# Degree 3 extension does not admit “new squares”

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 $K$.

Thanks!

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 $K$.