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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 $D \in F$!

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    $\begingroup$ I'm confused. If $D \not \in F$ then what does $F(\sqrt{D} )$ mean? $\endgroup$ – Ethan Bolker Mar 9 at 22:57
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    $\begingroup$ @Ethan You don't know what $\,\Bbb Q(\sqrt{1+\sqrt 3})\,$ means? $\endgroup$ – Bill Dubuque Mar 9 at 23:04
  • $\begingroup$ @BillDubuque Now I'm no longer confused. Thanks. $\endgroup$ – Ethan Bolker Mar 10 at 1:10
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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 $\Bbb{Q}(\sqrt D)=\Bbb{Q}(D)=\Bbb{Q}(\sqrt2)$.

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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.

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    $\begingroup$ If $\operatorname{char} F = 2$, then $a+b\sqrt D\mapsto a-b\sqrt D$ is the trivial automorphism. $\endgroup$ – FredH Mar 10 at 9:40
  • $\begingroup$ Ah! @JyrkiLahtonen is right; I mentioned near the end of my question that in the minimal polynomial of $\sqrt D$, $a \neq 0$. $\endgroup$ – fauxefox Mar 11 at 1:10
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    $\begingroup$ @JyrkiLahtonen, I fixed my answer. $\endgroup$ – Andrea Mori Mar 11 at 8:52

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