I am trying to work through a hw problem to show that the ring of gaussian integers $G=\{a+bi:a,b\in\mathbb{Z}\}$ is principal. To make it concrete I picked the ideal $I$ generated by $a=5$ and $b=3+4i$. I believe $I=\{xa+yb:x,y\in G\}$ since this is closed under multiplication by gaussian integers and is an additive group and is contained in any ideal containing $a$ and $b$.
An element in $I$ has squared modulus $(ax+by)(\overline{ax}+\overline{by})=25|x|^2+25|y|^2+5ax\overline{y}+5y\overline{ax}.$ So every element in $I$ has a squared modulus divisible by $5$, so $I$ does not contain the units $\pm, 1\pm i$.
So $I$ is a proper ideal generated by some gaussian integer $u+iv, u,v\in\mathbb{Z}$ with squared modulus divisible by 5. From $b=5=(x+iy)(u+iv), x,y\in\mathbb{Z}$, setting real and imaginary parts equal, I get either $v=0$ or $-y(v^2+u^2)/v=5$. This can only hold for $0\le v\le 5$ and working through the cases the only possibilities for $(u,v)$ one of $(1,0),(5,0),(0,0),(0,1),(1,2),(4,2),(0,5)$. The only ones with squared modulus divisible by 5 are $(5,0),(1,2),(0,5)$. But for none of these does $a=3+4i=(x'+iy')(u+iv)$ have a solution with $x,y\in \mathbb{Z}$.
Best Answer
Let us work through applying the Euclidean algorithm to find the gcd in $\mathbb{Z}[i]$ of $3+4i$ and $5$. Now, $|3+4i|^2 = |5|^2 = 25$, so neither element is really "smaller" than the other one. So, let us take the simpler quotient, and divide $3+4i$ by $5$. In fact, $\frac{3+4i}{5} = \frac{3}{5} + \frac{4}{5} i$, and the nearest Gaussian integer is $1 + i$. So, the remainder is $(3+4i) - (1 + i) \cdot 5 = -2 - i$; and $\gcd(3+4i, 5) = \gcd(5, -2 - i)$.
Similarly, in this step, $\frac{5}{-2 - i} = -2 + i$; thus, the remainder at this step has reached 0, and we conclude $\gcd(3+4i, 5) = -2 - i$. Therefore, $-2 - i$ is a generator of the ideal.
(In fact, since the gcd is only defined up to multiplication by a unit of $\mathbb{Z}[i]$, we could just as well have said that the gcd is $2 + i$, or $1 - 2i$, or $-1 + 2i$. You could also apply some sort of normalization on the remainders at each step of the Euclidean algorithm if you wanted - for example, I tend to like to normalize to the region of the complex plane with argument in $(-\frac{\pi}{4}, \frac{\pi}{4}]$.)