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Find the degree of $\mathbb Q(\sqrt 2)$ over $\mathbb Q$

We have to find an irreducible polynomial $p(x)$ of $\mathbb Q[x]$ such that $p(\sqrt 2)=0$ and the degree of this polynomial is the degree of $\mathbb Q(\sqrt 2)$ over $\mathbb Q$.

The problem we can find more than one polynomial with these properties, for example $p(x)=x^2-2$ and $q(x)=x^4-4$.

Which polynomial I have to choose and why? Sorry I'm a really beginner, I need help. thanks

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up vote 3 down vote accepted

$x^4-4=(x^2-2)(x^2+2)$ is not irreducible. The minimal polynomial is $x^2-2$ so the degree is 2. The minimal polynomial is unique up to a constant factor.

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@user42912 Note that the minimal polynomial is always irreducible, so there is only one choice, namley $p(x)=x^2-2$. The way to calculate minimal polynomial is: Say $\alpha=\sqrt{2}.$ Then $\alpha^2=2$. This means $\alpha^2-2=0$. Hence $\alpha$ is a root of $f(x)=x^2-2$. We also note that $f$ is Eisenstein at $p=2$, hence irreducible over $\mathbb Q$. This means $f$ is the minimal poynomial.In this way you always come to minimal polynomial. – Reader Oct 31 '12 at 15:23

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