I'm trying to prove that if a polynomial over a commutative ring is nilpotent, then all its coefficients are nilpotent.

Let $f= \sum_{i=0}^n a_i x^i$.

I'm using induction on the degree, and then by subtracting $f-\sum_{i=0}^{n-1} a_i x^i = a_n x^n$, we get that $a_n x^n$ is nilpotent.

I'm having a hard time showing this implies that $a_n$ is nilpotent, and the only thing i can think of is showing that $x$ is a unit in a larger ring of the formal polynomials of the form $\sum_{i=-m}^n a_i x^i$.

Am i missing some super-simple proof?

  • 3
    $\begingroup$ What is the leading coefficient of $f^m$? $\endgroup$ – Mariano Suárez-Álvarez Aug 25 '16 at 6:01
  • $\begingroup$ What do you mean by nilpotent? Do you mean "squares to 0", for both the polynomial and coefficients? Or are you weakening one and/or the other to "becomes 0 when raised to some positive integer power"? $\endgroup$ – J.G. Aug 25 '16 at 6:23
  • $\begingroup$ @J.G. the second. $\endgroup$ – user160823 Aug 25 '16 at 15:41
  • $\begingroup$ @MarianoSuárez-Álvarez thanks for reminding the obvious. $\endgroup$ – user160823 Aug 25 '16 at 15:44

More generally:

Proposition. Let $A$ be a commutative ring, and $A[[x]]$ be the ring of formal power series in $x$ over $A$. Assume that some power series $\sum_{i=0}^{\infty} a_ix^i \in A[[x]]$ is nilpotent. Then, all its coefficients $a_0, a_1, \ldots ,a_k, \ldots $ are nilpotent.

Proof. Let $P=\sum_ia_i x^i$, assume $P^k=0$.

The $0$-coefficient of $(\sum_i a_ix^i)^k=0$ is $a_0^k$, so $a_0$ is nilpotent.

Therefore $P-a_0$ is nilpotent as well, and since $x$ is a regular element of $A[[x]]$ it follows that $\frac{P-a_0}{x}$ is nilpotent. Thus $a_1$ is nilpotent. By induction...

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