Let $x,y$ be transcendental over $\mathbb{F}_3$ with $y^2+x^4-x^2+1=0$. Prove that the set of elements in $\mathbb{F}_3(x,y)$ which are algebraic over $\mathbb{F}_3$ has exactely $9$ elements.

Here's the first thing that came into my mind:

Let $u\in\mathbb{F}_3(x,y)$ be algebraic over $\mathbb{F}_3$ and $p(T)\in\mathbb{F}_3[T]$ with $p(u)=0$. Since $y^2$ can be written as a polynomial on $x$, then $u$ can be written as $\frac{ay+b}{cy+d}$ where $a,b,c,d\in\mathbb{F}_3[x]$.

Notice that if $ad-bc\neq 0$, then we have an automorphism $\sigma\in\text{Aut}(\mathbb{F}_3(x,y)\mid \mathbb{F}_3(x))$ given by $\sigma(y)=\frac{ay+b}{cy+d}=u$. Since $p(T)\in\mathbb{F}_3[T]$ is fixed by $\sigma$, we have $p(y)=p(\sigma^{-1}(u))=\sigma^{-1}(p(u))=0$ (absurd, since $y$ is transcendental over $\mathbb{F}_3$).

So we must have $ad-bc=0$, which means $u$ doesn't depend on $y$, so $u\in \mathbb{F}_3(x)$. But since $x$ is transcendental and $p(u)=0$, we conclude $u\in \mathbb{F}_3$. (so the only algebraic elements are in $\mathbb{F}_3$)

Obviously I did something wrong, because $0=y^2+x^4-x^2+1=y^2+(x^2+1)^2$, therefore $\left(\frac{x^2+1}{y}\right)^2+1=0$, so $\frac{x^2+1}{y}\notin\mathbb{F}_3$ is algebraic over $\mathbb{F}_3$.

What am I doing wrong?

Going back to the problem, let $\lambda:=\frac{y}{x^2+1}$. Then we have $\lambda^2+1=0$, so $[\mathbb{F}_3(\lambda):\mathbb{F}_3]=2$ and $\#\mathbb{F}_3(\lambda)=9$. How can I prove that there is no other algebraic element outside of $\mathbb{F}_3(\lambda)$?

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    $\begingroup$ Well you can try to go through your argument with that specific element. Here $ad-bc = -(x^2+1)$, so you're in the first scenario. Clearly $y^2+1 \neq 0$ (as it is algebraic), so there is no such $\sigma$. Can you try to make explicit your argument for why $\sigma$ should exist ? $\endgroup$ – Max Sep 20 at 21:33
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    $\begingroup$ Oh, I see. We can say $\sigma$ exists if $y$ is transcendental over $\mathbb{F}_3(x)$, which is not the case. Right? $\endgroup$ – rmdmc89 Sep 20 at 21:43
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    $\begingroup$ Yes, that's it. $\endgroup$ – Max Sep 20 at 21:45
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    $\begingroup$ Watch out, you made a mistake at some point : at the beginning you have $x^4-x+1$ and when you find the element you have $x^4-x^2+1$ $\endgroup$ – Max Sep 21 at 8:38
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    $\begingroup$ Then you can try to prove that your field is isomorphic to $\mathbb F_3 (x) [i] = \mathbb F_3[i](x)$ where $i^2+1=0$ $\endgroup$ – Max Sep 21 at 14:35

Let $K\subset \mathbb{F}_3(x,y)$ be the field of algebraic elmenent over $\mathbb{F}$. $\left(\frac{x^2+1}{y}\right)^2 +1 = 0$ provides $[K:\mathbb{F}_3]\geq 2$. $y^2 = x^4 -x^2+ 1$ indicates $[\mathbb{F}_3(x,y):\mathbb{F}_3(x)]\leq 2$.

Let's assume $[K:\mathbb{F}_3]> 2$. Because there is only one field extension with $[L:\mathbb{F}_3] = 2$ we can choose $u\in K$ with minimal polynom $f\in\mathbb{F}_3[T]$ and $\deg(f)\geq 3$. Because $x$ is transcendetial $f\in \mathbb{F}_3(x)[T]$ is irreducible, too, therefore $[\mathbb{F}_3(x,u):\mathbb{F}_3(x)]\geq 3$. This contradicts $[\mathbb{F}_3(x,y):\mathbb{F}_3(x)]\leq 2$


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