0
$\begingroup$

I am working on the following question:

Let $\alpha \in \mathbb{C}$ and let $\mathbb{Z}[\alpha]$ be the smallest subring of $\mathbb{C}$ containing $\alpha$. That is, $\mathbb{Z}[\alpha] := \bigcap S$ for all $S$ which are subrings of $\mathbb{C}$ that contain $\alpha$. Prove that $\mathbb{Z}[\alpha] = \{f(\alpha) | f(x) \in \mathbb{Z}[x]\}$.

My thoughts:

I'm mostly confused by the notation. I think that $\mathbb{Z}[x]$ represents all integer-coefficient polynomials of one variable. So am I to prove that $\mathbb{Z}[\alpha]$ is the set of evaluations of all those polynomials at $x=\alpha$? If so, I don't see how this is itself a subring of $\mathbb{C}$.We've also used the square bracket notation to refer to the ring of Gaussian integers, so $$\mathbb{Z}[\alpha] = \{x+\alpha y | x,y \in \mathbb{Z}\}?$$

If someone could help me understand more precisely what I'm being asked to show it would be much appreciated.

2 Answers 2

3

Your guesses are correct. $\mathbb{Z}[x]$ is the ring of all polynomials in the variable $x$ with coefficients in $\mathbb{Z}$. The proof that $\{f(\alpha):f(x)\in\mathbb{Z}[x]\}$ is a subring of $\mathbb{C}$ is very straightforward. For instance, if you have two elements $f(\alpha)$ and $g(\alpha)$ in this set, then their sum is just $f(\alpha)+g(\alpha)=(f+g)(\alpha)$, which is another element of this set (since $f+g$ is another polynomial in $\mathbb{Z}[x]$).

Once you've shown that this set is a subring, that proves that it contains $\mathbb{Z}[\alpha]$. For the reverse inclusion, the idea is to prove that for any $f(x)\in\mathbb{Z}[x]$ $f(\alpha)$ is forced to be an element of $S$ if $S$ is a subring containing $\alpha$. For example, if $f(x)=x^2+3x+4$, then $$f(\alpha)=\alpha^2+3\alpha+4=\alpha\cdot\alpha+(1+1+1)\cdot\alpha+1+1+1+1$$ must in any such subring $S$ because $S$ contains $\alpha$ and $1$ and is closed under addition and multiplication.

(The case of Gaussian integers is a special case--it turns out that for $\alpha=i$, every element of $\mathbb{Z}[i]$ can be written in the form $a+ib$ for $a,b\in\mathbb{Z}$. In other words, you only need to use linear polynomials $f(x)$ in order to get all the elements of $\mathbb{Z}[i]$. But this is not true for general $\alpha$, and is not the definition of $\mathbb{Z}[\alpha]$ in general.)

0

Indeed, all you have to prove is the set set of all evaluations of polynomials with integer coefficients at α is ring. This results from the observations that $$f(\alpha)+g(\alpha)=(f+g)(\alpha), quad f(\alpha)g(\alpha)=(fg)(\alpha).$$ If $\alpha$ is algebraic of degree $d$, Euclidean division shows any element of $\mathbf Z[\alpha]$ is represented uniquely as a polynomial in $\alpha$, of degree $

As to the ring of Gaussian integers, it is the smallest subring of $\mathbf C$ containing $i$, and as $i$ is a root of the polynomial $X^1+1$, its elements are uniquely represented by linear polynomials in $i$.