Degree of $a+b$ over a field $k$, where $a$ and $b$ are distinct roots of the same polynomial

Question

Let $K/k$ be an algebraic extension of fields with $a$ and $b$ distinct roots in $K$ of the same irreducible polynomial $f(x) \in k[x]$ of degree $n$. Show that the degree of $k(a+b)/k$ is less than or equal to $\frac{n(n-1)}{2}$.

Also, how does one construct fields $k$ and $K$ together with roots $a,b\in K$ so that the preceding inequality is actually an equality?

I'm pretty sure I can get that $k(a+b)/k$ has degree less than or equal to $n(n-1)$ since the minimal polynomial of $b$ over $k(a)$ has degree less than or equal to $n-1$, but I'm not sure how to reduce this by a factor of $1/2$. I've also seen that there are computational techniques for computing the minimal polynomial of a sum, but a proof that avoids things such as resolvents would be ideal.

Answer 1

If $f$ is inseparable, then $k$ has prime characteristic $p$ and $f(X) = X^{p^l} - t$ for some $t\in k$ and $l\geq 0$. Then $a+b$ satisfies the polynomial $X^{p^l} - 2t$, which has degree $p^l=n$, which is $\leq {n \choose 2}$ when $n\geq 3$. $n=2$ is a special case, since $p=2$ and so $a+b=0$, while $n=1$ is impossible.

On the other hand, suppose that $f$ is separable. Then every conjugate of $a+b$ is a sum of two distinct roots of $f$, and there are ${n\choose 2}$ of these. Since $\prod_{\gamma \sim a+b} (X-\gamma)$ is fixed by the Galois group, it has coefficients in $k$, and it follows that $a+b$ satisfies a polynomial of degree $\leq {n\choose 2}$.

To see that the inequality is tight, let $f$ be any polynomial with doubly-transitive Galois group, e.g. $S_n$, such that the pairwise sums of roots are distinct. (These should be common, but is there a good way to construct them?) Then $a+b$ has exactly ${n\choose 2}$ conjugates.