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.

  • $\begingroup$ It seems that $k=\bar k$ is not needed. If you can factor $f(X,Y)$ in $k[X,Y]$, you can also factor it in $\bar k[X,Y]$. $\endgroup$ – Hagen von Eitzen Oct 21 '14 at 17:00
  • $\begingroup$ Regarding use of $\lambda$: I suspect that the intended proof was something like this: $\lambda\ne0$ or$\lambda\ne 1$, hence at least one root of $X(X-1)(X-\lambda)$ must be simple, hence $X(X-1)(X-\lambda)$ cannot be a square. - Or: In the light of the obvios factorization of $\alpha\beta$, no choice leads to $\alpha+\beta=0$. $\endgroup$ – Hagen von Eitzen Oct 21 '14 at 17:03
  • $\begingroup$ @HagenvonEitzen I couldn't assume $\lambda \not = 1,0$, for the problem specifies for all $\lambda \in k$. $\endgroup$ – FireGarden Oct 21 '14 at 17:08
  • $\begingroup$ What I meant was that you cannot have $\lambda=0$ and $\lambda=1$ $\endgroup$ – Hagen von Eitzen Oct 21 '14 at 17:12
  • $\begingroup$ Oh, sure. That makes a lot more sense! $\endgroup$ – FireGarden Oct 21 '14 at 17:17

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