Is $K $ a $K [X]$ injective module? Let $K $ be a field.  $K $ is a $K [X] $-module with $P\cdot a:= P (0)\cdot a , \forall P\in K [X], a\in K $.
Is $K $ a $K [X]$ injective module?
I know that $K $ is an injective $K [X]$- module iff $Ext^1_{K [X]}(K [X]/I,K)=0$ for any $I $ ideal of $K [X] $.
 A: A module $M$ over a principal ideal domain $R$ is injective iff it is divisible, i.e. for all $m \in M$ and all $ r \in R \backslash \{ 0 \}$, there exists an $m' \in M$ such that $rm' = m$. As egreg points out in the comments below, the forward implication actually holds under the weaker assumption that $R$ is a domain.
In this particular example, if we take $m = 1$ and $r = X$, then we cannot find such an $m'$, so the module is not injective.

Edit: If we want to prove non-injectivity of $k$ by computing an Ext module, we can certainly do so. I think it's worth working through this so that we can see the connection between the "divisibility argument" and the "vanishing Ext-module argument", which is quite striking.
First, I suggest we write $k$ as $ k[X]/(X)$, to emphasise its $k[X]$-module structure. 
A good Ext module to compute is $${\rm Ext}_{k[X]}^1(k[X]/(X), k[X]/(X)).$$
[I'll explain why when the dust settles.]
To compute this, we need to find a projective resolution of the left-hand $k[X]/(X)$. This is easy:
$$  \dots \to 0 \to  0 \to k[X] \overset{\times X}{\longrightarrow} k[X] $$
Next, we apply the ${\rm Hom}_{k[X]} (., k[X]/(X))$ functor to the projective resolution, giving:
$$ k[X]/(X) \overset{\times X}\longrightarrow k[X]/(X) \to 0 \to 0 \to \dots.$$
But the $\times X$ morphism is not surjective: its image is $0$. Hence
$$ {\rm Ext}_{k[X]}^1 (k[X]/(X), k[X]/(X)) = k[X]/(X) \neq 0.$$
Now here is the interesting point: The image of the $\times X$ morphism is precisely the submodule of elements of $k[X]/(X)$ that are divisible by $X$.  The fact that not all elements of $k[X]/(X)$ are divisible by $X$ is the reason why the Ext group is non-trivial. So that is the connection between the two arguments!
And how can we tell in advance that taking $I = (X)$ in ${\rm Ext}_{k[X]}^1(k[X]/I, k[X]/(X))$ would be a good choice for the purposes of proving that $k[X]/(X)$ is not injective? Because we know in advance that not every element in $k[X]/(X)$ is divisible by $X$!
