Prove that $y^2 = x(x-1)(x- \lambda)$ is irreducible for all $\lambda \in k$

Question

I wish to prove that $y^2 = x(x-1)(x- \lambda)$ is irreducible for all $\lambda \in k$. It seems like this follows from the fact that $x(x-1)(x- \lambda)$ cannot be written as the square of any polynomial in $x$, but I'm curious whether there is a more direct/rigorous way to prove this.

Answer 1

Your reasoning is sufficiently rigorous: recall that if $f(T)\in D[T]$, where $D$ is a domain, and $f$ is monic of degree $2$ or $3$, then $f$ is irreducible in $D[T]$ iff $f$ has no roots in $D$. In your case you can see your polynomial as a polynomial in $y$ with coefficients in the domain $k[x]$, and considerations on the degree on $x$ show that $x(x-1)(x-\lambda)$ is never a square in $k[x]$.