Enlightening proof that the algebraic numbers form a field

Question

The proof I'm familiar with that the algebraic numbers $\mathbb A$ form a field uses the fact that the resultant of two polynomials $p,q\in\mathbb Q[x]$ satisfies the following properties:

• It is $0$ iff $p$ and $q$ have a common factor.
• It is a polynomial in the coefficients of $p$ and $q$.

We then introduce a new variable and cleverly manipulate $p$ and $q$ to get polynomials which vanish at the sums and products of their roots. This is in some ways a nice proof, e.g. it is constructive and so can be converted into an algorithm to find such polynomials (which I in fact just finished doing in C). But I don't find it very enlightening; it seems like the fact that $\mathbb A$ is a field is simply an accident. Is there a more enlightening proof of this fact?

Answer 1

My favourite proof of this goes through matrices. A complex number is an eigenvalue of a square matrix of rational numbers if and only if it is algebraic (e.g. any monic polynomial has a companion matrix of which it is the characteristic polynomial).
If $A$ is an invertible matrix with eigenvalue $\alpha$, then $A^{-1}$ has eigenvalue $1/\alpha$. If $A$ and $B$ are square matrices with rational entries and eigenvalues $\alpha$ and $\beta$ for eigenvectors $u$ and $v$ respectively, then $A \otimes B$ and $A \otimes I + I \otimes B$ have eigenvalues $\alpha \beta$ and $\alpha + \beta$ for eigenvector $u \otimes v$.

This also shows, BTW, that the algebraic integers are closed under addition and multiplication: they are the eigenvalues of matrices with integer entries.

Answer 2

If $a,b$ are algebraic then $[F(a,b):F]$ is finite so $a+b, ab \in F(a,b)$ are algebraic too. I find this proof to feel the most enlightening.

If you specifically wanted to see why the constructive proof is not merely an accident, then see this link- after proving that the sum of algebraic numbers is algebraic in the usual way, they add:

Now let us analyse the above argument a little. We see that what made it work was that the vector space of all rational combinations of powers of x+y was finite-dimensional. We then observe that we deduced this from the fact that the corresponding vector spaces for powers of x and powers of y were also finite-dimensional - and also that we could build a spanning set for the (x+y)-space out of the spanning sets for the x-space and the y-space.

Answer 3

I find another charming proof in Lectures on the theory of algebraic numbers by Hecke.

It is constructive in appearance but is impractical in fact.
Let $p(x)$ and $q(x)$ be two polynomials with rational coefficients and roots $\alpha$ and $\beta$, respectively. Let $\{\alpha_i\mid i=0,\ldots,m\}$ and $\{\beta_j\mid j=0,\ldots,n\}$ be their conjugates, i.e. roots of $p$ and $q$ respectively. Then set $$r(x)=\prod_{i=0}^m\prod_{j=0}^n(x-\alpha_i-\beta_j).$$ We know that $\alpha+\beta$ is a root of $r(x)$ and that coefficients of $r(x)$ are symmetric functions of roots of $p$ and $q$, and hence could be written as polynomials of coefficients of the two polynomials, thus is rational. Therefore $\alpha+\beta$ is also algebraic. A similar proof goes for products. Finally, write $$p(x)=\sum_0^ma_ix^i.$$Then $$\sum_0^ma_i\alpha^i=0,$$ so that $$\sum_0^ma_{m-i}\alpha^{-i}=0,$$ and hence $\alpha^{-1}$ also is algebraic.
Inform me of any ambiguity or error, thanks.

Answer 4

I am surprised that no one has suggested the following approach, which circumvents the verification of field axioms (more precisely, the job is delegated to the proof of some other proposition):
It suffices to show

Claim. Let $K$ be an extension field of a field $F$ and let $E$ be the set of all elements in $K$ that are algebraic over $F$. Then
$E$=$F(E)$.

Proof. By definition $E\subset F(E)$. Because every element of $E$ is algebraic over $F$, $F(E)$ is an algebraic extension of $F$. Therefore, $F(E)\subset E$ by the definition of $E$. $\square$