# 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?

Related Calculus and Beyond Homework Help News on Phys.org
HallsofIvy
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
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
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
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.