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

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?

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.

• Do you mean $a$ and $b$ are not squares? Otherwise take $a=b=1$ for example: lots of polynomials have $2$ as a root but not $0$. – Greg Martin Jan 29 at 6:54
• The roots of $x^4-10x^2+1$ are $\pm\sqrt2\pm\sqrt3$ (all four sign combinations) in any field where the required square roots exist. – Jyrki Lahtonen Jan 29 at 8:20
• All: This question was probably spawned by the OP's comment under this answer of mine that I was unavailable to address in time. – Jyrki Lahtonen Jan 29 at 8:22

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 $$a=b$$.

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.

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.