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Be $K$ of characteristic zero and $f(x) \in K[x]$. Prove that $f'(x) = 0$ then $f(x)$ is a constant polynomial.

I know that the field of zero characteristic is a field where any sum of multiplicative identity element with itself, $1 + 1 + ... + 1$ may not result in the additive neutral element $0$. I doubt I can extract being $f'(x) = 0$.

These issues Algebra drive me crazy!

ajotatxe
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Danilo Santana
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  • Hint: If $f$ is not constant, say of degree $n > 0$, what will be the coefficient of the degree $n-1$ term of the derivative? – Tobias Kildetoft May 06 '15 at 13:17
  • If the characteristic of $K$ was $3$, then the derivative of $x^3$ is zero, but $x^3$ is not a constant polynomial. That is why they insist on zero characteristic. – Arthur May 06 '15 at 13:18

2 Answers2

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Let $f$ be a nonzero polynomial. Write $f(x)= \sum_{j=0}^n a_j x^j$ and assume $a_n \neq 0$. That is let $n$ be the degree.

Then $f'(x)= \sum_{j=1}^n ja_j x^{j-1}$. Now, if $n \neq 0$ then $na_n \neq 0$ (here you use that the characteristic is zero). The claim follows.

The argument can be expanded to show that $f'=0$ if and only if the only non-zero coefficients $f$ are those whose index is a multiple of the characteristic.

quid
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1

HINT:

Write

$$f(x)=a_0+a_1x+\cdots +a_nx^n$$

We have that

$$f'(x)=a_1+2a_2x+\cdots +na_nx^{n-1}=0$$

This is for all $x\in K$ so, in particular it holds for $x=0$, which gives $a_1=0$. You can iterate this process and argue (since $m\neq 0$ for every $m$) that every coefficient of index greater than zero equals zero.

Daniel
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  • While this is correct, it does mean the OP would also need to argue that polynomials over fields of characteristic $0$ are determined by their values after substitution. – Tobias Kildetoft May 06 '15 at 13:27
  • @TobiasKildetoft You're right. However I can't imagine another way to prove it without using substitution. (Comparing degrees?). – Daniel May 06 '15 at 13:31
  • Just look at the answer by quid. Polynomials are always determined by their coefficients (by definition). – Tobias Kildetoft May 06 '15 at 13:32
  • That's what I did: Determine the coefficients. However, quid's answer is far more elegant (degree comparison, at the end). – Daniel May 06 '15 at 13:34