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Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

An elliptic curve in characteristic zero admits no nonconstant rational map of projective varieties from the projective line. Over $\mathbb{C}$, write its equation as $y^2={\prod }_{j=1}^3(x-e_j)$ with distinct $e_j$. Substituting $x=u/v$ with coprime polynomials gives $(y{v}^2{)}^2=v{\prod }_j(u-e_{j}v)$. These four pairwise coprime factors must all be squares. The polynomial pencil with four square members forces $u,v$ constant, ruling out a nonconstant rational parametrization of an algebraic curve.

Here is an elementary polynomial pencil with four square members argument. Suppose first that $u,v$ are linearly dependent. Their coprimality of polynomials then forces both to be constant. Otherwise write the four distinct members as $L_j={\alpha }_{j}u+{\beta }_{j}v=s_j^2$. They are nonzero and pairwise coprime polynomials: a common nonconstant factor of two members would divide both $u$ and $v$.

Put $d=\max (\mathrm{deg}u,\mathrm{deg}v)>0$. At most one member of the pencil has degree of a polynomial smaller than $d$, since cancellation of its leading coefficient determines a unique projective pair. Choose a member $L_k$ of minimal degree $e$ and any independent member $L_l$ of degree $d$. The polynomialis nonzero: otherwise the rational function $L_k/L_l$ would have zero derivative, hence would be constant in characteristic zero. Its degree is at most $d+e-1$; if both members have degree zero the original $d>0$ assumption has already failed.

Replacing this pair by any other independent pair changes $W$ only by a nonzero scalar. Since $L_j=s_j^2$, each $s_j$ divides $W$. The pairwise coprimality of polynomials therefore gives ${\prod }_j{s}_j\mid W$. But three members have degree $d$, soa contradiction. Consequently $u$ and $v$ are constant. The possibility that a member is zero was already covered by linear dependence.

To apply this to an elliptic curve, complete the square in its Weierstrass equation of an elliptic curve and work over $\mathbb{C}$. Nonsingularity gives three distinct roots $e_1,e_2,e_3$, so the equation becomes $y^2={\prod }_{j=1}^3(x-e_j)$. A nonconstant rational parametrization of an algebraic curve would have $x=u/v$ with coprime polynomials. Clearing denominators givesThe four factors are pairwise coprime polynomials. A rational function whose square is a polynomial is itself a polynomial, by comparing numerator and denominator in lowest terms. Unique factorization, and the fact that every nonzero complex constant has a square root, make each of these four factors a square in $\mathbb{C}[t]$. They correspond to four distinct projective pairs. The result just proved forces $u,v$ to be constant, and the equation then forces $y$ to be constant as well. This proves the nonparametrizability of an elliptic curve.

Let $u,v\in \mathbb{C}[t]$ be coprime polynomials. If $\alpha u+\beta v$ is a square for four distinct $(\alpha :\beta )$, then $u,v$ are constant. For independent $u,v$, the four square members $L_j=s_j^2$ are pairwise coprime polynomials. Let $d$ and $e$ be their maximal and minimal degrees; at least three have degree $d$. For independent members of degrees $e,d$, the nonzero polynomial $W=L_k^{\prime }{L}_l-L_k{L}_l^{\prime }$ has degree at most $d+e-1$. Each $s_j$ divides $W$, giving $\mathrm{deg}W\ge (3d+e)/2>d+e-1$, a contradiction. The dependent case follows directly from coprimality of polynomials.