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The original question states:

Let $K$ be an extension of $F$, $[K:F] = 3$. Show that for every $\alpha \in K, \beta \in K \backslash F$, $\alpha = \frac{a+b\beta}{c+d\beta}$ for some $a,b,c,d \in F$.

For every such $\beta$, $K = F(\beta)$. So there are $a_0, a_1, a_2$ such that $\alpha = a_0 + a_1 \beta + a_2 \beta^2$. How to continue from here?

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    $\begingroup$ Do you mean $[K:F] = 3$? Or has the notation commuted itself since the days I studied this stuff? $\endgroup$ – John Hughes Apr 3 at 17:00
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    $\begingroup$ You've got four free parameters, $a,b,c,d$ (although it's really only $3$, because everything's homogeneous). Try multiplying through by the denominator. $\endgroup$ – John Hughes Apr 3 at 17:02
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    $\begingroup$ @JohnHughes Yup, $[K :F]$. My bad. $\endgroup$ – Bary12 Apr 3 at 17:12
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    $\begingroup$ Multiplying by the denominator gives some incoherent equality, from which I can't seem to solve it $\endgroup$ – Bary12 Apr 3 at 17:14
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    $\begingroup$ Write out that incoherent equality for us, and move everything to one side. The result is a polynomial in $\beta$, a polynomial that's equal to the zero polynomial. $\endgroup$ – John Hughes Apr 3 at 17:17
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Write $\alpha=a_0+a_1\beta+a_2\beta^2$. If $a_2=0$, there is nothing to prove. Otherwise, write $\alpha= a_2(a_0'+a_1'\beta+\beta^2)= a_2(a_0'+(\frac{a_1'}{2}+\beta)^2-\frac{a_1'^2}{4})= a_2(a_0''+(a_1''+\beta)^2)$, where $a_0',a_1',a_0'',a_1''$ are obviously defined. (Now I see that I have to assume $char F\neq 2$, I don't see now how to deal with this case.)

Consider $\gamma=a_1''+\beta$. Obviously, $K=F(\gamma)$ and minimal polynomial of $\gamma$ over $F$ is of degree $3$; let it be $X^3+b_2X^2+b_1X+b_0$. So $\gamma^3+b_2\gamma^2+b_1\gamma+b_0=0$, wherefrom $\gamma^2=-\frac{b_1\gamma+b_0}{\gamma+b_2}$. By returning, $\alpha= a_2(a_0''+\gamma^2)= a_2(a_0''-\frac{b_1\gamma+b_0}{\gamma+b_2})= \frac{c_0\gamma+c_1}{\gamma+b_2}$, where again $c_i$'s are obviously defined. Since $\beta=\gamma-a_1''$, $\alpha=\frac{c_0\beta+d_0}{\beta+d_1}$, where $d_i$'s are obviously defined.

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A friend of mine came with a very elegant solution, which I ended up using.

$[K : F] = 3$ implies that $\{1, \alpha, \beta, \alpha\beta\}$ is linearly dependent. So there are $a_1, a_2, a_3, a_4 \in F$, at least one of which is not equal to $0$, such that $a_1 + a_2 \alpha + a_3 \beta + a_4 \alpha \beta = 0$. A quick rearrangement gives $\alpha = \frac{-a_1 -a_3 \beta}{a_2+a_4\beta}$.

The case where $a_2+a_4\beta = 0$ can be easily ruled out. Since $F(\beta) = K$, $[F(\beta):F] = 3$, so $a_2 = a_4 = 0$. Using that in the first equation we also get $a_1 + a_3 \beta = 0$ so $a_1 = a_2 = a_3 = a_4 = 0$.

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