# Finding the algebraic elements of $\mathbb{F}_3(x,y)$ over $\mathbb{F}_3$, where $y^2+x^4-x^2+1=0$.

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 $$\mathbb{F}_3(\lambda)$$?

• 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 ? – Max Sep 20 at 21:33
• Oh, I see. We can say $\sigma$ exists if $y$ is transcendental over $\mathbb{F}_3(x)$, which is not the case. Right? – rmdmc89 Sep 20 at 21:43
• Yes, that's it. – Max Sep 20 at 21:45
• 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 $x^4-x^2+1$ – Max Sep 21 at 8:38
• Then you can try to prove that your field is isomorphic to $\mathbb F_3 (x) [i] = \mathbb F_3[i](x)$ where $i^2+1=0$ – 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 $$[\mathbb{F}_3(x,y):\mathbb{F}_3(x)]\leq 2$$