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Let $\alpha=\sqrt3+\sqrt5$, I want to calculate the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

The Idea seems to be to determine the smallest $n \in \mathbb{N}$ such that $\{1,\alpha,...,\alpha^{n-1},\alpha^n \}$ is linearly dependent over $\mathbb{Q}$.

So lets try it for $\alpha$ as above.

$\alpha^0=1$

$\alpha^1=\sqrt3+\sqrt5$

$\alpha^2=8+2 \sqrt15$

$\alpha^3=18\sqrt3 +14 \sqrt5$

$\alpha^4=124+32\sqrt15$

In that case $n=3$. My questions is the following: How do I know how many powers I should calculate? Do I have to check for every power and solve a linear system? Or is it possible to "see" when to stop?

Edit: Dietrich Burde did suggest a method to finding the minimal polynomial. A candidate for the minimal polynomial seems to be $x^4-16x^2+4$. To confirm this we need to show that $x^4-16x^2+4$ is irreducible.

I tried Eisenstein but it didn't work.

That's were I am stuck at the moment

Sigi
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    No need to compute the powers. Write $\sqrt{5}=\alpha-\sqrt{3}$ and square. Then rearrange and square again. Then the minimal polynomial over $\mathbb Q$ is $x^4-16x^2+4$. This gives you $n=3$ for your powers. – Dietrich Burde Jun 19 '23 at 18:48
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    @DietrichBurde It's clear that this polynomial is a multiple of the minimal of $\alpha$, so we have an upper bound on $n$. I think it's important to answer this: how can we be sure that there is no smaller polynomial that will have $\alpha$ as a zero? Is there something about your process that ensures that this polynomial is irreducible? – Ben Grossmann Jun 19 '23 at 18:56
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    @BenGrossmann One has to calculate the quadratic factors, see the duplicate, namely Lubin's answer. I don't think that one can skip this step (but it is well known in this case). – Dietrich Burde Jun 19 '23 at 19:04

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