# Finding a minimal polynomial in char $2$.

[Some (useless) context: the following problem comes from a problem in Algebraic Geometry, where I have to show that a certain morphism $\textbf P^2\to \textbf A^2$ is inseparable of degree $2$.]

Let $s,t,x,y$ be indeterminates and let $k$ be a field (algebraically closed if it helps) of char $2$. I define a monomorphism of fields $j:k(s,t)\to k(x,y)=:L$ by sending

$$s\mapsto j(s)=\frac{x^2+xy}{y+1} \\ t\mapsto j(t)=\frac{x^2+x}{y(y+1)}.$$

Let us call $K$ the image of $j$, i.e. $K:=k(j(s),j(t))\subset L$. I would like to show that $L/K$ is inseparable of degree $2$.

To do so I just observed that $x\in K(y)$, so it remains to find the minimal polynomial of $y$ and check it is inseparable of degree $2$. Does anyone have a hint? I tried to write $y^2=\alpha y+\beta$ for some $\alpha,\beta\in K$ but I did not succeed (and of course this would not be enough).

Hint#1: Prove that $x=yj(t)+j(s).$
Hint#2: Show that $y$ satisfies a relation $y^2=z$ with $z\in K$ by eliminating $x$ from the definition of $j(s)$.
I see that the first equality does hold but I cannot use it to write $y^2=z$. For example if I square your hint 1 I just fall into a circular argument... Also, writing $y^2=z$ would ensure that $T^2-z$ is the minimal polynomial of $y$ over $K$? – Brenin May 17 '12 at 17:43
@atricolf: Just eliminate $x$ from the first equation defining $j(s)$ (probably the second will do just as well) by replacing it with $yj(s)+j(t)$. You get an equation involving just $y, j(s)$ and $j(t)$ which is what you want, right. – Jyrki Lahtonen May 17 '12 at 17:49