Prove that the irreducible polynomial for $\sqrt2+\sqrt3$ over $\mathbb{Q}$ is reducible modulo $p$ for every prime $p$

Question

I am attempting to solve part (d) of the following problem from Artin's algebra book. I have already solved parts (a) through (c).

(a) Let $p$ be an odd prime. Prove that exactly half of the elements of $\mathbb{F}_p^\times$ are squares and that if $α$ and $β$ are nonsquares, then $αβ$ is a square.

(b) Prove the same assertion for any finite field of odd order.

(c) Prove that in a finite field of even order, every element is a square.

(d) Prove that the irreducible polynomial for $γ=\sqrt2+\sqrt3$ over $\mathbb{Q}$ is redubile modulo $p$ for every prime $p$.

Now, I have shown that $g(x)=x^4-10x^2+1\in \mathbb{Q}[x]$ is the irreducible polynomial for $γ$. Let $p$ be a prime. Let $f(x)=x^4-10x^2+1 \in \mathbb{F}_p[x]$.

I want to show $f$ is reducible. I see that either 2, 4, or 6 is equal to $s^2$ for some $s\in\mathbb{F}_p$. I think it will be useful to proceed by cases here. It someone could show me a complete, elementary, simple proof for the $s^2=2$ case, I should be able to finish the rest on my own.

Answer 1

I think you tried to say that $2, 3$ or $6$ are squares in $\mathbb F_p$ for all $p$, this follows either using Legendre's symbol or using the fact that there is only one field of size $p^2$.

Now if either $2$ or $3$ have a square root in $\mathbb F_p$ the polynomial is obviously reducible mod $p$. Otherwise $6$ has a root modulo $p$ and as $$(\sqrt 2+\sqrt 3)^2=5+2\sqrt 6$$

this implies that your polynomial is reducible mod $p$ in this case too.

Alternatively, as there is only one field of size $p^2$ all square roots of elements of $\mathbb F_p$ lie in $GF(p^2)$ and this implies automatically that your polynomial is reducible mod $p$ for all prime $p$.