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The minimal polynomial of a root is dependent on the field it lies in. Correct?

I was thinking that to find the minimal polynomial of $\sqrt{2}$ over $\Bbb Q$ meant to do the following:

$$\alpha =\sqrt2\implies \alpha^2=2\implies \alpha^2-2=0$$

So we get the minimal polynomial of $\sqrt{2}$ is $f(x)=x^2-2$ over $\Bbb Q[x]$.

But if we were looking for the minimal polynomial of $\sqrt{2}$ over $\Bbb Q(\sqrt{2})$ we would just take $\alpha=\sqrt{2}\implies \alpha-\sqrt{2}=0$ and we have the minimal polynomial $f(x)=x-\sqrt{2}$ over $\Bbb Q(\sqrt{2})$.

Is this correct?

(see my self answer below that occurred after asking this)

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  • $\begingroup$ Did you understand my answer? $\endgroup$ – user21820 Dec 13 '15 at 4:02
  • $\begingroup$ @user21820 Not yet, but I will read over it again soon, thanks for your answer. $\endgroup$ – Oceans Bleed Dec 13 '15 at 6:37
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    $\begingroup$ Okay sure, feel free to ask for clarification on any point! $\endgroup$ – user21820 Dec 13 '15 at 6:46
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To see that the second paragraph of your answer is false, here is one counter-example. $\def\ff{\mathbb{F}}$

Let $f(x) = x^3+x+1$ for any $x \in \ff_2$. Let $r$ be a root of $f$ in some extension of $\ff_2$. Then you can verify that the roots of $f$ in $\ff_2(r)$ are $r,r^2,r^2+r$. So $f(x) = (x+r)(x+r^2)(x+r^2+1)$ for any $x \in \ff_2(r)$. But the minimal polynomial of $r^2$ over $\ff_2(r)$ is just $x \mapsto x + r^2$ since $r^2 \in \ff_2(r)$, and not $x \mapsto (x+r^2)(x+r^2+1)$.

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Straight after asking this, it made sense:

$x^2-2$ is irreducible over $\Bbb Q$ and has roots $\pm\sqrt{2}$, but over $\Bbb Q(\sqrt{2})$ it is not irreducible, and in fact splits as $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ so the minimal polynomial is the linear polynomial $x-\sqrt{2}$.

This will be the case in general, since if $\alpha$ is a root, we form the minimal polynomial over $K$ and obtain some $f(x)$. If $L/K$ is an algebraic field extension with $L=K(\beta)$ and $\beta$ is also a root of $f(x)$, we get $f(x)=(x-\beta)^kg(x)$ over $L$, and then the minimal polynomial of $\alpha$ over $L$ must be $g(x)$.

(If $K$ is of characteristic $0$, we have $k=1$.)

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  • $\begingroup$ Your first paragraph is correct. That is why in textbooks that are precise, they use notation that specifies the base field, such as $m_F(r)$ for the minimal polynomial of $r$ over $F$, for any $r$ in a field extension of $F$. (Of course, one has to first prove that it is unique regardless of the field extension.) But your second paragraph doesn't make sense and is actually false. $\endgroup$ – user21820 Dec 12 '15 at 8:52

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