I want to find the extension degree of the minimal splitting field $L$ of $X^4+2tX^2+t$ over $\mathbb{C}(t)$.
What I know are followings.
Let $\alpha$ be a root of $f(X) := X^4+2tX^2+t$ and $K = \mathbb{C}(t)$. Since $f$ is irreducible over $K$, $[K(\alpha):K] = 4$. If $K(\alpha) = L$, then $\beta$, a root of $f$ which is not $\pm \alpha$ (this exists as the character of $K$ is zero and $f$ is separable) is contained in $K(\alpha)$. Also, since the roots of $X^2+2tX+t$ is $\alpha^2, \beta^2$, we have $\alpha^2\beta^2=t$, and $\alpha\beta=\pm \sqrt{t} \in K(\alpha)$. Therefore, we have $L = K(\alpha, \beta) = K(\alpha, \sqrt{t})$ or $L=K(\alpha)$. This implies $[L:K] = 8$ or $[L:K]=4$.
I think $[L:K] = 8$, but cannot show $\sqrt{t} \notin K(\alpha)$.
I would appreciate any help. Thank you!
Note that $(\alpha^2 + t)^2 = t^2 - t$; and similarly, as you observed, $(\alpha \beta)^2 = t$. Therefore, if we set $u := \frac{\alpha^2 + t}{\alpha\beta} \in L$, then $u^2 = t-1$.
Now, the original polynomial $X^4 + 2t X^2 + t = X^4 + 2(u^2+1) X^2 + (u^2+1)$ is still irreducible over $\mathbb{C}[u]$, applying Eisenstein's criterion with the prime ideal $\langle u + i \rangle$. Therefore, by Gauss' lemma, $X^4 + 2t X^2 + t$ is also irreducible over $\mathbb{C}(u)$. It follows that $[L : \mathbb{C}(t)] \ge [\mathbb{C}(u, \alpha) : \mathbb{C}(u)] [\mathbb{C}(u) : \mathbb{C}(t)] = 4 \cdot 2 = 8$. Since you have also already shown that $[L : K] \le 8$, it follows that $[L : K] = 8$.