# With $[K: F] = 3, \alpha \in K, \beta \in K \backslash F$, show that there are $a,b,c,d \in F$ such that $\alpha = \frac{a+b\beta}{c+d\beta}$

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?

• Do you mean $[K:F] = 3$? Or has the notation commuted itself since the days I studied this stuff? – John Hughes Apr 3 at 17:00
• 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. – John Hughes Apr 3 at 17:02
• @JohnHughes Yup, $[K :F]$. My bad. – Bary12 Apr 3 at 17:12
• Multiplying by the denominator gives some incoherent equality, from which I can't seem to solve it – Bary12 Apr 3 at 17:14
• 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. – John Hughes Apr 3 at 17:17

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.

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 $$a_1 = a_2 = a_3 = a_4 = 0$$.