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Why degrees are equal if polynomial are equal? 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?

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Take nth derivatives of each side and let $x$ tend to zero, if there equal all there coefficients will match. – Ethan Jun 16 '13 at 7:16
It depends on your definition of equality of polynomials. One requires the equality of all coefficients, in which case there should be no difficulty in seeing why the statement holds. The other is to require that the values at all $x$ are equal. Now it depends upon the field you are working in. For axample, in a finite field of $p$ elements, where $p$ is a prime, the two polynomials $x^p-x$ and $0$ are equal, since their values coincide. – awllower Jun 16 '13 at 7:17
The "correct" definition of equality of polynomials is that their coefficients are all equal. Then the statement should be clear. – Qiaochu Yuan Jun 16 '13 at 7:24

First assume our field is infinite, like $\mathbb{R}$ or $\mathbb{C}$. 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 $\mathbb{Z}/p\mathbb{Z}$ , $p$ prime, we have $x^p - a = x - a$, since $x^p = x$.

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If $f(x) = g(x)$ for all $x$ then the polinomial $f(x)- g(x)$ has infinitely many roots, hence it equals $0$.

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But OP did not specify what field he is working in! Maybe he is referring to a finite field? – awllower Jun 16 '13 at 7:18
The Op is asking questions about basic calculus, I don't think he is referring to a finite field – Ethan Jun 16 '13 at 7:19
OK. Per chance I mis-understood something, haha. – awllower Jun 16 '13 at 7:21
@ Ethan: Thank you, I thought the same. – Boris Novikov Jun 16 '13 at 7:21

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