If you have algebraic numbers $x$ and $y$, and you know the polynomials of least degree of which each is a solution (written as a vectors of coefficients), then how is the vector of coefficients of the polynomial of least degree of which $x y$ is a solution found; and likewise for $x+y$?
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Absolutely not! I'm asking for a recipe for explicitly computing the vector of coefficients of the polynomial is a solution - not for a proof that it exists. – AmbretteOrrisey Dec 14 '18 at 09:05
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@Hans Lundmark -- However ... now that I look at the answers more carefully, I see that there is one that might possibly give the solution. So thankyou for signposting it for me. I'm not sure it does give the answer yet ... but it kindof looks like it might. – AmbretteOrrisey Dec 14 '18 at 09:07
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The accepted answer definitely says how to do it. (You have to look up how to compute resultants, and the formulas that you get aren't going to be simple, but that's just how it is.) – Hans Lundmark Dec 14 '18 at 09:12
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And anyway the question is a duplicate, since that other question also asks for an explicit formula for a polynomial having $xy$ or $x+y$ as a root. – Hans Lundmark Dec 14 '18 at 09:14
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@Hans Lundmark -- I do think I might be able to extract a solution (or maybe even more than one!) from it. It's extremely interesting post! But I could do with something abitt more explicit & specifically tuned to my question. Like ... what's this resultant? But at least that gives me a handle on it that I didn't previously have. – AmbretteOrrisey Dec 14 '18 at 09:16
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You can find a definition of the resultant function here which tells you how to compute it, but note that it gives you a polynomial which is a product of what you want, so it might not be the actual minimal polynomial. Say if $xy$ has degree lesser than the product of the individual degrees, or if $x,y$ are not algebraic integers. For example if $x=\sqrt{2}$ and $y=3\sqrt{2}$ you're going to get a linear polynomial. However if you expand out the functions it's actually an explicit function (of sums and products). – Yong Hao Ng Dec 14 '18 at 09:48
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@Yong Hao Ng -- Please kindly accept my thanks for that signposting. I've got a handle on the matter, now, certainly ... but I can see that it's not as simple as just having someone dispense me ___the answer___! But what you said about the method of 'resultants' yielding a linear polynomial in the case of x=√2 & y=√3 !!? That doesn't seem so good on the face of it! But the 'ball is in my court' now, as is said ... & I'm not expecting anyone to serve me a real-time coaching to the extent of ironing-out every wrinkle for me! – AmbretteOrrisey Dec 14 '18 at 16:09