I would like to know if there is an elementary proof (without Galois theory, i.e. using the fact that conjugates are images by the base field automorphisms) of the fact that the conjugates of a sum of two algebraic numbers $x+y$ are the $x'+y'$ with $x'$ conjugate to $x$ and $y'$ conjugate to $y$. (If need be, restricting to $\mathbb{Q}$ is perfectly fine)
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In the absence of Galois theory, conjugate being defined as "root of the same irreducible polynomial"? – Hagen von Eitzen Mar 19 '20 at 07:03
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@HagenvonEitzen Yes! – Amomentum Mar 19 '20 at 07:09
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If you’re familiar with resultants, following link indirectly provides you with an answer: https://math.stackexchange.com/questions/155122/how-to-prove-that-the-sum-and-product-of-two-algebraic-numbers-is-algebraic – mathcounterexamples.net Mar 19 '20 at 07:24
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Let $f,g$ be polynomials $\in K[X]$ of degrees $n,m$ and let $x_1,\ldots,x_n$ and $y_1,\ldots y_m$ be their roots. Let $h\in K[X,Y]$ and $z_{i,j}=h(x_i,y_j)$. Thne the elementary symmetric polynomials in the $nm$ numbers $z_{i,j}$ are symmetric polynomials in the $x_i$ with coefficients from $K[y_1,\ldots, y_m]$ and these coefficient-polynomials themselves symmetric in the $y_j$. Hence the coefficient-polynomials can be expressed in therms of the elementary symmetric polynomials in the $y_j$, i.e., in terms of the coefficients of $g$. But then the $z_{i,j}$ are symmetric polynomials in the $x_i$ with coefficients in $K$, which means they can be expressed in terms of the coefficients of $f$, so are ultimately $\in K$. Ultimately, this gives us the coefficients $\in K$ for a polynomial with all $h(x_i,y_j)$ as its roots. The irreducible factor of that polynomial that happens to have $h(x_1,y_1)$ among its roots, say, has all its other roots also of the form $h(x_i,y_j)$, as was to be shown (in your specific question, for $h(X,Y)=X+Y$).
Hagen von Eitzen
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