Let $a,b$ is not a square element in finite field.
If polynomial over the finite field has a root $\sqrt a+\sqrt b$, then the polynomial has $\sqrt a-\sqrt b$ as a root too?
Thank you for your help.
This question arises from concrete problem, $x^4-10x^2+1$（this is a minimal polynomial of $\sqrt{2}+\sqrt{3}$ over rational field） is reducible over arbitrary finite field.
$\sqrt{2}+\sqrt{3}$ is root over finite field, then $\sqrt{2}-\sqrt{3}$ is also a root（why?）.And $\sqrt2+\sqrt3$ and $\sqrt2-\sqrt3$ exists in $2$-degree extension of the finite field, so the minimal polynomial spilts. This leads minimal polynomial is at most $2$ degree, so $x^4-10x^2+1$ is reducible.
Best Answer
As the multiplicative group of a finite field $F$ is cyclic, it follows that $b=c^2a$ with come $c\in F^\times$ whenever $a,b$ are both non-squares. Then wlog. $\sqrt b=c\sqrt a$ and so $\sqrt a+\sqrt b$ is a root of one of $X^2-(1+ c)^2a$ whereas $\sqrt a-\sqrt b$ is a root of $X^2-(1- c)^2a$. These minimal$^1$ polynomials of $\sqrt a+\sqrt b$ and $\sqrt a-\sqrt b$ are different except in characteristic $2$.
$^1$ Okay, one of them is not minimal when $c=\pm1$, i.a., when $a=b$.