# Proving Characteristics of Projections

Can anyone think of a trivial proof for the fact that the projection of any polynomial onto the xy-plane itself gives a polynomial? My professor and I were speculating about this, but could not discover a trivial proof for the fact. Does such a thing exist?

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
First, it would help if you would explain what you are talking about. A "polynomial" is not a geometric figure and so has no "projection". Do you mean the graph of a polynomial in x, y, and z.

Yes, that's exactly what I meant.

Bacle2
Science Advisor
I am not sure I understand: you have , I guess, a polynomial P(x,y,z) . Then evaluate P(x,y,0). This is a polynomial in (x,y).

But a projection onto the xy-plane isn't the same as letting the z-coordinate equal zero. It may even be that the polynomial doesn't intersect with the xy-plane.

Bacle2
Science Advisor
I'm sorry, it may be just me, but I'm not sure I understand your layout. Maybe a formal definition and/or some examples may help, if you could provide them. I understand, e.g., the projection of the unit sphere x2+y2+z=1 into the xy-plane to be the circle x2+y2=1 , but I don't understand well what you're doing.

HallsofIvy
Science Advisor
Homework Helper
No, the projection of the unit sphere, $x^2+ y^2+ z^2= 1$ onto the xy-plane is NOT the circle, $x^2+ y^2= 1$. It is the disk $x^2+ y^2\le 1$. That is NOT a polynomial function, it is not even a function.

The only interpretation I can put on this, so that it is true, is that a polynomial path in three dimensions, given by parametric equations, x= p(t), y= q(t), z= r(t), where p, q, and r are polynomials, projected to the xy-plane, which would indeed be x= p(t), y= q(t), z= 0, is a polynomial path in the xy-plane since p and q are still polynomials.

Bacle2
Science Advisor
Yes, my bad, H of Ivy, i was obviously wrong, I was thinking about something else, also trying to make sense of the OP.