An ideal in $\Bbb Z[i]$ is $(1)$ if its generators have coprime norms

Question

Let $a$, $b$, and $c$ be three elements of $\mathbb{Z}[i]$, the ring of Gaussian integers. We define the following to be subset of $\mathbb{Z}[i]$:

$$(a,b,c)=\{ax+by+cz:\ x,y,z \in\mathbb{Z}[i]\}.$$

We know that $(a,b,c)$ is an ideal of $\mathbb{Z}[i]$. Show that the ideal $(-1+3i, -1+5i,-4+7i$) is equal to $\mathbb{Z}[i]$ .

What I have done so far:

I have factorised the three elements into a product of irreducible elments in $\mathbb{Z}[i]$:

$(-1+3i)=(1+i)(1+2i)$, $(-1+5i)=(1+i)(2+3i)$,$(-4+7i)=(2+3i)(1+2i)$

However I don't know how to show that the ideal is equal to $\mathbb{Z}[i]$. I think it might have something to do with the norms of the three elements in the ideal...

Answer 1

There are very many ways of handling a specific question like this. Let me show you the technique I most often use, which has nothing to do with the fact that $\Bbb Z[i]$ has unique factorization. \begin{align} (−1+3,−1+5,−4+7)&=(-1+3i,2i,-4+7i)\\ &=(-1+3i,2,7i)\\ &=(-1+3i,2,7)=(-1+3i,1)=(1)\,. \end{align} If it’s not self-explanatory, I can elaborate.

EDIT: Elaboration.
Let’s call the three Gaussian numbers $U, V, W$, in the order you gave. Step 1 (first line) replaces $(U,V,W)$ by $(U,V-U,W)$. Do you see why the two triples generate the same ideal? (If you don’t see this immediately, show that every number of form $aU+bV+cW$ is also of the form $a'U+b'(V-U)+c'W$, and also show the converse. This understanding is the most important part of the deal. Make sure you absorb it.)

Second step combines two steps, replacing the generator $2i$ by a number that is $2i$ times the unit $-i$. Do you see why this ideal is the same as the preceding? Once you’ve done that, do the same thing as in step 1, by keeping the second generator but adding twice it to the third generator. Do you see why $(-1+3i,2,-4+7i)$ is the same ideal as $(-1+3i,2,7i)$?

Next step replaces $7i$ by $7$, same as the first part of step 2.

Final step uses the fact that if $2$ and $7$ are in the ideal, so is $1$, which is enough to make the ideal the whole ring.

I hope it’s clear now.

Answer 2

As it was pointed out in the comments: In general, if $R$ is a commutative ring with identity and $I\subset R$ is an ideal, then $I=R$ if and only if $1\in I$.

So it suffices to show that there exist $a,b,c\in\Bbb{Z}[i]$ such that $$a(-1+3i)+b(-1+5i)+c(-4+7i)=1.$$

Answer 3

Easy way: $ $ ideal $I\,$ contains the norms $\,w\bar w\,$ of every generator $\,w\,$ so by Bezout it also contains the gcd of the norms = $(10,2\color{#c00}6,5\color{#0a0}5)=(10,\color{#c00}6,\color{#0a0}5)=(10,1,5)\!=\!1\,$ by Euclid. So $\,1\!\in\! I\Rightarrow I\! =\! \Bbb Z[i]$.

Remark $ $ As explained here this can be viewed an instance of the method of simpler multiples (or, equivalently, the correspondence or Third Isomorphism Theorem, from a structural viewpoint). See here for an analogous example using the norm and trace.

Because the norm map is multiplicative it preserves many properties related to factorization. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For further interesting examples see the papers of Bumby, Dade, Lettl, Coykendall cited in this answer.

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Alternatively working $\!\bmod I\,$ (or, equivalently, in $\,\Bbb Z[i]/I)\,$ simplifies ideal arithmetic (as in Lubin's answer) to more intuitive arithmetic of numbers (ring elements), e.g. here we find by subtracting neighboring stacked congruences that

$$\begin{align} \color{#c00}{3i\equiv 1}\\ 5i\equiv 1\\ 7i\equiv 4\end{align}\Rightarrow \begin{array}{} 2i\equiv \color{#0a0}0\\ 2i\equiv \color{#0a0}3\end{array}\Rightarrow\, \color{#0a0}{0\equiv 3}\,\Rightarrow\, 0\equiv \color{#c00}{3i\equiv 1}\qquad$$

While it may seem ad hoc at first glance, as long as we eventually keep finding smaller (norm) elements this will eventually yield the gcd, due to the fact that $\,\Bbb Z[i]\:\!$ is (norm) Euclidean. Thus, instead of mechanically applying the Euclidean algorithm, we can often use our intuition to step outside the algorithm to optimize it (one way is as above - using "simpler multiples", where "simpler" need not always imply smaller norm).