# Nilpotent and invertible polynomials over noncommutative rings

Let $R$ be a noncommutative ring.

1) Prove or disprove: $a_0+a_1 x+\cdots+a_n x^n\in R[x]$ is nilpotent iff $a_i$ is nilpotent $\forall i$.

2) Prove or disprove: $a_0+a_1 x+\cdots+a_n x^n\in R[x]$ is a unit iff $a_i$ is nilpotent $\forall i\geq 1$ and $a_0$ is a unit.

I think this will be disproved. Can anyone give some counterexample? Or if this case is discussed before then give me the link.

1) Let $R=M_{2\times 2}(\mathbb Z)$. Then $a:=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $b:=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ are nilpotent, but $ab=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ is not (in fact is idempotent). Consider $f(X)=a+bX\in R[X]$. Since $a,b$ have nonnegative entries, this is also true for sums and products, i.e. all coefficients of $f(X)^n$ have nonnegative entries. Among others, the coefficient of $X^n$ in $f(X)^{2n}$ is obtained from one summand $(ab)^n\ne 0$ and other nonnegative summands, hence is nonzero. We conclude that $f$ is not nilpotent, disproving the first statement.
2) With the same $R$ let $g(X)=1-f(X)X=1-aX-bX^2$. If $g$ is a unit, then its inverse - if it exists - is of the form $h(X)=1+a_1X+a_2X^2+\ldots$ and the coefficients can be computed recursively: Assuming we have already found $a_1,\ldots, a_{n-1}$, we consider the coefficient of $X^n$ in the product $g(X)h(X)$, which should be $0$, to obtain: $$\tag1 a_n=aa_{n-1}+ba_{n-2}$$ (with $a_0=1$ and $a_{-1}=0$ understood). At least this shows that $h$ is unique, but it may fail to be a polynomial (i.e. it might happen that $a_n\ne 0$ infinitely often. Indeed, what we find is also the unique inverse of $g$ in the ring of formal power series $R[[X]]$. But in the ring of power series we also have that $$k(X)=\sum_{n=0}^\infty f(X)X^n$$ is a multiplicative inverse of $g(X)$. As in the first part, all $f(X)^n$ are nonzero and have nonnegative coefficients, hence there is no cancellation, i.e. $k(X)$ is a proper power series and not a polynomial. By uniqueness, $h(X)=k(X)$ is not a polynomial, i.e. $g(X)$ is not a unit in $R[X]$, thus disproving also the second statement.
• Why is $K(x)$ multiplicative inverse of $g(x)$? – Ri-Li Oct 27 '14 at 18:57
• 1) $a^2=b^2=0$, $ab+ba=1$, so $(a+bX)^2=X$, and therefore $(a+bX)^{2n}=X^n$ for $n\ge 1$. – user26857 Nov 26 '14 at 19:58
• 2) $a_1=a$, $a_2=b$, $a_3=1$, $a_4=a$, $a_5=b$, $a_6=1$, and so on. Moreover, $ba_n=0$, so $a_n=b$. On the other side, $aa_n+ba_{n-1}=0$, a contradiction. – user26857 Nov 26 '14 at 20:12