Let E be an extension of a field F of degree 2. Show there is an element with $\beta = x^2 - \alpha$.
I wanna know if my approach is correct.
If it's an extension of degree 2, then $F(\alpha)$ for our extension has a degree of 2, thus it has a basis than has 2 elements: $[1,\alpha]$ And any element can be written as a linear combo of these 2 guys. In math language, that be:
$\omega = a + b\alpha$
What we want to show is that there exists an element of this extension we shall calls it $y$ such that $irr(y,F) = x^2 - \alpha$
But $irr(y,F)$ is the monic polynomial with y as it's root in and it's coefficients in our field F.
Is it correct to say, because any element $\beta$ can be written as:
$\beta = a + b\beta$
that if we let that $b$ equal $\beta$ and we let that $a$ equal $-\alpha$ That we have found our element?
And also, this is alledgedly not supposed to work for extension 2. What's an example of a field this might not work for?
Thanks a lot yall
Start with the minimal polynomial of $\alpha$, which has degree $2$, and consider the quadratic formula. How can you modify $\alpha$ so that it is a square root of an element of $F$?
This approach does not work in a field where $1+1=0$, since the quadratic formula requires division by $2$.
For a counterexample, consider the field $\mathbb{F}_2$. It already has all its square roots, so the extension $\mathbb{F}_2\subset\mathbb{F}_4$ cannot contain any new ones.