Why $a/b$ is a square and $\sqrt{b}$ is a root of $p(cx)$ in a finite field

Question

This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise F2.

Let $F$ be a finite field. If $a, b \in F$, let $p(x) = x^2-a$ and $q(x) = x^2 - b$ be irreducible in $F[x]$, and let $\sqrt{a}$ and $\sqrt{b}$ denote roots of $p(x)$ and $q(x)$ in an extension of $F$. Explain why $a/b$ is a square, say $a/b = c^2$ for some $c \in F$. Prove that $\sqrt{b}$ is a root of $p(cx)$.

Let $F(\sqrt{a},\sqrt{b})$ be the smallest field containing $F$, $\sqrt{a}$ and $\sqrt{b}$, and let $F(\sqrt{a},\sqrt{b})^*$ be the multiplicative group of nonzero elements of $F(\sqrt{a},\sqrt{b})$.

1. Let $c=\sqrt{a}/\sqrt{b}\in F(\sqrt{a},\sqrt{b})^*$, so $c^2=a/b$. This shows $a/b$ is a square.
2. $c=\sqrt{a}/\sqrt{b}\implies \sqrt{a} = c\sqrt{b} \implies p(c\sqrt{b}) = p(\sqrt{a}) = 0$. Hence $\sqrt{b}$ is a root of $p(cx)$.

Correct?

Answer 1

1 is not correct. You need to show $c\in F$ as pointed out by Brian in the comment. For 2, we don't know $c=\sqrt{a}/\sqrt{b}$. You can plug in $\sqrt{b}$ directly: $p(c\sqrt{b})=(c\sqrt{b})^{2}-a=c^{2}b-a=0$. So $\sqrt{b}$ is a root of $p(cx)$.

Edit: For (1), I looked at the book you mentioned. It has this paragraph before this question.

Let $F$ be a finite field, and $F^{*}$ the multiplicative group of nonzero elements of $F$. Obviously $H=\{x^{2}:x\in F^{*}\}$ is a subgroup of $F^{*}$; since every square $x^{2}$ in $F^{*}$ is the square of only two different elements, namely $\pm x$, exactly half the elements of $F^{*}$ are in $H$. Thus, $H$ has exactly two cosets: $H$ itself, containing all the squares, and $aH$ (where $a\notin H$), containing all the nonsquares. If $a$ and $b$ are nonsquares, then by Chapter 15, Theorem 5(i), $ab^{-1}=\frac{a}{b}\in H$. Thus: if $a$ and $b$ are nonsquares, $a/b$ is a square. Use these remarks in the following:

The above paragraph seems to assume $ch(F)\neq 2$. So let's show $ch(F)\neq 2$.

Lemma. If $ch(F)=2$, then every element of $F^{*}$ is a square.

Proof. Let $f:F^{*}\to F^{*}$ be $f(x)=x^{2}$. Then $f$ is a group homomorphism. Let $x\in \ker(f)$. Then $1=f(x)=x^{2}$. So $x=1$. $f$ is injective. Hence $f$ is bijective. So every element of $F^{*}$ is a square.

Here $a$ and $b$ are nonsquares because otherwise suppose $a=d^{2}$ for some $d\in F$. Then $p(x)=x^{2}-d^{2}$ would be reducible in $F[x]$. Therefore, $ch(F)\neq 2$. So by the above result, $a/b$ is a square.