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,


1 Answer 1


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.

  • 1
    $\begingroup$ That looks good to me. Thank you very much. :* $\endgroup$
    – user49685
    May 2, 2013 at 15:57
  • 1
    $\begingroup$ Yes, your answer is absolutely correct, but some details are missing. For example you define $g$ by giving the images of generators that do not form a basis of $\langle 2,x\rangle$ (namely $2$ and $x$) , but then you have to mention that the map is well-defined, i.e. that $u(x)\cdot 2+v(x)\cdot x=0\implies u(x)[0]+v(x)[1/2]=0$. $\endgroup$ Sep 29, 2019 at 16:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .