1

Let $\mathbb{F}$ be a field and let $P:\mathbb{F} \to \mathbb{F}$ be a polynomial (i.e, there exists a finite sequence $a_{0},\dots ,a_{n}$ of scalars in $\mathbb{F}$ such that $P(z)=\sum_{i=0}^{n}a_{i}z^{i}$ for all $z$ in $\mathbb{F}$). How can one prove that the degree of polynomial is unique or in other words ($a_{n} \neq 0,b_{m} \neq 0$ and $\sum_{i=0}^{n}a_{i}z^{i}=\sum_{i=0}^{m}b_{i}z^{i}$ for all $z \in \mathbb{F} \implies n=m $ )

Problem that I am facing:

If $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ then it's quite easy because each of these fields are infinite so the polynomial $\sum_{i=0}^{n}a_{i}z^{i}-\sum_{i=0}^{m}b_{i}z^{i}$ cannot have infinite roots due to Fundamental Theorem of Algebra but what if $\mathbb{F}$ is finite.

Anonymous
  • 75

1 Answers1

1

You don't, because $x^p\equiv x \bmod p$ in the field with $p$ elements, so the polynomials $x^p$ and $x$ are equivalent in $\mathbb F_p[x]$ using the test you have given (though the polynomials are distinct as polynomials and have different degrees as polynomials).

So how are you defining the degree of a polynomial?

Mark Bennet
  • 100,194
  • Thanks a lot. Ok, I get it so we define the degree of a polynomial for only infinite fields. – Anonymous May 11 '20 at 17:32
  • @Anonymous No, the degree of a polynomial is well-defined over a field - how it is defined depends on how polynomials are constructed - but any sensible constructions and definitions are equivalent. What is happening here is that different polynomials over a finite field can represent identical functions, And that is obvious because there are only a finite number of functions from a finite field to itself, while there are infinitely many polynomials. But a polynomial is not just a function. – Mark Bennet May 11 '20 at 17:40