subfields of transcendental field extension

Suppose $t$ is transcendental over field $\mathbb F_2$. How can I show the two fields $\mathbb F_2(\frac{(t^3+t+1)(t^3+t^2+1)}{(t^2+t)^2})$ and $\mathbb F_2(\frac{t^6+t^3+1}{t(t^2+1)(t^2+t+1)})$ are equal or not ?

I know they are contained in $\mathbb F_2(t)$ and the degree of extensions $\mathbb F_2(t)$ over the two fields above are both $6$ , however, that's not enough.

Edit: maybe one way is to let $u=\frac{(t^3+t+1)(t^3+t^2+1)}{(t^2+t)^2}$, and check if $\frac{t^6+t^3+1}{t(t^2+1)(t^2+t+1)}$ can be writen as $\frac{au+b}{cu+d}$, where $a,b,c,d\in\mathbb F_2$ and at least one of $a,c$ is non -zero. That would make it easier to compute. Is this approach accessible?

Solution 1:

I don't see any smart way of doing it, so I put it into Sage and the result is that they are not equal.

More precisely, I write $k = \Bbb F_2$ and $u = \frac{(t^3 + t + 1)(t^3 + t^2 + 1)}{(t^2 + t)^2}$, so that $k(t)$ is an extension of $k(u)$ of degree $6$, with minimal polynomial $(t^3 + t + 1)(t^3 + t^2 + 1) - u(t^2 + t)^2$.

I then ask Sage to compute $\frac{t^6 + t^3 + 1}{t(t^2 + 1)(t^2 + t + 1)}$. The result is: $$\frac u {u + 1}t^5 + \frac{u^2}{u + 1}t^3 + ut^2 + \frac{u^2}{u + 1}t + 1$$ which is not in $k(u)$ because $1, t, \dots, t^5$ form a $k(u)$-basis of $k(t)$.

The Sage code, which can be tested here:

k = GF(2)
L.<u> = k[]
F = L.fraction_field()
S.<tt> = F[]
T.<t> = S.quotient((tt^3 + tt^2 + 1)*(tt^3 + tt + 1) - u*(tt^2 + tt)^2)
(t^6 + t^3 + 1) / (t*(t^2 + 1)*(t^2 + t + 1))