Zero polynomials, null polynomials and characteristic 2.

Question

The canonical example that a non-null polynomial might be identically zero is to take $x^2 - x$ over a field of characteristic 2.

I wonder if this feature is exclusive to characteristic 2, i.e., is it true that if char$F \neq 2$ then the only identically zero polynomial is the null polynomial?

Thanks in advance.

Answer 1

As long as a field is finite, say $F=\{x_1,\dots,x_n\}$, you can use the polynomial $$P(X) = (X-x_1)(X-x_2)\dots (X-x_n).$$

It is easy to see that each $x_i$ is a root of $P$, so as a function it is zero on $F$, but it has degree $n$.

Answer 2

First, your distinction between null and identically zero is nonstandard. If you told me a polynomial was either null or identically zero, my first thought would be what you mean by "null," i.e., all coefficients zero.

Instead, for what you are calling "identically zero," I would use $f(\alpha)=0$ for all $\alpha\in \Bbb{F}_2$.

Note (though you seem to be aware of this) that there is a distinction between a polynomial and a function, and that a polynomial defines many functions depending on the domain and the embedding of the coefficients into the domain. Thus it's weird to say $x^2+x$ is identically zero in $\Bbb{F}_2[x]$, because (for example) the induced function $\Bbb{F}_4\to \Bbb{F}_4$ is nonzero. $\Bbb{F}_4 \cong \Bbb{F}_2[x]/(x^2+x+1)$, so if $\theta = \overline{x} \in \Bbb{F}_4$, we see that $\theta^2+\theta = \overline{x^2+x} = 1 \in \Bbb{F}_4$.

Finally, for any finite commutative ring $R$, there is always a polynomial $g\in R[x]$ such that $g(\alpha) = 0$ for all $\alpha \in R$. I'll leave it to you to come up with the construction.