Examples of non-constant polynomials $f(x)^3+g(x)^2=c$ for some non-zero constant $c$

Question

Let $k$ be a field (feel free to take $k=\mathbb{C}$), does there exist non-constant polynomials $f(x),g(x)\in k[x]$ s.t. $$f(x)^3+g(x)^2=c$$ for some non-zero constant $c\in k^*$.

The context comes from that the descrimant of an elliptic curve $y^2=x^3+ax+b$ over $\mathbb{A}^1_k$ (with $a,b\in k[t]$) is $\Delta=-16(4a^3+27b^2)\in k[t]$. The set of points with bad reduction is a closed subset of $\mathbb{A}^1_k$ defined by $4a^3+27b^2=0$ (assuming $\mathop{\mathrm{char}}k\neq 2,3$), but it will be an empty set if $4a^3+27b^2=c$ for some non-zero constant $c$.

Answer 1

This is possible in characteristic $2$ or $3$. For instance, in characteristic $2$, you could have $c=1$, $f(x)=x^2$, and $g(x)=x^3+1$ in characteristic $2$. More generally, if $f(x)$ is a polynomial in $x^2$, then $c-f(x)^3$ is also a polynomial in $x^2$, so as long as its coefficients have square roots then it has a square root. Similarly, in characteristic $3$, if $g(x)$ is a polynomial in $x^3$ then $c-g(x)^2$ will have a cube root as long as its coefficients have cube roots.

On the other hand, it is impossible if $\operatorname{char}(k)\neq 2,3$. Here is one simple way to see it. We may extend $k$ to assume it is algebraically closed. We then have $-f(x)^3=g(x)^2-c=(g(x)-\sqrt{c})(g(x)+\sqrt{c})$. Since $c\neq 0$ and $\operatorname{char}(k)\neq 2$, $g(x)-\sqrt{c}$ and $g(x)+\sqrt{c}$ are relatively prime, so this implies they are both cubes, say $g(x)-\sqrt{c}=h(x)^3$ and $g(x)+\sqrt{c}=i(x)^3$. Then $h(x)^3-i(x)^3$ is a constant so its factor $h(x)-i(x)$ is a constant, say $h(x)=i(x)+a$. But then $h(x)^3-i(x)^3=3ai(x)^2+3a^2i(x)+a^3$ is not a constant since $\operatorname{char}(k)\neq 3$, a contradiction.

Answer 2

Another way to see that this is possible in characteristic 2 or 3 but nowhere else is by using a little more algebraic geometry.

If we have polynomials $f(x),g(x)$ satisfying $f^3+g^2=c$ for $c\neq 0$, then this is a morphism $\Bbb A^1 \to V(t^3+u^2=c)\subset\Bbb A^2$ by sending $x\mapsto (f(x),g(x))$. This extends to a morphism $\Bbb P^1\to V(t^3+u^2v=cv^3)\subset\Bbb P^2$, and if $V(t^3+u^2v=cv^3)$ is smooth, then it's an elliptic curve and the map must be constant by Riemann-Hurwitz. So the only way this can happen is if $V(t^3+u^2v=cv^3)$ is singular, and indeed since every singular cubic is rational we can find such a morphism.

Taking the Jacobian matrix of $V(t^3+u^2v=cv^3)$, we get $$\begin{pmatrix} 3t^2 & 2uv & u^2-3cv^2 \end{pmatrix}$$ and we can split by characteristic:

• characteristic $2$: the Jacobian becomes $\begin{pmatrix} t^2 & 0 & u^2-cv^2 \end{pmatrix}$, so $[0:\sqrt{c}:1]$ is a singular point on the curve.
• characteristic $3$: the Jacobian becomes $\begin{pmatrix} 0 & 2uv & u^2 \end{pmatrix}$, so $[\sqrt[3]{c}:0:1]$ is a singular point on the curve.
• characteristic $\neq 2,3$: for there to be a singular point, we must have $t=0$ and at least one of $u$ or $v$ must be zero. But then $u^2-3cv^2=0$ implies $t=u=v=0$ and we have no singular points.