2
$\begingroup$

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

$\endgroup$
5
$\begingroup$

$x^4-4=(x^2-2)(x^2+2)$ 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.

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – Reader Oct 31 '12 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.