$Q(\mathbb{Z}[t]) / \mathbb{Z}[t]$ is not injective

Question

I am doing this exercise:

Let $R = \mathbb{Z}[t]$ and let $K$ be its fraction field. Show that the $R$ module $K/R$ is divisible but not injective.

I have done the divisible part, but I am stuck on the injective part, any hint ?

Answer 1

Divisible implies injective only if R is a PID. Here that is not the case, so the counterexample should reflect that.

Because the submodule $J_1=\langle 2\rangle\subset R$ is free, we have a homomorphism $f:J\to K/R$ defined by $f(2)=\dfrac1t+R$. We can glue this together with the zero $R$-module homomorphism from $J_2=\langle t\rangle$ to $K/R$, because $J_2\cap J_1=\langle 2t\rangle$ and $f(J_2\cap J_1)=0$. Thus we get a homomorphism from $I=J_1+J_2$ to $K/R$ that I denote by $f$ as well. Note that $I$ is not a principal ideal.

Non-injectivity of $K/R$ will follow, if we can show that $f:I\to K/R$ cannot be extended to a homomorphism of $R$-modules $\tilde{f}:R\to K/R$. $R$ is also a free module, so it suffices to prove that no compatible choice for $\tilde{f}(1)$ exists.

Can you do that?