I need to prove the polynomial $f(X,Y) = Y^2-X(X-1)(X-\lambda)$ is irreducible over any algebraically closed field $k$, for any $\lambda \in k$. I've not had any instruction in how one usually goes about doing this - only the frankly unhelpful comment that there is no known criterion for irreducibility in several variables.
Here is my attempt, the question is whether it sufficiently proves the claim.
Instead, we will consider $Y^2-X(X-1)(X-\lambda) \in k(X)[Y]$. Irredcuibility here will imply irreducibility in $k[X,Y]$ by Gauss' lemma for UFD's.
Suppose it is reducible, so that $f=(Y-\alpha)(Y-\beta)$ for some $\alpha ,\beta \in k(X)$.
Expanding and equating coefficients gives $\alpha + \beta = 0, \quad \alpha \beta = -X(X-1)(X-\lambda) \implies \alpha^2 = X(X-1)(X-\lambda)$
We will now consider degrees. Write $\alpha^2 = \frac{\gamma^2}{\delta^2}$, where $ \gamma ,\delta \in k[X]$ Now;
$3 = \deg(X(X-1)(X-\lambda)) = \deg(\alpha^2) = \deg(\gamma^2) - \deg(\delta^2) = 2(\deg(\gamma)-\deg(\delta)) \implies \deg(\gamma)-\deg(\delta) \not\in \mathbb{Z}$ which is impossible.
Therefore no such polynomials exist, and $f$ is irreducible.
My issues are that I haven't apparently needed that $k$ is algebraically closed (perhaps necessary to invoke Guass' lemma..?) nor made any mention of $\lambda$, which makes me suspicious.