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

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.

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.
• @leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $s\in \mathbb{Q}$. See here for a discussion of why this has full Galois group. Dec 25, 2018 at 21:52
• Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,\ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=\prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-) Dec 26, 2018 at 9:27