Simplify the ideal $(5+23i, 341)$ in $\mathbb Z[i]$

Question

This is possible because $\mathbb Z[i]$ is a Euclidean domain, hence a PID.

First I checked that $341^2=d(341) > d(5+23i)=5^2+23^2$.

I used the division algorithm.

$341= \left(5+23i\right)\left(3-14i\right)\:+4+i$

$5+23i=\left(3+5i\right)\left(4+i\right)-2$

$4+i=-2(-2)+i$

$-2=(2i)i$. $i$ divides $2$, so it is the $\gcd(5+23i, 341)$. So the answer is $(i)$.

Is this method correct?

Answer 1

As I pointed out in the comments, I would personally rather end on $(1)$ than on $(i)$. Apart from that, this proof looks complete to me.

As a little polish, I would make it clear what all the calculations are supposed to signify. Specifically, after each of your calculations, I suggest you clarify which two ideals you have just shown to be equal.

So something like this:

• $341 = (5+23i)(3-14i) + 4 + i$ gives us $(341, 5+23i) = (4+i, 5+23i)$

and so on. This makes it so that people who read your proof don't have to keep as much mental track, and it makes it easier to follow your logic one step at a time.