$\sqrt a+\sqrt b$ is a root of polynomial, then $\sqrt a -\sqrt b$ is so over finite field

Question

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.

Answer 1

Consider $\Bbb F_5$ and consider $a=2$, $b=3$ and $\sqrt b=2\sqrt a$. Then, $x^2+2$ has roots $\pm(\sqrt a+\sqrt b)$, and neither $\sqrt a-\sqrt b$ nor $\sqrt b-\sqrt a$ are among them.

Answer 2

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 $2$.

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$^1$ Okay, one of them is not minimal when $c=\pm1$, i.a., when $a=b$.

Answer 3

Let's look at the concrete example, to see what is going on, and assume we are not in characteristic $2$.

Then, using the normal quadratic formula if $$x^4-10x^2+1=0$$ then $$x^2=\frac {10\pm \sqrt{100-4}}{2}=5\pm 2\sqrt 6$$ and if $6$ has a square root in your finite field, then the original equation is reducible.

The roots in a splitting field are $\pm \sqrt 2\pm \sqrt 3$ and the factorisation already obtained corresponds to one pairing of the roots: $$(x^2-5-2\sqrt 6)(x^2-5+2\sqrt 6)=$$$$=(x-\sqrt 2-\sqrt 3)(x+\sqrt 2+\sqrt 3)\cdot(x+\sqrt 2-\sqrt 3)(x-\sqrt 2+\sqrt 3)$$

If we were instead to choose other pairings we'd get $$(x-\sqrt 2+\sqrt 3)(x+\sqrt 2+\sqrt 3)\cdot(x-\sqrt 2-\sqrt 3)(x+\sqrt 2-\sqrt 3)=$$$$=(x^2+2\sqrt 3 x+1)(x^2-2\sqrt 3x+1)$$ or $$(x-\sqrt 2+\sqrt 3)(x-\sqrt 2-\sqrt 3)\cdot(x+\sqrt 2-\sqrt 3)(x+\sqrt 2+\sqrt 3)=$$$$=(x^2-2\sqrt 2 x-1)(x^2+2\sqrt 2x-1)$$ and these give factorisations in the case that $\sqrt 3$ or $\sqrt 2$ exist in your finite field - ie that $3$ or $2$ are squares.

Now, you may also know that the product of two non quadratic residues modulo $p$ is a quadratic residue. From this it is easy to deduce that at least one of $2,3, 6$ has a square root in a field of characteristic $p$. So the polynomial is reducible. If two of the three have roots then the third does and the polynomial splits into linear factors.

Modulo $5$ we have that $6$ is a square while $2$ and $3$ are not. Modulo $7$ we have that $2$ is a square while $3$ and $6$ are not. Modulo $11$ we have that $3$ is a square and $2$ and $6$ are not. Modulo $23$ we have that $2, 3, 6$ are all squares. So all four possibilities occur.

Finally it is easy to show that in characteristic $2$ we have $x^4-10x+1=x^4+1=(x+1)^4$.

So I've worked this in some detail in case it helps you to see what is going on. You can generalise to $a$ and $b$. Obviously there are more efficient expositions, but sometimes longhand helps.