1
$\begingroup$

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?

$\endgroup$
  • 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
5
$\begingroup$

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...

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.