Divisible module which is not injective On searching for some example of divisible module but not injective, I come across one in T.Y.Lam, Lectures on Modules and Rings.

He considers the $\mathbb{Z}[x]$-module $M=\mathbb{Q}(x)/\mathbb{Z}[x]$, where $\mathbb{Q}(x)$ denotes the quotient fields of $\mathbb{Z}[x]$.

It's clear that $M$ is divisible, but how can I go about proving it's not injective? Can someone please give me a little push on this?
Thanks a lot,
And have a good day,
 A: Consider two $\mathbb Z[x]$-modules. The first is $Y = \mathbb Z[x]$ itself, and the second is the ideal $X = (2, x) \subseteq \mathbb Z[x]$. There's a natural injection $f: X \to Y$.
Now, define a $\mathbb Z[x]$-homomorphism $g: X \to \mathbb{Q}(x)/\mathbb{Z}[x]$ by:
$$
\begin{array}{rcl}
g(2) & = & [0],\\
g(x) & = & [1/2].
\end{array}
$$
If $\mathbb{Q}(x)/\mathbb{Z}[x]$ were injective, then there would exist a $\mathbb Z[x]$-homomorphism $h: Y \to \mathbb{Q}(x)/\mathbb{Z}[x]$ such that $g = h \circ f$. Then we have:
$$
\begin{array}{l}
[0] = g(2) = h(f(2)) = h(2) = 2h(1), \\
[1/2] = g(x) = h(f(x)) = h(x) = xh(1).\end{array}
$$
So, $h(1) \in \mathbb{Q}(x)/\mathbb{Z}[x]$ has the property that $2h(1)=[0]$ and $xh(1)=[1/2]$. It can be checked that there's no such element in $\mathbb{Q}(x)/\mathbb{Z}[x]$. This is a contradiction, so $\mathbb{Q}(x)/\mathbb{Z}[x]$ is not an injective $\mathbb{Z}[x]$-module.
Please, check this carefully, because I've already made a couple of stupid mistakes today.
