Why degrees are equal if polynomial are equal?

1. Jun 16, 2013

Avichal

Two polynomial f(x) and g(x) are equal then their degrees are equal.
This is a very trivial statement and it shouldn't worry me much but it is.

I get an intuitive idea why they should be equal. Their graphs wouldn't coincide for unequal degrees.
But what if somehow the coefficients make f(x) = g(x) for all values of x?

Is there a more rigorous proof for this statement?

2. Jun 16, 2013

HomogenousCow

??? if f(x)=g(x) then of course they are of the same degree, the two functions are equal.
I'm not quite sure what you're asking, if two functions are the same then everything about them are equal.

3. Jun 16, 2013

rattanjot14

First assume our field is infinite, like R or C. The difference between two polynomials is a polynomial, so f−g is a polynomial. Since f(x)=g(x) for all x, this means that f−g have infinitely many zeros, whence f−g=0.

Now, for Z/pZ , p prime, we have x^p−a=x−a, since x^p=x

4. Jun 16, 2013

Avichal

Yes, I know its obvious. But I was looking for a proof.
Thanks for the proof.
By your last line are you suggesting that this isn't always the case.

5. Jun 16, 2013

jbriggs444

If you are using the two-element field (i.e. GF[2]) then the polynomial function f(x) = x3 + 1 and the polynomial function g(x) = x + 1 both have the same graph:

0: 1
1: 0

As formal polynomials they have different degrees. But as polynomial functions they have the same graph.

Last edited: Jun 16, 2013
6. Jun 16, 2013

Bacle2

Assuming fields of characteristic zero , re the cases brought up above by jbriggs and rattan, maybe you can use the Fundamental Theorem of Algebra (considering complex roots ), to argue that both must have the same roots and therefore the samenumber of roots.

Otherwise,how about this: if f=g, then f-g =0 is the zero polynomial ( not just the number zero). But every
non-zero polynomial has only finitely-many roots, while the zero polynomial (over char. zero ) does not. Then
f-g must be the 0 polynomial:

0*x^n +0*x^{n-1)+....+0*x+0.