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If $K$ is a field of characteristic zero, then a polynomial $f(t) = \sum_{k = 0}^n a_k t^k0 \in K[t]$ with zero derivative $Df(t) = 0$ must be constant, because $Df(t) = \sum_{k = 1}^nka_kt^{k-1} = 0 \implies a_k = 0$, since $k \neq 0$. The book I'm reading (Algebraic Number Theory by Stewart and Tall) mentions that this is not true for finite fields. So I tried to make up a counterexample: in $\mathbb{F}_2$, the derivative of $f(t) = t^2$ is $Df(t) = 2 t = 0$. My question is: is $f(t)$ constant? I think the definition of polynomials as sequences gives $f(t) = (0, 0, 1, 0, \dots)$, but I am confused about the exponent $2$: if read modulo $2$, wouldn't it make $f(t) = t^2 = 1$?

hdecristo
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    The standard example is the polynomial $f=t^p-1$ with derivative $pt^{p-1}\equiv 0$, since $p=0$ in a field $K$ of characteristic $p>0$. – Dietrich Burde May 18 '25 at 17:00
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    The exponent is not "read modulo $2$": $t^2$ means $t \cdot t$, and this is not the same as $1$ – Robert Israel May 18 '25 at 17:06
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    While the answerers are probably aware of it, a distinction should be made between polynomials and polynomial functions. See for example this old thread or this for more discussion. The distinction is particularly crucial when the natural domain is a finite field. In that case it is best to think that we can also evaluate the polynomials at every element of every extension field. Much like we happily evaluate real polynomials at complex inputs. – Jyrki Lahtonen May 20 '25 at 10:56

2 Answers2

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A (univariate) polynomial over a commutative ring $K$ is an expression of the form $$ a_0 + a_1 t + a_2 t^2 + \cdots + a_n t^n $$ where the coefficients $a_0, a_1, \ldots, a_n$ are elements of $K$, and $t, t^2, \ldots, t^n$ are symbols which follow the usual rules of exponentiation.

So to address your confusion: The exponents are nonnegative integers and not elements of $K$. In $\Bbb F_2[t]$ is $f(t) = t^2 = t \cdot t $ and that is different from $t^0 = 1$. Therefore $f(0) =0$, $f(1) = 1$, and $f$ is not constant.

Martin R
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Your example is correct and it is also non-constant. Look at $f(0)$ and $f(1)$.

Paramanand Singh
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Utsabraj Sarkar
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    A non-constant polynomial can define a constant function, so this is not really necessary. – tomasz May 19 '25 at 16:00