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So if $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}$ are algebraic numbers and their minimal polynomials are $f$ and $g$ in $\mathbb{Q}[x]$ respectively, then it is claimed that $h(z)=\text{Res}(f(x),g(z-x))$ is the minimal polynomial of $z=\alpha+\beta$. Now I understand why $\alpha+\beta$ is a root, but I don't understand why the resultant is irreducible. I know its an irreducible in the intermediates corresponding to the coefficients of $f$ and $g$ when they are arbitrary but I don't understand why we get irreducibility in the variable $z$, this was claimed but doesn't seem obvious or true in generality. Can someone clarify this for me please.

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    `it is claimed` What is the source, and exact wording, of that claim? – 2017-01-23
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    This is from lecture, and I couldn't find any written source on this either which is why I'm doubting the claim. More than likely what the professor meant was that this gives us a polynomial with $\alpha+\beta$ as a root, but we must bound the degree by the degree of the extension $\mathbb{Q}(\alpha+\beta)$ to claim that $h$ is irreducible. Am I correct in my reasoning in that the claim in my original post isn't necessarily true. – 2017-01-23
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    As @dxiv mentioned, this is not true in general. For example, let $\alpha = \sqrt{2}$ and $\beta = -\sqrt{2}$. Perhaps a way to fix the claim would be to require $\mathbf Q(\alpha)$ and $\mathbf Q(\beta)$ linearly disjoint over $\mathbf Q$. Note that, in this case, $$\mathbf Q(\alpha + \beta) = \mathbf Q(\alpha, \beta)$$ has degree $[\mathbf Q(\alpha) : \mathbf Q][\mathbf Q(\beta) : \mathbf Q]$ over $\mathbf Q$ and the resultant $h(z)$ has the same degree. – 2017-01-23
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    @Alex Thanks. That fixes the claim. I think that's what I might have missed from the lecture. – 2017-01-23

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