2

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...

user26857
  • 51,328
  • 13
  • 66
  • 136
  • 1
    To show that the ideal $(a, b, c)$ is equal to $\mathbb{Z}[i]$ you just need to show that $1\in (a,b,c)$. This is true in general by the way. Any ideal $I\subset R$ is equal to the ring $R$ if it contains the multiplicative identity. – SeraPhim Nov 05 '20 at 14:50

3 Answers3

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.

Lubin
  • 61,457
  • 5
  • 68
  • 138
  • 1
    @Rox It's essentially the Euclidean algorithm. Generally applying [an invertible ("unimodular") linear transformation](https://math.stackexchange.com/a/3533044/242) to the ideal generators ("basis") preserves the ideal, so we aim to do this in a way that "simplifies" the generators, e.g. reduces their norm (cf. [norm Euclidean number rings](https://en.wikipedia.org/wiki/Euclidean_domain#Norm-Euclidean_fields)). But here it suffices to work only with the norms since they are already coprime - see my answer. – Bill Dubuque Nov 05 '20 at 23:02
  • Everything you say is true, @BillDubuque , but that’s not the way I look at things. I just plow through it, looking to simplify the set of generators down to something transparent. – Lubin Nov 10 '20 at 18:57
  • @Rox I added an alternative proof to my answer explaining an equivalent *equational* view of the ideal arithmetic methods used above (which you will likely find more intuitive; exercise: do Lubin's exact computation this way). – Bill Dubuque Nov 10 '20 at 22:06
  • @Lubin Btw, do you know if there is any "standard" name [for this?](https://math.stackexchange.com/a/3387196/242) – Bill Dubuque Nov 11 '20 at 00:58
  • Sorry, @BillDubuque , I’m notoriously ignorant of standard terminologies. – Lubin Nov 11 '20 at 04:50
0

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 $$a(-1+3i)+b(-1+5i)+c(-4+7i)=1.$$

Darsen
  • 3,488
  • 1
  • 8
  • 25
Servaes
  • 60,235
  • 7
  • 71
  • 159
0

Easy way: the 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 is an instance of the method of simpler multiples (or, equivalently, the correspondence or Third Isomorphism Theorem, from a more structural viewpoint).

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.

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

$\qquad\quad\qquad\begin{align} \color{#c00}{3i\equiv 1}\\ 5i\equiv 1\\ 7i\equiv 4\end{align}\,$ $\Rightarrow\ \begin{align}2i\equiv\color{#0a0}0\\[.4em] 2i\equiv \color{#0a0}3\end{align}\,$ $\Rightarrow\ \color{#0a0}{0\equiv 3}\,\Rightarrow\, 0\equiv \color{#c00}{3i\equiv 1}$

While it seems 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 use our intuition to step outside the algorithm to optimize it when the occasion arises (one way is as above - using "simpler multiples", where "simpler" need not always imply smaller norm).

Bill Dubuque
  • 265,059
  • 37
  • 279
  • 908