Question

Show that from: if I is a nil ideal in any ring R, then Mn(I) is nil for any n, follows: if I is a nil ideal in any ring R, then I[t] ⊆ rad R[t].

Answer 1

It suffices to show that, for any $a_i \in I$, $1 + a_1t + \dots + a_nt^n$ is invertible in $R[t]$. Let $\sum_{i} b_i t^i$ be the "formal" inverse of $1 + a_1t + \dots + a_nt^n$ in the power series ring $R[[t]]$. Then we have: $b_0 = 1$, $b_1 = -a_1$, $b_2 = -(b_1a_1 + a_2)$, $b_3 = -(b_2a_1 + b_1a_2 + a_3)$, etc. Let $A$ be the companion matrix.

By direct matrix multiplication, we see that, for $e = (0, \dots, 0, 1)$:

$$eA = (0, \dots, 1, -a_1) = (0, \dots, 0, b_0, b_1)$$ $$eA^2 = (0, \dots, 0, b_0, b_1, -b_0a_2 - b_1a_1) = (0, \dots, 0, b_0, b_1, b_2),$$ $$eA^n = (b_1, \dots, b_{n-1}, b_n)$$ $$eA^{n+1} = (b_2, \dots, b_n, b_{n+1}), \text{ etc.}$$

Also, we can check that the entries of $A^n$ "involve" only the $a_i$'s and not the element 1, so $A^n \in \mathbf{M}n(I)$. By the hypothesis, $A^n$ is nilpotent, so the equations for $eA^k$ above show that $b_k = 0$ for sufficiently large $k$. Therefore, the formal inverse for $1 + \sum_i a_i t^i$ lies in $R[t]$, as desired.