# If coefficients are algebraic then the roots are algebraic

I would appreciate some help on this problem: If $\alpha$ and $\beta$ are algebraic integers, prove that any solution of $x^2+\alpha x+ \beta=0$ is an algebraic integer. I think some manipulations with symmetric polynomials are necessary but I did not make it work.

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Integrality is a transitive property: If there is a ring inclusion $R \subset S \subset T$, where $S$ is integral over $R$ and $T$ is integral over $S$, then $T$ is integral over $R$. In this case, consider the chain $\mathbb{Z} \subset \mathbb{Z}[\alpha, \beta] \subset \mathbb{Z}[\alpha, \beta, x]$.

If you want explicit equations (which seems possible from the way you phrase your question), consider $p(x), q(x) \in \mathbb{Z}[x]$ monic of degree $m$ and $n$ respectively such that $p(\alpha) = 0$, and $q(\beta) = 0$. Notice that $\mathbb{Z}[\alpha,\beta,x]$ has a $\mathbb{Z}$-basis $\{\alpha^r \beta^s x^t\}$ where $0 \leq r < m$, $0 \leq s < n$, $0 \leq t < 2$. Then a Cayley-Hamilton argument (similar to the proof of $x$ integral over $\mathbb{Z}$ iff $\mathbb{Z}[x]$ is a finite $\mathbb{Z}$-module) would give you an equation.

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That $\alpha$ and $\beta$ are algebraic, means that $\mathbb{Q}[\alpha, \beta]$ is finitely generated over $\mathbb{Q}$. That a solution $x$ of the polynomial $x^2+\alpha x + \beta=0$ is algebraic, means that $\mathbb{Q}[\alpha,\beta][x]$ is finitely generated over $\mathbb{Q}[\alpha, \beta]$.

So your problem is reduced to proving that if we have a tower of field extension $\mathbb{Q} \subset K \subset L$, with $K/\mathbb{Q}$ finitely generated and $L/K$ finitely generated, then $L/\mathbb{Q}$ is finitely generated. (hint to show this: write a $K$-basis for $L$ and a $\mathbb{Q}$-basis for $K$, and try to use this to write a $\mathbb{Q}$-basis for $L$).

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Here is an alternative explanation involving symmetric functions.

First let $P_\alpha(x)$ be the minimal polynomial of $\alpha$ over $\mathbb Q$, and let $\alpha, \alpha', \alpha'', \dots$ be it's roots.

The symmetric functions theorem tells us that any symmetric polynomial on the $\alpha$'s can be written as polynomial on the coefficients of $P_\alpha(x)$ and as a consequence will be a rational number.

Now consider the polynomial $$G(x,\beta) = (x^2+\alpha x + \beta)(x^2+\alpha' x + \beta)(x^2+\alpha'' x + \beta)\dots$$ (seen as a polynomial in $x$ and $\beta$). Clearly it's coefficients are symmetric polynomials in the $\alpha$'s and so are rational numbers.

Repeat the same argument with $G(x,\beta)$ instead of the quadratic polynomial and with the conjugates of $\beta$, and you'll obtain a polynomial with rational coefficients having $x^2+\alpha x + \beta$ as roots and you're done.

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