Gaussian Integer is Integral Over Integer

Question

I am reading from Wikipedia's article on Integral Element here, where it says that,

... an element $b$ of a commutative ring $B$ is said to be integral over $A$, a subring of $B$, if there are $n \geq 1$ and $a_j \in A$ such that $$b^n + a_{n-1}b^{n-1} + ... + a_1b + a_0 = 0.$$ This is to say, $b$ is a root of a monic polynomial over $A.$

It goes on stating that the Gaussian Integers, a complex numbers of the form $a + b\sqrt {-1},$ with $a, b \in \mathbb Z,$ are integral over $\mathbb Z.$ Here are my two questions:

(a) I have been tinkering with this example for more than one hour but could not come up with a monic polynomial having $a + b\sqrt {-1}$ as its root. Could anybody give me a hand? Do I convert it to Euler Formula $e^{ix}$ first?

(b) Do you have any other simple example of this integrality over $\mathbb Z$ befitting a beginner's knowledge?

Thank you for your time and help.

Answer 1

For (1), take the polynomial $x^2-2ax+(a^2+b^2)$. You will get this very easily by plugging $a+bi$ into an arbitrary polynomial $x^2+\alpha x+\beta=0$ and solving for $\alpha$ and $\beta$.

For (2), you may try this technique for $\mathbb{Z}[\sqrt{2}]$.

Answer 2

The monic polynomial $f(x)=x^2-2ax+b^2+a^2$ is in $\mathbb{Z}[x]$ for $a,b\in \mathbb{Z}$ and satisfies $f(a+bi)=0$. Secondly, it is instructive to show that $\frac{1}{2}$ is not integral over $\mathbb{Z}$. Suppose we would have a monic integer polynomial $f(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_0$ such that $f(\frac{1}{2})=0$. Then we obtain a Diophantine equation, which gives a contradiction modulo $2$ (you should try to insert the details here).

Answer 3

More generally, if you have something that looks like $\alpha = a_1\theta_1 + a_2\theta_2 \dots + a_n\theta_n +a_{n+1}$ and the $a_i's$ are integers and you know that $\theta_i$ is a root of the integer polynomial $p_i(x)$ with other roots $\theta_i = \theta_{i,1} \dots \theta_{i,d_i}$, the polynomial:$$p(x) = \prod_k(x-\alpha_k)$$ where the $\alpha_k$ are all possible ways of replacing the $\theta_{i,j}$ with some other $\theta_{i,j'}$ will be an integer polynomial with root $\alpha.$

In your specific example, $a_1 = b, \theta_{1,1} = i, \theta_{1,2} = -i, p_1(x) = x^2 + 1$ and the polynomial $p(x) = (x-(a+bi))(x-(a-bi)) =(x-a)^2 + b^2 $. You could replace integer by rational throughout and everything will still hold.