# proving that if a polynomial is nilpotent, then all the coefficients are nilpotent

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?

• What is the leading coefficient of $f^m$? – Mariano Suárez-Álvarez Aug 25 '16 at 6:01
• 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"? – J.G. Aug 25 '16 at 6:23
• @J.G. the second. – user160823 Aug 25 '16 at 15:41
• @MarianoSuárez-Álvarez thanks for reminding the obvious. – user160823 Aug 25 '16 at 15:44

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...