# The degree of an algebraic element over a field extension

Let $L/K$ be a field extension and let $\alpha$ be an algebraic element of prime degree over $K$, i.e $[K(\alpha) : K] = p$ for some prime $p$. Is it always the case that we have $[L(\alpha) : L] = 1$ or $p$?

If $[L : K]$ is not divisible by $p$, we have that $[L(\alpha) : K] = [L(\alpha) : L] [L : K]$ is divisible by $p$ since it contains the subfield $K(\alpha)$, which implies $[L(\alpha) : L] = p$. Can we also obtain this result when $[L : K]$ is divisible by $p$, or is there a counterexample?

No. Consider the polynomial $f=x^3-2$ over the rationals. Each root generates a degree 3 extension, so let's take $L=\mathbb{Q}(2^{1/3})$. If we adjoin another root, like $2^{1/3}\zeta_3$, we get a degree 2, not 3, extension over $L$.
• What if we also require that the normal closure of $L$ does not contain $\alpha$? Commented May 30, 2016 at 23:19
It is true if $\alpha$ is not contained in the normal closure of $L$. In this case we can assume that $L/K$ was normal to start with and $\alpha$ is not contained in $L$. Using the statement from this question - $f$ is factored into many same degree irreducible polynomials. - we know that the minimal polynomial of $\alpha$ over $K$ splits over $L$ into factors of the same degree. By the degree being prime, we deduce that the factors are either linear factors or there is only one factor. The first case is ruled out by the assumption that $\alpha \notin L$. This shows that the minimal polynomial over $L$ is still the same as the minimal polynomial over $K$, hence of degree $p$.