Proving that Algebraic Numbers Form a Field

as mentioned in this article,  I will prove that $Q[\alpha]$ is a field. The other axioms of fields are rather trivial here, so I will just do the following two.

Suppose there exists $\alpha$ such that $P(\alpha)=0$ for some polynomial $P$ of degree $n, \ n >0$. Define $Q[\alpha]=${ $\sum^{n-1}_{i=0} k_{i} \alpha^{i}$ | $k_{i} \in Q$}.

    1. If $x \in Q[\alpha]$ and $y \in Q[\alpha]$, then $xy \in Q[\alpha]$

       

In the following, I use $C$ to denote any rational number for convenience. They are different numbers, but the numeric values are irrelevant to the discussion.

Let $n=k+1$. 

$P(\alpha)=\sum^{k+1}_{i=0} C \alpha^{i}=0 \iff \alpha^{k+1}=\sum^{k}_{i=0} C \alpha^{i}$ 

For $x, y \in Q[\alpha]$, let $x =\sum^{k}_{i=0} C \alpha^{i},\ y=\sum^{k}_{i=0} C \alpha^{i}$. We need to show that $xy \in Q[\alpha]$.

$xy=(\sum^{k}_{i=0} C \alpha^{i}) ( \sum^{k}_{i=0} C \alpha^{i})=\sum^{2k}_{i=0} C \alpha^{i}$

$=\sum^{2k}_{i=k+1} C \alpha^{i}+\sum^{k}_{i=0} C \alpha^{i}$

Since $\sum^{k}_{i=0} C \alpha^{i}$ is already in $Q[\alpha]$, we only need to consider $\sum^{2k}_{i=k+1} C \alpha^{i}$ to determine if $xy$ is in $Q[\alpha]$. $\sum^{2k}_{i=k+1} C \alpha^{i}=\alpha^{k+1} \sum^{k-1}_{i=0} C \alpha^{i}= (\sum^{k}_{i=0} C \alpha^{i}) (\sum^{k-1}_{i=0} C \alpha^{i})=\sum^{2k-1}_{i=0} C \alpha^{i}$

$=\sum^{2k-1}_{i=k+1} C \alpha^{i}+\sum^{k}_{i=0} C \alpha^{i}$

Substitude back in the equation, we have 

$xy=(\sum^{2k-1}_{i=k+1} C \alpha^{i}+\sum^{k}_{i=0} C \alpha^{i})+\sum^{k}_{i=0} C \alpha^{i}=\sum^{2k-1}_{i=k+1} C \alpha^{i}+\sum^{k}_{i=0} C \alpha^{i}$

Notice that we reduce the term $\sum^{2k}_{i=k+1} C \alpha^{i}$ to $\sum^{2k-1}_{i=k+1} C \alpha^{i}$. Repeat the same process $k-1$ times, $\sum^{2k}_{i=k+1} C \alpha^{i}$ will be reduce to $\sum^{k+1}_{i=k+1} C \alpha^{i}=C \alpha^{k+1}=\sum^{k}_{i=0} C \alpha^{i}$

Therefore, $xy=\sum^{k}_{i=0} C \alpha^{i}+\sum^{k}_{i=0} C \alpha^{i}=\sum^{k}_{i=0} C \alpha^{i} \in Q[\alpha]$

    2. If $x \in Q[\alpha]$ and $x \neq 0$, then $\frac{1}{x}\in Q[\alpha]$

        

Let $n=k+1$

$P(\alpha)=\sum^{k+1}_{i=0} C_{i} \alpha^{i}=0 \iff \alpha^{k+1}=\sum^{k}_{i=0} C_{i} \alpha^{i}$

Let $x =\sum_{i=0}^{j} C \alpha^{i}$, $0\leq j \leq k$, so that $x \in Q[\alpha]$.

If $j=0$, then $x=C$, $\frac{1}{x}=\frac{1}{C} \in Q \subset Q[\alpha]$

If $j>0$, then 

If $x$ divides $P(\alpha)$, then it's equivalent that $P(\alpha)=x(\sum^{k+1-j}_{i=0}C_{i} \alpha^{i})=0$. Since $x \neq 0$, we have $\sum^{k+1-j}_{i=0}C_{i} \alpha^{i}=0$.

If $k+1-j=1$, then we have $C_1\alpha+C_2=0$, $\alpha=-\frac{C_1}{C_2}$, $x \in Q$, and $\frac{1}{x}\in Q \subset Q[\alpha]$.

If $k+1-j>1$, then we have a new $P(\alpha)=\sum^{k+1-j}_{i=0}C_{i} \alpha^{i}=0$ of degree $n=k+1-j$. Repeat the previous process- divide $P(\alpha)$ by $x$, until $k+1-j=1$ which gives $\frac{1}{x}\in Q \subset Q[\alpha]$, or until $P(\alpha)$ can't be divided by $x$, which we should refer to the next section.

If $x$ doesn't divide $P(\alpha)$, Let $\frac{P(\alpha)}{x}=Q+\frac{R}{x}$, $Q \in Q[\alpha]$ and $R$ is the remainder with degree $r \leq n$.

Since $P(\alpha)=0$ and $x\neq 0$, $\frac{P(\alpha)}{x}=0$; 

$Q+\frac{R}{x}=0$, $\frac{1}{x}= \frac{Q}{-R}$

Supose r=0, then $R \in Q$, $R\neq0$.