For which $f,g \in k[t]$, $k[f,g]$ is integrally closed?

Question

Let $f,g \in k[t]$, $k$ is a field of characteristic zero, $\deg(f)=n \geq 2$, $\deg(g)=m \geq 2$.

Is it possible to characterize all such $f$ and $g$ for which $k[f,g]$ is integrally closed in its field of fractions $k(f,g)$?

I do not mind to further assume that $k(f,g)=k(t)$ (but I do not want to assume that $k[f,g]=k[t]$, which makes my question trivial).

Thank you very much!

Edit: (i) This question is relevant.

(ii) If also $k(f,g)=k(t)$, then being integrally closed is equivalent to $k[f,g]=k[t]$. Indeed: (a) If $k[f,g]=k[t]$ then $k[f,g]$ is integrally closed (since $k[t]$ is). (b) If $k[f,g]$ is integrally closed then, in particular, since $t \in k(t)=k(f,g)$ (= the field of fractions of $k[f,g]$) and obviously $t$ is integral over $k[f,g]$ (for example, $t$ is a root of $f(T)-f \in k[f,g][T]$) hence $t \in k[f,g]$, so $k[f,g]=k[t]$.

For example: $f(t)=t^2+1$, $g(t)=t^3-4t$. We have, $t= \frac{t(t^2-4)}{t^2+1-5}=\frac{g}{f-5}$, so $k(f,g)=k(t)$. By Abhyankar-Moh-Suzuki theorem, $k[f,g] \neq k[t]$, so $k[f,g]$ is not integrally closed. Similarly, for every $f,g$ such that $k(f,g)=k(t)$ and one of the degrees does not divides the other, then $k[f,g]$ is not integrally closed.

(iii) If also $k(f,g)=k(t)$ and $k$ is algebraically closed, then being integrally closed is equivalent to $H(t)=(f(t),g(t))$ is injective and $H'(t)=(f'(t),g'(t)) \neq (0,0)$ for every $t \in k$. Indeed, $k[f,g]=k[t]$ is equivalent to the new conditions (it is important that $k$ is algebraically closed!!).

Edit: What if we replace $k$ by an integral domain $D$ (which is a $k$-algebra)? See also this question.

Answer 1

Here is an answer in the case where $k$ is algebraically closed. The parametric plane curve $(x,y) = (f(t), g(t))$ can be given implicitly as the vanishing locus of a polynomial $F \in k[x,y]$. Then $k[f,g]$ is integrally closed if and only if the affine plane curve $C: F(x,y) = 0$ is nonsingular. This occurs iff $F$, $F_x$, and $F_y$ don't vanish simultaneously, i.e., there is no $x_0, y_0 \in k$ such that $F(x_0, y_0) = F_x(x_0, y_0) = F_y(x_0, y_0) = 0$, where $F_x$ and $F_y$ denote the partials of $F$.

$k[f,g]$ is the coordinate ring for the plane curve $C: F(x,y) = 0$, and it is a standard result that nonsingularity corresponds to being integrally closed. (Cf., Corollary $13$ of $\S16.2$ of Dummit and Foote or Theorem I.5.1 of Hartshorne.)

Addendum: As you request in the comments, let $f(t) = t^2 +1$ and $g(t) = t^3 - 4t$. We form the ideal $I = (x - f(t), y - g(t))$ in the polynomial ring $k[t,x,y]$ with the elimination ordering $t > x > y$. Computing a Gröbner basis for $I$, we find that the elimination ideal $I \cap k[x,y]$ is generated by the polynomial $$ F = x^3 - 11 x^2 + 35 x - y^2 - 25 \, . $$ To determine if the curve $C: F = 0$ has any singular points, we compute a Gröbner basis for the ideal $(F, F_x, F_y)$ of $k[x,y]$, which produces $\{x - 5, y\}$. Thus the point $(5,0)$ is a singular point of $C$, so $k[f,g]$ is not integrally closed.

I used SageMath to compute all the above using the following code.

R.<t,x,y> = PolynomialRing(QQ,3,order="lex")
I = ideal([x - (t^2+1), y - (t^3-4*t)])
G = I.groebner_basis()
show(G)
F = G[-1]
J = ideal([F, F.derivative(x), F.derivative(y)])
show(J)
J.groebner_basis()

You can try it yourself in this SageMathCell.