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I have to find the minimal polynomial of $\sqrt{5}+i$ over the field $\mathbb{Q}$ and to verify it has integers as coefficients.

I tried:
$$z=\sqrt{5}+i\;\Rightarrow\; z-i=\sqrt{5}\;\Rightarrow\; z^{2}-2iz-1=5$$ So I found a polynomial, $h(x)=x^{2}-2ix-6$, which has $z$ has a root and I know $f(x)|\;h(x)$ where $f(x)$ is the minimal polynomial of $z$ over $\mathbb{Q}$. I checked the possible rational roots of $h(x)$ with the formula: $R_{r}=\{\frac{\alpha}{\beta}:\alpha|(-6),\beta|1\}$ but it has none. My definition of the minimal polynomial of an element over a field is: let $F\subseteq K$ be fields and $\alpha \in K$ be algebric over $K$; the m.p. of $\alpha$ over $K$ is A (it follows is unique) monic generator of $ker(\psi_{z})$, where $\psi_{z}:K[x]\rightarrow K[z]:p(x)\mapsto p(z)$ is the evaluation function on $z$. This has no operative meaning so I read that if someone finds a $(i)$ monic $(ii)$ irreducible polynomial $\mu(x)$ in $K[x]$ it has to be the m.p.. Still I don't know how to continue.

Math Attack
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    Does this answer your question? Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ – Anne Bauval May 07 '23 at 06:26
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    Your polynomial $z^2-2iz-1$ is not over $\Bbb Q$. It has an $i$ in the linear coefficient, so it is over $\Bbb Q(i)$. – Arthur May 07 '23 at 06:26
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    You need to continue: $z^2-6=2iz$ so $(z^2-6)^2=-4z^2$ i.e. $z^4-8z^2+36=0$. Note the zeros of $x^4-8x^2+36$ are $\pm\sqrt{5}\pm i$ which can help you prove that it is irreducible. –  May 07 '23 at 06:39
  • In fact, I write $f(x)|h(x)=x^2-2ix-6$ where $f(x)$ is the minimal polynomial of $z=(\sqrt{5}+i)$. I think if we call $z=(\sqrt{5}+1)$ by "notation issues" $z^{2}-2iz-1$ is not even a polynomial in this context but a numerical value. –  May 07 '23 at 06:42
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    @matteoC1729 Sorry, but that last comment makes no sense. You cannot write $f(x) \mid h(x)$ because you have no basis to claim that the minimal polynomial over $\mathbb Q$ divides the minimal polynomial over $\mathbb Q(i)$. – dxiv May 07 '23 at 06:45
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    You're right in saying that the minimal polynomial over $\mathbb{Q}$ does not divide the minimal polynomial over $\mathbb{Q}(i)$, and I understand now the "algebraic process" has to be completed before bringing in divisibilty. Thanks to "Stinking bishop" it was possibile for me to calculate $\mu_z^{\mathbb{Q}}(x)$. What i tried to say is that it was clear to me that $h(x)=x^{2}-2ix-6$ was not in $\mathbb{Q}[x]$ but in $\mathbb{Q}(i)[x]$, because it is the reason of the question in the first place. It war also strange to see $z$ in the place of $x$ as the book was clear about this. –  May 07 '23 at 07:03

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I am unsure about my reasoning, but I believe that because $$\left[\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}(i)\right] = 2$$ and $$\left[\mathbb{Q}(i):\mathbb{Q}\right] = 2$$ (by degree considerations), we can apply the 'Tower Law' for field extensions, which states that $$[F:K]=[F:L][L:K].$$

As a result, the field extension degree of $$\left[\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}\right] = 4.$$ Consequently, the minimal polynomial should have a degree of at least 4.

To determine this minimal polynomial, I propose the polynomial $$x^4 - 8x^2 + 36.$$ and because of $i + \sqrt{5}$ is a solution to this polynomial, so it should be the minimal polynomial.

Gor Melkumyan
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