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I was able to show that $x^4-4x^2+16$ is the minimal polynomial of $\sqrt3 + i$ over $\mathbb{Q}$.

Does this imply that this is also the mininal polynomial over $\mathbb{R}$ and $\mathbb{C}$?

3 Answers 3

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No. Hint: Consider $(x-\sqrt{3}-i)(x-\sqrt{3}+i)$ over $\mathbb{R}$. For over $\mathbb{C}$, it will have smaller degree.

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    Then, over $\mathbb{C}$ it should be $x + (-\sqrt3-i)$, right? – 2017-01-17
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    @Gorid Yes. Over $\mathbb{C}$ all minimal polynomials are degree one. – 2017-01-17
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    So $(1 / (2*\sqrt3))*x^2 - x + (-(1/\sqrt3)+\sqrt3)$ should then be the minimal polynomial over $\mathbb{R}$, right? – 2017-01-17
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    @Gorid, your polynomial should have a leading coefficient 1. – 2017-01-17
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    Right thank you, I've forgot about that. So it should be $x^2-(2*\sqrt3)x+4$ then, right? – 2017-01-17
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    yes. that looks to be correct. – 2017-01-17
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No. Every polynomyal in $\mathbb R[X]$ can be factored in linear factors and quadratic factors depending on how many roots are real and how many are not, thus the minimum polynomial has degree at most $2$. In this case it is of degree 2, ie $(x-\alpha)(x-\overline{\alpha})$. Note that when there is a real solution $\beta$ you can factor out $(x-\beta)$ and repeat, when there is a not real root $\gamma$ ie $p(\gamma)=0$ you can see that $p(\overline{\gamma})=\overline{p(\gamma)}=0$ also, thus you can factor out $(x-\gamma)(x-\overline{\gamma})$ which is of course in $\mathbb R[X]$.

Every polynomial in $\mathbb C[X]$ can be factored in linear factors because every polinomyal of degree at least $1$ has a solution via the Fundamental Algebra Theorem. Thus the minimum polynomial has degree $1$, ie $(x- \alpha) \in \mathbb C[X]$.

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let $x=\sqrt3+i\Rightarrow x-\sqrt3=i\Rightarrow (x-\sqrt3)^2=-1\Rightarrow x^2-2\sqrt3 x+4=0$ then minimal polynomial of $\sqrt3+i$ over $\mathbb R[x]$ is $x^2-2\sqrt3 x+4=0$

While $\sqrt3+i \in \mathbb C$ then minimal polynomial of $\sqrt3+i$ over $\mathbb C[x]$ is $x-(\sqrt3 +i)=0$