For tower fields, am I allowed to use the modulo equations recursively?

Question

If I build up $\mathbb F_{q^4}\cong\mathbb F_{q^2}[Y]/(Y^2-X)$ on $\mathbb F_{q^2}\cong \mathbb F_q[X]/(X^2+1)$ am I allowed to use $-1 = X^2 = Y^4$? I do not see any reason or rule, that would be critical here.

Answer 1

There is the following problem in your construction. Not necessarily in the notation (I endorse sharding4's comment), but in thinking that you get a field of order $q^4$ this way. This is not really your main question - just something I wanted to explain.

Your $X$ (or its coset) would be an element of multiplicative order $4$. This will, indeed, give you a quadratic extension of $\Bbb{F}_q$ provided that it did not contain a fourth root of unity already. In other words, for this to work you need $q\equiv-1\pmod4$. Also observe that if $q$ is even, then $X^2+1=(X+1)^2$.

But, at that point your luck runs out. You won't get $\Bbb{F}_{q^4}$ by adjoining an eighth root of unity (your $Y$). This is because $8\mid q^2-1$ for all odd $q$. As $\Bbb{F}_{q^2}^*$ is cyclic this means that it will already contain primitive eighth roots of unity. In other words, $Y^2-X$ will never be irreducible over $\Bbb{F}_q(X)$.