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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?

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3 Answers

up vote 9 down vote accepted

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.

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Ah, I see. I was missing the fact that $[F(a,b):F(a)]$ is finite hence $[F(a,b):F]$ is. I'm not sure why that didn't occur to me. –  Alex Becker Mar 15 '13 at 8:06
    
"we could build a spanning set for the (x+y)-space out of the spanning sets for the x-space and the y-space." How? –  Martin Brandenburg Mar 16 '13 at 20:14
    
In my opinion this standard proof is not constructive at all. It does not help us to find explicitly an algebraic equation for $a+b$ when we have found one for $a$ and $b$. It only shows the existence. –  Martin Brandenburg Mar 16 '13 at 20:17
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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.

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+1. This proof is constructive and also works for ring extensions, it shows that the set of integral elements is a subring. –  Martin Brandenburg Mar 16 '13 at 20:15
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I find another charming proof in Lecture 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 with roots $\alpha$ and $\beta$ respectively. Let $\{\alpha_i|i=0,\ldots,m\}$ and $\{\beta_j|j=0,\ldots,n\}$ 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 kow 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.

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If we replace $\mathbb{Q}$ by an arbitrary field, this proof only works for separable $\alpha,\beta$, right? Or can it be modified? –  Martin Brandenburg Mar 16 '13 at 20:13
    
@MartinBrandenburg I cannot see the reason it only works for separable numbers: it does not use Galois theory at all; rather, it makes use of the fundamental theorem of symmetric functions. Per chance this works only for separable extensions? In any case, thanks for your attention. –  awllower Mar 17 '13 at 5:35
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