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I'm doing exercise 1.35 in Fulton's Algebraic Curves, where I'm asked to show that $V(y^2-x(x-1)(x-\lambda)) \subseteq k^2$ is an irreducible curve for any algebraically closed field $k$ and any $\lambda \in k$.

I know that one way is to show that the ideal $(y^2-x(x-1)(x-\lambda)) \subseteq k[x,y]$ is prime, however, I'm all stucked.

Can anyone help me?

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    $\begingroup$ By Gauß' lemma, it suffices to note that $x(x-1)(x-\lambda)$ is not a square in $k[x]$ which is obvious because of the degree. $\endgroup$ – MooS Feb 27 '18 at 13:50
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Recall that the polynomial ring $k[x,y]$ is a UFD; in such a ring, every irreducible element is prime and that a prime element generates a prime ideal (caveat: it is not true that $n\geq 2$ prime elements generate a prime ideal in general).

Hence, it is sufficient to prove that the polynomial $y^2 - x(x-1)(x-\lambda)$ is irreducible in $k[x,y]$ for every $\lambda$. To do that use the Eisenstein's criterion.


Theorem. Le $R$ be an integral domain and let $f=a_0+a_1T+\ldots+a_nT^n\in R[T]$ be a polynomial. Suppose that there exists a prime ideal $\mathfrak p$ of $R$ such that

(1) $a_i\in \mathfrak p$ for $i=0,\ldots,n-1$;

(2) $a_n\notin \mathfrak p$;

(3) $a_0\notin \mathfrak{p}^2$.

Then $f$ is irreducible in $R[T]$.


Now, let us distinguish two cases. In our setting we have $R=k[x]$

  1. Suppose that $\lambda\neq 0$ and take $\mathfrak{p}=(x)$. Then $a_2 = 1\notin (x)$ and $a_0=x(x-1)(x-\lambda)\in (x)$ while $a_0\notin (x^2)$ as $\lambda \neq 0$. The thesis follows.
  2. Suppose that $\lambda = 0$. In this case, repeat the same argument with $\mathfrak{p}=(x-1)$.
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