Injectivity of $R \to R[t]/(f)$ for non-constant $f\in R[t]$ 
Question: Let $R$ be a (unital commutative) ring and $f = a_0 t^n + \cdots + a_n \in R[t]$ a non-constant polynomial. What are (necessary and sufficient) conditions on the coefficients $a_0,\ldots,a_n \in R$ for the injectivity of the canonical morphism $R \to R[t]/(f)$?

In the book Algorithmic Algebraic Number Theory by Pohst and Zassenhaus (1997), the following is claimed:

1.4, Exercise 1. [...] Show that the natural homomorphism [$R \to R[t]/(f)$] is a monomorphism, if and only if the coefficients of $f$ satisfy the following condition: If $a_0z = 0$ for $z \in R$, then also $a_iz=0$ $(1 \leq i \leq n)$. [...]

However, this condition seems wrong to me. Consider $R=\mathbb{Q}[X,Y]/(XY)$, $f = Xt^2 - t - 1$. This polynomial does not fulfill the condition (for $z=Y$), because $a_0z = XY = 0_R \in R$, but $a_1z = -Y \neq 0_R \in R$. However, the morphism $$\mathbb{Q}[X,Y]/(XY) \to \mathbb{Q}[X,Y,t]/(XY,Xt^2-t-1)$$ is injective, because $\{Xt^2 -t-1, XY, Yt-Y\}$ is an Elimination basis for $(XY,Xt^2-t-1)$.
 A: Injectivity of the canonical morphism is clearly equivalent to $fg\in R$ $\Rightarrow$ $fg=0$ for all $g\in R[t]$. For this we need:
(1) $\,\,I:=\bigcap_{\,i<n}\text{ann}_R(a_i)\subseteq\text{ann}_R(a_n)$
For else any constant $g$ in the intersection that satisfies $ga_n\ne0$ would give a non-zero element $a_n g=f g$ in the kernel. Condition (1) will guarantee injectivity in case:
(2)$\,\,I$ is a radical ideal of $R$ (or even just $b^n \in I\Rightarrow b \in I$)
(Note that (2) holds automatically if the ring $R$ is reduced.)
Indeed, let $g=b_0 t^m + \cdots + b_m\in R[t]$ with $b_0\ne0$ and $fg\in R$. If $m \geq 1$, one finds from the equations $\Sigma\,_{i+j=k}\,a_i b_j$ $=0$ for $0\leq k \leq n$ (putting $b_j:=0$ for $j>m$), representing the vanishing of $t^{n+m}$ through $t^m$, inductively that $a_i b_0^{i+1}=0$ for $0\leq i \leq n$, so that $b_0^{n} \in I$, hence $b_0 \in I$ by (2).
But then we can replace $g$ by $g-b_0 t^m$, which is of lower degree and has the same product with $f$.
By induction, there is a constant polynomial having the same product with $f$ as $g$, and by (1) that product must be zero.
