I was reading another post on here (unfortunately, I couldn't comment on the post due to my low reputation), one poser wrote "that a degree 3 extension of a field does not admit new square elements". I am just confused at what this statement means and why this is.

To give contexts, I want to show that an irreducible polynomial $x^3 + px + q$ over a finite field $K$ characteristic not 2 or 3, then $-4p^3 - 27q^2$ is square in $K$. I know that $-4p^3 - 27q^2$ is the discriminant of the said polynomial and that the discriminant is a square in the splitting field, which has degree 3. Now, I just want to conclude that it is also a square in $K$.



Let $L/K$ be an extension of fields with finite odd degree $d$. I claim that any element of $K$ that's a square in $L$ is already a square in $K$.

If $a=b^2$ with $a\in K$ and $b\in L$ then $K\subseteq K(b)\subseteq L$ and so $d=|L:K|=|L:K(b)||K(b):K|$. As $d$ is odd, then $|K(b):K|$ is odd. As $b^2\in K$ then $|K(b):K|\le2$ and so $|K(b):K|=1$. Therefore $b\in K$, so that $a$ is already a square in $K$.

  • $\begingroup$ Thank you so much! This clears a lot of things up $\endgroup$ – useranon May 30 '18 at 19:36

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.