Suppose a plane curve is given parametrically by $x=p(t),y=q(t)$, where $p,q$ are rational functions. I originally assumed that this means that the parametrized curve is algebraic, i.e. that it is the zero set of a polynomial in $x$ and $y$, but now I have doubts. Take for starters the case where $p,q$ are polynomials of degrees $m$ and $n$ respectively. In simple examples like $x=t^2+2,y=t^3-1$ the algebraic equation is easy to obtain, $(x-2)^3=(y+1)^2$, and the degree of the curve appears to be $\text{max}(m,n)$ generically (of course one can make up degenerate examples like $x=t^3,y=t^3$, where it drops). But what about $x=t^5+t^2+1, y=t^6-t+1$? It's not like we can solve for $t$ in radicals, plug in, and take powers to eliminate the radicals to get an algebraic equation in terms of $x$ and $y$. And even when that's possible it is not clear what the degree will come out to be after taking all the powers.

So my questions are: when are polynomially (or more generally rationally) parametrized plane curves algebraic? Is there an algorithm for finding the algebraic equation when there is one? Is there a way to find the algebraic degree without finding the equation? Is the degree generically $\text{max}(m,n)$ when the parametrizing polynomials have degrees $m,n$? I looked in Gibson's Elementary Geometry of Algebraic Curves, and Reid's Undergraduate Algebraic Geometry, but they only deal with converse questions, from equation to parametrization.

EDIT: After searching I found out that converting parametric equations into implicit ones is called implicitization (the opposite of parametrization), and apparently it is a big thing in computational geometry and applications because it provides an efficient way of determining if a given point lies on a given curve (or surface). Sederberg, Anderson and Goldman give an overview of elimination theory that in particular implicitizes rationally parametrized plane curves, and explain why the degree does not increase under implicitization (p.78).


1 Answer 1


Yes, such curves are algebraic. There are several ways to see this via abstract algebra (field extensions, transcendence degrees, etc.); an explicit equation can be found by elimination theory, specifically by writing $x=p(t)$, $y=q(t)$ as polynomial relations between $t$ and $x,y$ respectively, and computing the resultant of these two polynomials with respect to $t$.

For example, for $x = t^5 + t^2 + 1$, $y = t^6 - t + 1$ the polresultant command in gp soon gives

-x^6 + 11*x^5 - 50*x^4 + (6*y^2 - 10*y + 124)*x^3 + (-31*y^2 + 57*y - 186)*x^2 +

(52*y^2 - 103*y + 164)*x + (y^5 - 8*y^4 + 25*y^3 - 66*y^2 + 86*y - 71)

  • $\begingroup$ Thank you. Is the degree always the maximum of polynomial degrees (or less in degenerate cases), any simpler way to find it than going through the elimination process? Any elementary book on this that you can recommend? $\endgroup$
    – Conifold
    Commented Jan 15, 2016 at 2:03
  • $\begingroup$ You're welcome. Yes, for polynomials the degree is bounded as you surmised. "The elimination process" is basically either solving simultaneous equations or carrying out the Euclidean algorithm for polynomials in $t$; both are conceptually simple (and the former leads to the determinant formula), though I think there are also more intricate techniques that do such computations more effiicently in large degrees. Sorry, I rarely remember specific texts, and this is no exception, but the Wikipedia pages or other online sources my give you useful pointers for sources to learn more about this. $\endgroup$ Commented Jan 15, 2016 at 2:13
  • $\begingroup$ (I meant "solving simultaneous linear equations" [for the coefficients of the polynomials relation between $x$ and $y$], whence the determinant connection.) $\endgroup$ Commented Jan 15, 2016 at 2:48

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