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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    $\begingroup$ I am not enthusiastic about the possibility of a simple answer. Consider the case $R = \mathcal{M}_n(F)$ for $F$ a field; we need the determinant of $f$ to be invertible (as a matrix of elements of $F[t]$) and this doesn't seem like an easy condition to generalize. $\endgroup$ – Qiaochu Yuan May 21 '12 at 19:15
  • $\begingroup$ Alright. I agree that this somehow shows that there is no simple answer. But perhaps at least some sufficient conditions can be found. $\endgroup$ – Martin Brandenburg May 21 '12 at 20:19
  • $\begingroup$ It might be true that $a_n$ needs to be nilpotent. $\endgroup$ – Qiaochu Yuan May 21 '12 at 20:54
  • $\begingroup$ Martin, @Qiaochu Yuan: Is the claim still true when $f\in R[x_i; i\!\in\!I]$ (i.e. $f$ has several variables) and $R$ is commutative unital? $\endgroup$ – Leo Apr 5 '13 at 18:16
  • $\begingroup$ @Leon: which claim? Perhaps you should ask this as a separate question. $\endgroup$ – Qiaochu Yuan Apr 5 '13 at 20:47

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.

  • $\begingroup$ Thank you. This convinces me that there won't be any nice criterion. $\endgroup$ – Martin Brandenburg Oct 25 '12 at 8:26
  • $\begingroup$ What do you mean by "under conjugation"? thanks. $\endgroup$ – Matemáticos Chibchas Jan 1 '17 at 20:17

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