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

  • $\begingroup$ So what you want is to find, if possible, $\beta \in F(\alpha)$ with $\beta^2 = \alpha$? If so, you might consider editing your title. Cheers? $\endgroup$ – Robert Lewis May 15 '15 at 20:07

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.

  • $\begingroup$ I still don't understand this question. I'm reviewing for finals and it makes like no sense to me. What exactly is this title saying? $\endgroup$ – user121615 May 24 '15 at 20:24
  • $\begingroup$ @user121615 The title doesn't make sense. See Robert Lewis's comment. $\endgroup$ – Slade May 24 '15 at 23:11

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