The curve $y^2=x^3+Ax+B$ can be parameterized if and only if $x^3+Ax+B$ has a repeated root.

Question

I'm working through Shafarevich's Basic Algebraic Geometry, and one of the problems asks the reader to prove the problem in the title. I found the "if" direction fairly straightforward, but I can't finish my proof of the converse. If anyone can help me finish it, that'd be greatly appreciated! Suppose we have the rational parametrization $$(\frac{f(t)}{g(t)})^2=(\frac{p(t)}{q(t)})^3+A\frac{p(t)}{q(t)}+B,$$ yielding the identity $$f(t)^2q(t)^3=g(t)^2[p(t)^3+Aq(t)^2p(t)+Bq(t)^3].$$ Assuming $f$ and $g$ are coprime, we find that $p(t)^3+Aq(t)^2p(t)+Bq(t)^3$ must have a square factor.

From here, I know that either two of the $(p(t)-aq(t)), (p(t)-bq(t)),$ and $(p(t)+(a+b)a(t)$ are the same, or, as @Aphelli pointed out, there are repeated factors. I cannot, for the life of me, figure out how to show this second case isn't so. If anyone can help me finish this, or give an alternative solution, that'd be greatly appreciated!

Edit: Completely changed the question to account for the fact that the proof is, in fact, incomplete, and I cannot finish it.

Answer 1

Andrea Mori gives a good abstract argument (that reflects what is really going on). Here’s an elementary one.

Lemma: let $k$ be an algebraically closed field with $2 \in k^{\times}$. Let $V \subset k[x]$ be a plane. Then $V$ does not contain four nonzero, pairwise non-associated squares.

Proof: by induction on the maximal degree $\delta$ of a polynomial $P \in V$. Clearly $\delta \geq 2$. So let $\delta \geq 2$ be minimal contradicting the statement. Then we can find $V$ and squares $p,q,p-q,p+\lambda q \in V$ with $\lambda \neq 0,-1$. Clearly, the gcd of $p,q$ is $1$, because it is a square and we could divide everything by it, contradicting the minimality of $\delta$.

Write $p=u^2,q=v^2$, then $u$ and $v$ have degree less than $\delta$, and are coprime. Moreover, $u^2-v^2=(u-v)(u+v)$ is a square and $u-v,u+v$ are coprime, so $u+v,u-v$ are squares. Write $\lambda=-\mu^2$, then $p+\lambda q=(u-\mu v)(u+\mu v)$ with $\mu\neq 0,\pm 1$. For the same reason, $u-\mu v,u+\mu v$ are squares.

Thus $ku+kv$ contains the four nonzero squares $u\pm v, u\pm \mu v$, contradicting the minimality of $\delta$.

Corollary: the equation $y^2=(x-a)(x-b)(x-c)$ has no rational parametrization when $a,b,c$ are pairwise distinct.

Proof: assume that we have $(f/g)^2=(p/q-a)(p/q-b)(p/q-c)$ where $f,g,p,q$ are polynomials with $gq\neq 0$ and $(f,g)=(p,q)=1$. Then we have $f^2q^3=g^2(p-aq)(p-bq)(p-cq)$.

By my comments, it follows that $(p-aq)(p-bq)(p-cq)$ (resp. $g^2$) is associated to $f^2$ (resp. $q^3$). Since $a,b,c$ are pairwise disjoint, $p-aq,p-bq,p-cq$ are pairwise coprime, hence every one of them is a square; since $q^3$ is associated to $g^2$, $q$ is a square. Hence the plane $kp+kq$ contradicts the lemma.

Answer 2

If the cubic $E$ is singular, after a suitable change of coordinates you can always assume that the origin $O=(0,0)$ is the singular point. Then the parametrization is obtained by associating each line $\ell$ in the pencil through $O$ to the third point $P$ of intersection in $\ell\cap E$. Since a pencil of line through a point is a copy of the projective line $\mathbb{P}^1$, this geometric construction defines a birational isomorphism $$ \mathbb{P}^1\longleftrightarrow E. $$ If the cubic is smooth, by taking any point $P\in E$ and doing the same construction we obtain a $2$-to-$1$ map $$ E\longrightarrow\mathbb{P}^1 $$ whose ramification points correspond to the lines through $P$ that are tangent to $E$. Of these there are $4$, for instance if we take as $P$ the point at infinity for the standard Weierstrass model of $E$ the $4$ tangents are the $3$ vertical lines that go through the points of $E$ with $x$-coordinate equal to $0$, i.e. the points of $2$-torsion of $E$, and the line at infinity itself.

From this situation one obtains that $E$ must be of genus $1$ and in particular that its function field is not a purely trascendental extension of the base field with degree $1$.

Therefore the curve $E$ does not admit a rational parametrization.