Find $n$ such that $J=n\mathbb Z$

Question

Let the ideal $I=(4+3i)\mathbb Z[i]$ and $J=I\cap\mathbb Z$. Prove that $J$ is an ideal in $\mathbb Z,$ and find $n$ such that $J=n\mathbb Z$.

The first part of the task is to use the definition of an ideal by checking if it meets the appropriate properties. However I don't know what the ideal $J$ looks like because I must find part of the common $\mathbb Z$ and $\mathbb Z[i]$. Also, I don't know exactly how to find $n$?

Answer 1

We have $$I=\{(4+3i) (a+bi):a,b\in\mathbb Z\} = \{(4a-3b) + (3a+4b)i:a,b\in\mathbb Z\}.$$ Now you want $I\cap \mathbb Z$, this will be the set of numbers of the form above with imaginary part zero, i.e., integers of the form $4a-3b$ such that $3a+4b=0$, i.e., integers of the form $\frac{25a}{4}$.

It follows that $J=25\mathbb Z$, since $25a/4$ is an integer if and only if $a$ is a multiple of $4$.

Answer 2

Since $J$ is the intersection of an ideal and a subring, it must be an ideal of the subring.

Now, when is an integer divisible by $4+3i$ in the Gaussian integers? To answer this question, suppose that $(4+3i)(a+bi)=m \in \mathbb{Z}$. Then, $4a-3b=m$ and $3a+4b=0$. Multiplying the first equation by $4$ and the second one by $3$, we get $16a-12b=4m$ and $9a+12b=0$. Now, the coefficients of $b$ are opposites, so we can use the elimination method. Adding the two equations gives $25a=4m$. Since $4m$ is divisible by $25$, and $4$ and $25$ are coprime, $m$ must also be divisible by $25$. This proves that $J \subseteq 25\mathbb{Z}$.

Conversely, we also have $25\mathbb{Z} \subseteq J$, since $25=4^2+3^2=(4+3i)(4-3i) \in J$.

Hence, we in fact have that $J=25\mathbb{Z}$, so the two possible values of $n$ that generate the ideal $J$ are $25$ and $-25$.

Answer 3

Since $(4+3i)(a+bi)=(4a-3b) + (3a+4b)i$ we can see that to be in $\mathbb{Z}$ we must have $3a+4b=0$. Since $3$ and $4$ are relatively prime $4|a$ and $3|b$. Choosing $a$ and $b$ as close to $0$ as we can minimal $a$ and $b$ are $a=4$ and $b=-3$ (or $a=-4, b=3$) which together with $4a-3b$ gives us $4^2 + 3^2=25$. Since we chose the smallest $a$ and $b$ that we have, $n=25$, as it is the closest to zero in the intersection.

Answer 4

These algebraic answers are great, but I think there's also something to be said for seeing what's happening here. The ideal generated by $4+3i$ consists of all multiples of $4+3i$ by Gaussian integers, i.e., by $a+bi$, where $a,b\in\mathbb{Z}$. We rewrite the product: $$\begin{align}(4+3i)(a+bi) &=a[4+3i] + b[(4+3i)i] \\ &= a[4+3i] + b[-3+4i] \end{align}$$

So, elements of this ideal are just integer combinations (like "5 of this, and -2 of that") of the two complex numbers $u=4+3i$ and $v=-3+4i$. These numbers form a square grid in the complex plane; in fact, this is what all ideals in $\mathbb{Z}[i]$ look like.

If you look at this grid, you can ask the geometric question: What is the first grid point that falls on the positive real line? Since each step in the "$u$ direction" gives us $3i$, and each step in the "$v$ direction" gives us $4i$, they'll cancel when you have the coefficients: $$4u-3v = 4(4+3i) -3(-3-4i) = (16+12i)-(-9+12i) = 25$$.

Algebraically, we're really thinking of $\mathbb{Z}$ as a rank-2 $\mathbb{Z}$-module, and looking at an ideal as a submodule, also with rank 2. That's why its picture is a grid that is similar to the regular grid that includes all points with integer coordinates. In some rings, such as $\mathbb{Z}[\sqrt{-6}]$, there are actually different "shapes" of ideals available, but in $\mathbb{Z}[i]$, they're all the same.

Answer 5

$$\overbrace{\dfrac{n}{4+3i} = \dfrac{(4-3i)n}{25}}^{\large {\rm rationalize\ denominator }_{\phantom |}\!\!}\in\Bbb Z[i]\iff 25\mid 4n,3n\iff 25\mid n = 4n\!-\!3n\qquad$$