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Show that the minimal polynomial of $\alpha =\sqrt{2} - i$ over $\mathbb{Q}(i)$ is $x^2 -2ix -3$.

Is this similar to showing that the minimal polynomial over $\mathbb{Q}$, because I know the minimal polynomial of $\alpha =\sqrt{2} - i$ over $\mathbb{Q}$ is $x^4-2x^2+9$.

I am confused on what $\mathbb{Q}(i)$ means.

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$K = \mathbb{Q}(i)$ is the subfield of $\mathbb{C}$ consisting of the elements $a + bi$ with $a,b \in \mathbb{Q}$. It is not difficult to see that it is in fact a field.

You may compute the minimal polynomial of $\alpha$ over $K$ by writing $\alpha = x$ and manipulating both sides in a such a way that you end up only with elements in $K$. Namely

\begin{align} \sqrt2 - i = x \Leftrightarrow \sqrt2 = x + i \end{align} and after squaring both sides to get rid of the root, we get $2 = x^2 + 2ix -1$, so $x^2+2ix-3 = 0$ if $x = \alpha$. The minimal polynomial of $\alpha$ must have degree at least two, since $\alpha\notin K$, thus this is in deed its minimal polynomial over $K$.

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$\mathbb{Q}(i)=\{a+bi\,\vert\,a,b\in\mathbb{Q}\}$

So the minimum polynomial of $\alpha=\sqrt{2}-i$ is as you said because all the coefficients are in $\mathbb{Q}(i)$ and the minimum polynomial cannot be linear since $\alpha\not\in\mathbb{Q}(i)$.

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