Proving the algebraic Integers are closed under addition

algebraic-number-theorypolynomials

I'm using the following definition of an algebraic Integer:

An Algebraic Integer is the solution to a monic integer polynomial.

I know that is a lot simpler to use advanced linear algebra and such to make the proof shorter, but I'd prefer a more elementary proof.
Proving it is closed under addition:
I have been able to prove that it is true for polynomials up to quadratic, but I really didn't find a way to solve it for higher degree polynomials.
To be more specific, I'm looking for some way, given a pair of polynomials to construct a third polynomial whose root is the sum of the pair's roots. I think this has a connection to the resultant, but I wasn't able to understand how.

Best Answer

The following theorem and corollary are in chapter 2 of Marcus's book Number Fields:

Theorem 2. The following are equivalent:

  1. $\alpha$ is algebraic integer.
  2. The additive group $\left(\mathbb{Z}[\alpha],+\right)$ is finitely generated.
  3. $\alpha$ belongs to some subring of $\mathbb{C}$ whose additive group is finitely generated.
  4. $\alpha A \subset A$ for some finitely generated subgroup $A \leq \mathbb{C}$.

The proof of $4 \implies 1$ goes as follows: Let $a_1 , \dots a_n$ generate $A$. Expressing each $\alpha a_i$ as a linear combination of $a_1 , \dots a_n$ with coefficients in $\mathbb{Z}$, we obtain: $$ \begin{pmatrix} \alpha a_1\\ \dots\\ \alpha a_n \\ \end{pmatrix} = M \begin{pmatrix} a_1\\ \dots\\ a_n \\ \end{pmatrix} $$

where $M \in \mathcal{M}_n (\mathbb{Z)}$. Equivalently, $$ (\alpha I -M)\begin{pmatrix} a_1\\ \dots\\ a_n \\ \end{pmatrix} $$ is the zero vector. Since the $a_i$ are not all zero, it follows that $\alpha I - M$ has determinant $0$. Expressing this determinant in terms of the $n^2$ coordinates of $\alpha I - M$, we obtain $$ \alpha^n + \text{lower degree terms}=0 $$ thus we have produced a monic polynomial over $\mathbb{Z}$ having $\alpha$ as a root.

Corollary 1. If $\alpha$ and $\beta$ are algebraic integers, then so are $\alpha + \beta$ and $\alpha \beta$

Proof: We know $\mathbb{Z}[\alpha]$ and $\mathbb{Z}[\beta]$ have finitely generated additive groups. Then so does the ring $\mathbb{Z}[\alpha, \beta]$. (If $\alpha_1 , \dots , \alpha_m$ generate $\mathbb{Z}[\alpha]$ and $\beta_1 , \dots , \beta_n$ generate $\mathbb{Z}[\beta]$, then the $mn$ products $\alpha_i \beta_j$ generate $\alpha_1 , \dots , \alpha_m$ generate $\mathbb{Z}[\alpha, \beta]$.) Finally $\alpha_1 , \dots , \alpha_m$ generate $\mathbb{Z}[\alpha, \beta]$ contains $\alpha + \beta$ and $\alpha \beta$, so by characterization 3 this implies thet they are algebraic integers.

You can apply this method to produce monic polynomials for $\alpha \beta$ and $\alpha + \beta$ from the minimal polynomials of $\alpha$ and $\beta$. For example,let $\alpha = \sqrt 3$ and $\beta = \sqrt{2}$ and let's apply the determinant method to find the monic polynomial associated to $\alpha+\beta$. The minimal polynomials of $\alpha$ and $\beta$ are, respectively, $f(x)=x^2 - 3$ and $g(x)=x^2 - 2$. Since $\deg(f)=\deg(g)=2$, then $\{1, \alpha\}$ generates $\left( \mathbb{Z}[\alpha],+ \right)$ and $\{1, \beta\}$ generates $\left( \mathbb{Z}[\beta],+ \right)$, and thus $A=\{1, \alpha, \beta, \alpha \beta\}$ generates $\left(\mathbb{Z}[\alpha, \beta],+ \right)$, and since: $$ \mathbb{Z}[\alpha + \beta] \subset \mathbb{Z}[\alpha, \beta]$$ $$\mathbb{Z}[\alpha \beta] \subset \mathbb{Z}[\alpha, \beta]$$ then $A$ generates these two subrings. We have the system of linear equations: $$ (\alpha + \beta) \begin{pmatrix} 1\\ \alpha \\ \beta \\ \alpha \beta \\ \end{pmatrix} = \begin{pmatrix} \alpha + \beta \\ 2 + \alpha \beta\\ \alpha \beta + 3 \\ 2 \beta + 3 \alpha \\ \end{pmatrix} = \begin{pmatrix} 0 & 1 & 1 & 0\\ 2 & 0 & 0 & 1\\ 3 & 0 & 0 & 1\\ 0 & 3 & 2 & 0\\ \end{pmatrix} \begin{pmatrix} 1 \\ \alpha \\ \beta \\ \alpha \beta \\ \end{pmatrix} $$ From where we obtain: $$ \det\begin{pmatrix} \alpha + \beta & -1 & -1 & 0\\ -2 & \alpha + \beta & 0 & -1\\ -3 & 0 & \alpha + \beta & -1\\ 0 & -3 & -2 & \alpha + \beta\\ \end{pmatrix}=0 $$ and thus, $\alpha + \beta$ is the root of the monic polynomial with coefficients in $\mathbb{Z}$: $$ p(x)=\det\begin{pmatrix} x & -1 & -1 & 0\\ -2 & x & 0 & -1\\ -3 & 0 & x & -1\\ 0 & -3 & -2 & x\\ \end{pmatrix} $$

Related Question