Let $F$ be a field. Is it true that if $[F(\sqrt{D}) : F] = 2$, then $D \in F$?

Question

In Dummit & Foote, problem 14.2.17(c), the authors hand us a quadratic extension of the form $F(\sqrt D)$. Now, while I am pretty sure you need that $D \in F$ to do this particular problem, I can't help but wonder if assuming this is even necessary. In other words, if the degree of the extension $F(\sqrt D)/F$ is $2$, then must $D$ belong to $F$?

My thoughts so far are as follows: If $D \not\in F$, then $[F(D) : F] > 1$. This puts $\sqrt D \in F(D)$, otherwise $$ 4 \le [F(\sqrt D) : F(D)][F(D):F] = [F(\sqrt D): F] = 2. $$ This tells us that $F(D) = F(\sqrt D)$. My next observation was that if $\sqrt D$ has minimal polynomial $x^2 + ax + b$, we can write $\sqrt D$ in terms of $D$ and elements of $F$ and use this to find the minimal polynomial of $D$ over $F$: $$ \sqrt D = -(D + b)/a, \ m_{D, F}(x) = x^2 + (2ab - a)x + b^2. $$ Notice here that $a \neq 0$, or else $D \in F$!

Answer 1

This is false. Consider the case of $F=\Bbb{Q}$, $D=(1+\sqrt2)^2=3+2\sqrt2.$

We have $\sqrt{D}=\pm(1+\sqrt2)$, so $\Bbb{Q}(\sqrt D)=\Bbb{Q}(D)=\Bbb{Q}(\sqrt2)$.

Answer 2

If $[F(\sqrt D):F]=2$ the extension is Galois with group $\{1,\sigma\}$ and $$ \sigma(\sqrt D)=a+b\sqrt D,\qquad a,b\in F. $$ Since $\sigma^2=1$ we must have $\sqrt D=\sigma^2(\sqrt D)=a+ab+b^2\sqrt D$, i.e. $$ a(b+1)=0\qquad\text{and}\qquad b^2=1. $$ If $a=0$ and $b=-1$ it readily follows that $D\in F$.

On the other hand, if $a\neq0$ and $b=-1$ we have $$ F\ni{\rm N}(\sqrt D)=\sqrt D\cdot\sigma(\sqrt D)=a\sqrt D-D $$ from which $D={\rm N}(\sqrt D)-a\sqrt D\notin F$. Jyrki Lahtonen's answer shows an instance of the latter situation happening.