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?
Best Answer
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.