then $N(u) \mid \gcd(N(2),N(4 + \sqrt{10}))=\gcd(4,6)=2$.
Therefore, $N(u)=\pm 1$ or $N(u)=\pm 2$.
You already know that the second case cannot happen, and so $u$ is a unit.