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I know that the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ is indeed $x^4-10x^2+1$ (as shown multiple times already) because of the fact that it is (1) monic and (2) irreducible in $\mathbb Q$.

Now i am asked to find the minimal polynomials for $\sqrt{2}+\sqrt{3}$ over $\mathbb R$ and $\mathbb C$ aswell.

I used the same technique as for $\mathbb Q$ but halted as soon as i reached the irreducible polynomial, which in this case is $p(x)=x-\sqrt{2}-\sqrt{3}$.

The polynomial $p(x)$ is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb R$ and $\mathbb C$. It should be trivial for $\mathbb R$ and $\mathbb C$ because $p(x)$ is indeed monic and irreducible by the definition of $\mathbb R$ and $\mathbb C$ (since all polynomials of $deg(1)$ are irreducible).

Is my thought process correct?

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    $\begingroup$ Yes. It is also trivial because for any field $F$ and any $\alpha\in F$ the minimal polynomial of $\alpha$ is $X - \alpha.$ $\endgroup$ – DeuzharNickens Jan 10 at 19:26
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    $\begingroup$ We may notice that $mathbb{C}$ is algebraically closed, so any polynomial splits in linear factors, so the minimal polynomial of any $\alpha\in\mathbb{C}$ is linear, i.e. is $x-\alpha$ $\endgroup$ – Qurultay Jan 10 at 19:29
  • $\begingroup$ @DietrichBurde Oh sure, it is just that i am quite new to field theory and the concept of minimal polynomials so i wanted to know if my idea is correct. It looked strange to me at first because $\sqrt{2}+\sqrt{3}$ is in $\mathbb R$ and from there on out we wanted to find a minimal polynomial in a class "above" so to say. But since $\mathbb R$ is in $\mathbb C$ i think my concerns were unnecessary. Thank you for all the answers! $\endgroup$ – null1 Jan 10 at 19:48
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You rightly noticed that $\sqrt2+\sqrt3\in\Bbb R$ (aswell as $\sqrt2+\sqrt3\in\Bbb C$). And, indeed, for every $\alpha\in L$, where $L$ is a field, the minimal polynomial of $\alpha$ over $L$ is just $f(x)=x-\alpha\in L[x]$. This is both (1) monic and (2) irreducible as you showed by yourself.

To answer your question directly: yes, your thought process is correct.

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  • $\begingroup$ A follow up question: Does that mean that technically speaking $p(x) = x - (\sqrt{2} + \sqrt{3})$ is also a minimal polynomial in $\mathbb Q$ but since $\sqrt{3}$ and $\sqrt{2}$ are not part of $\mathbb Q$ we first have to remove the square-roots? And would that imply since $p(x) = x - (\sqrt{2} + \sqrt{3})$ is a sort of minimal polynomial in $\mathbb Q$, we can be sure that $x^4−10x^2+1$ is a minimal polynomial in $\mathbb Q$? $\endgroup$ – null1 Jan 11 at 9:10
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    $\begingroup$ Yes, technically speaking $p(x)=x-(\sqrt2+\sqrt3)$ is a minimal polynomial, but not in $\Bbb Q[x]$ as the coefficients are not in $\Bbb Q$. That's why we have to remove the square roots by squaring until we obtain something in $\Bbb Q[x]$. Then, if you finnd a monic and irreducible polynomial with root $\sqrt2+\sqrt3$ as minimal polynomials are unique. $\endgroup$ – mrtaurho Jan 11 at 10:13
  • $\begingroup$ Okay, i understand now, thank you very much! $\endgroup$ – null1 Jan 11 at 10:18
  • $\begingroup$ @null1 Glad to help! :) $\endgroup$ – mrtaurho Jan 11 at 10:24

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