Proving Characteristics of Projections

[TranscendArcu]

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?

 

[HallsofIvy]

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.

 

[TranscendArcu]

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).

 

[TranscendArcu]

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.

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]

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