TITLE: How to prove that algebraic numbers form a field?
QUESTION [12 upvotes]: I'd like to know how to prove that algebraic numbers form a field by using Kronecker Product, but not sure exactly how to do it.
Edit: This question is different from the suggested duplicated one in that this question asks for an answer to prove that algebraic numbers form a field in unconventional ways such as using Kronecker Product. As the result, answers are very different from the one suggested duplicated. So this question deserves to remain open.

REPLY [2 votes]: Here is yet another proof of this well-known fact. This will be a rather elementary proof based on linear algebra.
First a little Lemma
Suppose $V\subset\mathbb{C}$ forms a finite dimensional vector space over $\mathbb{Q}$. If $a\in\mathbb{C}$ satisfies
$$ aV \subset V,$$
then $a$ is algebraic.
Let $\{e_1,\dots,e_n\}$ be a basis for $V$. Then there is an $n\times n$ matrix  $A=(\alpha_{ij})$ in $\mathbb{Q}$ such that
$$
\begin{align}
ae_1&= \alpha_{11}e_1 +\ldots +\alpha_{1n}e_n\\
\ldots&\ldots\\
ae_n&= \alpha_{n1}e_1 +\ldots + \alpha_{nn}e_n
\end{align}
$$
This means that $a$ is a solution to the equation
$$ p(z) = \operatorname{det}(A-zI) =0.$$
Since $p$ is a rational polynomial (i.e., $p$ has rational coefficients), $a$ is algebraic.

With this result at hand, we now prove that the set of algebraic numbers  is a subfield of $\mathbb{Q}$.
Suppose $a$ and $b$ are algebraic. Let $m$ and $n$ the smallest positive integers for which rational polynomials $p$ and $q$ of degree $m$ and $n$ exist such that $p(a)=0=q(b)$. Let $V$ the linear space over $\mathbb{Q}$ spanned by $\{a^jb^k: O\leq j < m, 0\leq k< n\}$.
It is easy to check that
$$ aV\subset V,\qquad bV\subset V $$
Consequently, $(a+b)V\subset V$, and $(ab)V\subset V$. From our little Lemma, we conclude that $ a+b$ as well as $ab$ are algebraic.
To conclude, notice that if $c\neq0$ solves the polynomial equation $p(x)=0$, then $-c$ solves $r(x)=p(-x)=0$ and $1/c$ solves $q(x)x^np(1/x)=0$ with $n$ being the degree of $p$. Clearly, if $p$ is a rational polynomial, so are $r$ and $q$.
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