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Let $f = a_0 + a_1 t + \dotsc + a_n t^n$ be a polynomial over some nontrivial, possibly noncommutative ring $R$. When is $f$ invertible in $R[t]$?

When $R$ is commutative, the answer is well-known: $a_0$ is a unit and $a_1,\dotsc,a_n$ are nilpotent. There are direct element proofs for this, but there is also very nice proof which reduces the claim to integral domains $R$ by using that the radical of $R$ is the intersection of all prime ideals of $R$.

What is known about the noncommutative case? Clearly $a_0$ has to be a unit (apply the homomorphism $t \mapsto 0$). Since $a_0^{-1} f$ is invertible iff $f$ is invertible, we may therefore assume that $a_0 = 1$. If $f = 1 + a_i t^i$ for some $i$, it is still true that $f$ is a unit iff $a_i$ is nilpotent.

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    @QiaochuYuan: The claim that $\sum_ka_kx^k\in R[x_i;i\!\in\!I]$ is invertible iff $a_0$ is a unit in $R$ and all other $a_k$ are nilpotents, when $R$ is commutative. If this is not known, I'll ask it in a separate question, but I hoped this is already stated somewhere in the literature.2013-04-06

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An expansion of my comment. No condition exists which can be stated only in terms of the $a_i$ separately and which is invariant under conjugation. To see this, take $R = \mathcal{M}_2(\mathbb{Q})$ and $a_0 = I, a_1 = \left[ \begin{array}{cc} 1 & 1 \\\ 1 & -1 \end{array} \right], a_2 = I, a_3 = \left[ \begin{array}{cc} 0 & 0 \\\ 1 & 0 \end{array} \right].$

Then $a_0 + a_1 t + a_2 t^2 + a_3 t^3 = \left[ \begin{array}{cc} 1 + t + t^2 & t \\\ t + t^3 & 1 - t + t^2 \end{array} \right]$

has determinant $1$ and so is invertible in $R[t]$. (And $a_1, a_2$ are not nilpotent!) However, letting $b_3 = \left[ \begin{array}{cc} 0 & 0 \\\ -1 & 0 \end{array} \right]$

which is conjugate to $a_3$ we have $a_0 + a_1 t + a_2 t^2 + b_3 t^3 = \left[ \begin{array}{cc} 1 + t + t^2 & t \\\ t - t^3 & 1 - t + t^2 \end{array} \right]$

which has non-invertible determinant.

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    What do you mean by "under conjugation"? thanks.2017-01-01