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We define an integer polynomial as polynomial that has only integer coefficients. Here I am only interested in polynomials in two variables.

Example:

  • $P = 5x^4 + 7 x^3y^4 + 4y$

Note that each polynomial P defines a curve by considering the set of points where it evaluates to zero. We will speak about this curve.

Example:

The circle can be described by

  • $x^2 + y^2 -1 = 0$

We say two polynomials $P,Q$ are touching in point $(a,b)$ if $P(a,b) = Q(a,b) = 0$ and the tangent at $(a,b)$ is the same. Or more geometrically, the curves of $P$ and $Q$ are not crossing.

enter image description here

(The Figure was created with IPE - drawing editor.)

We also need a further technical condition. For this let $D$ be a ''small enough'' disk around $(a,b)$. Then $Q$ and $P$ define two regions indicated green and yellow. Those regions must be interior disjoint. Without this condition for $P = y-x^3$ and $Q=y$ the point $(0,0)$ would be a touching point as well. See also the right side of the figure. (I know that I am not totally precise here, but I don't want to be too formal, so that I can reach a wide audience.) (Thanks for the comment from Jeppe Stig Nielsen.)

Example:

  • $P = y - x^2$ (Parabola)
  • $Q = y$ ($x$-axis)

They touch at the origin $(0,0)$.

My question:

Does there exist two integer polynomials $P,Q$ that touch in an irrational point $(a,b)$? (It would be fine for me if either $a$ or $b$ is irrational)

Many thanks for answers and comments. Till

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    $\begingroup$ I believe the proper terminology for touching is that both curves osculate in $(a,b)$. $\endgroup$
    – Iwillnotexist Idonotexist
    Oct 15, 2016 at 3:24
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    $\begingroup$ I think that another terminology is "having a contact of order 1". $\endgroup$
    – Taladris
    Oct 15, 2016 at 3:47
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    $\begingroup$ Of course having the same tangent point and the same tangent line is not enough to say the curves "are not crossing" at that point, for example $y-x^3=0$, and $y=0$, "touch" but also intersect in $(0,0)$. $\endgroup$
    – Jeppe Stig Nielsen
    Oct 15, 2016 at 8:08
  • $\begingroup$ Two curves osculate if their tangents and curvatures are equal. $\endgroup$
    – Anton Sherwood
    Oct 18, 2016 at 19:32

3 Answers 3

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What about $(x^2-2)^2$ and $0$?

quad and 0

If you want both coordinates to be irrational, you can add something like $x^3$ to both.

quad and cubic

I hope this helps $\ddot\smile$

Images courtesy of WolframAlpha.

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  • $\begingroup$ Taken as a polynomial in two variables, $(x^2-2)^2$, has as "curve" (asker's terminology) two parallel (vertical) lines on each side of the coordinate axis and at distance $\sqrt{2}$. And in the same interpretation, the polynomial $0$ has as "curve" the entire plane. Same problem with the two polynomials of the other example. $\endgroup$
    – Jeppe Stig Nielsen
    Oct 15, 2016 at 7:59
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    $\begingroup$ @JeppeStigNielsen: Well, make it $(x^2-2)^2-y=0$, then. $\endgroup$
    – Ilmari Karonen
    Oct 15, 2016 at 10:50
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If you have an irrational tangent point between your curves, it must have at least another conjugate, and since tangent points are double points, by Bézout's theorem, the product of the degrees of the curves must be at least $4$.

Since dtldarek gave an example of a degree $4$ curve and a degree $1$ curve, let me give an example between two degree $2$ curves :

The circle $x^2+y^2= 1$ and the ellipse $17x² + 8y² + 12x = 4$ are tangent at $(-2/3, \pm \sqrt {5}/3)$

enter image description here

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    $\begingroup$ Hi merico, many thanks for your answer. I actually find your first comment way more interesting than the example itself. Can you elaborate on: What is a conjugate. What is a double point? And doest the statement also hold for polynomial surfaces? $\endgroup$
    – Till
    Nov 13, 2017 at 16:29
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    $\begingroup$ here, a conjugate of something is its image by any field automorphism of $\Bbb R$. If the object is defined by rational numbers (coefficients of its equation or coordinates of points etc) then it doesn't have any conjugate other than itself (because field automorphisms have to fix $\Bbb Q$). If it has a finite number of conjugates then the object is algebraic, and if not it is transcendental. The nice thing about conjugates is that it preserves all the algebraic properties your objects have, and since being tangent at a point is one, all the conjugates are points where both curves are tangent. $\endgroup$
    – mercio
    Nov 13, 2017 at 18:14
  • $\begingroup$ When you look at the intersection of two curves of degree $d_1$ and $d_2$, they will have (over $\Bbb C$) $d_1d_2$ points counted with multiplicity. Tangent points count double because not only the curve go through the same point, but they also have the same direction there (and if more derivatives agree, then the multiplicity increases as well) geometrically, you can see that if you move one the curves slightly, the tangent points can "split" into two intersection point, so that's why they count as double points of intersection. $\endgroup$
    – mercio
    Nov 13, 2017 at 18:18
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Here's a general way to find such examples where both curves are of the form $y=f(x)$. Notice that $y=f(x)$ and $y=g(x)$ meet at a given value of $x$ iff that value of $x$ is a root of the polynomial $h(x)=f(x)-g(x)$, and they have the same tangent line iff that value is a root of $h(x)$ of multiplicity greater than $1$.

So this means that to find an example, we just need a polynomial $h(x)$ with integer coefficients that has a double root at some irrational value of $x$ (we can then take $g(x)$ to be any polynomial with integer coefficients at all, and $f(x)=h(x)+g(x)$). This is easy to do: just take any polynomial $p(x)$ with integer coefficients and an irrational root, and let $h(x)=p(x)^2$.

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  • $\begingroup$ This also shows that both coordinates must be algebraic. $\endgroup$
    – Kevin
    Oct 16, 2016 at 17:35

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