Let $R$ be the ring $\begin{pmatrix} {\mathbb Z }_{ p } & 0 \\ {\mathbb Z }_{ p } & {\mathbb Z }_{ p } \end{pmatrix}$ with $p$ prime. $R$ is an artinian ring and is also a $\mathbb Z_p$-$\mathbb Z_p$ bimodule in an obvious way.
The lattice of ideals of $R$ is (where arrows go from bigger to smaller submodules):
$$ \begin{array}{ccccc} R & \rightarrow & Soc(R)=\begin{pmatrix} 0 & 0 \\ {\mathbb Z }_{ p } & {\mathbb Z }_{ p } \end{pmatrix} & \rightarrow & I=\begin{pmatrix} 0 & 0 \\ { 0 } & {\mathbb Z }_{ p } \end{pmatrix}\\ \downarrow & & \downarrow & & \downarrow \\ \begin{pmatrix} {\mathbb Z }_{ p } & 0 \\ {\mathbb Z }_{ p } & { 0 } \end{pmatrix} & \rightarrow & Rad(R)=\begin{pmatrix} 0 & 0 \\ {\mathbb Z }_{ p } & { 0 } \end{pmatrix}={ S }_{ 1 } & \rightarrow & 0 \end{array} $$
We have that R-simp={${ S }_{ 1 }=\begin{pmatrix} 0 & 0 \\ {\mathbb Z }_{ p } & { 0 } \end{pmatrix}, { S }_{ 2 }=R/\begin{pmatrix} 0 & 0 \\ {\mathbb Z }_{ p } & {\mathbb Z }_{ p } \end{pmatrix}$} and that $I$ is isomorphic to $S_1$.
The question is: why is $E(S_1)=\begin{pmatrix} {\mathbb Z }_{ p } & 0 \\ {\mathbb Z }_{ p } & { 0 } \end{pmatrix}$ ($E(S_1)$ denotes the injective hull of $S_1$ in $R$-mod) and why is $S_2$ already injective?
It is important to say that I really do not want to use Baer´s test.
I noticed that R/Rad is isomorphic to the direct sum of $S_1$ and $S_2$, so I thought that computing the injective hull of $R/Rad(R)$ and decomposing it as the direct sum of the injective hulls I look for then I could finish via Krull-schmidt (I've already checked the hypothesis required).
Lam's book says that if we have a finite dimensional k-algebra R, then the injective hull of $R/Rad(R)$ in $R$-Mod is ${ Hom }_{ k }(R,k)$ viewed as $R$-module, so all is reduced to compute ${ Hom }_{ \mathbb{Z}_p }(R,\mathbb{Z}_p)$ as $R$-module but I couldn't see how.
Does anyone knows how to compute it or something that could help?
Other ways or hints are welcome.
Throughout, I assume you are looking at left $R$-modules.
Since $R$ is a finite-dimensional $\mathbb{Z}_p$-module, then using the duality $$ D={\rm Hom}_{\mathbb{Z}_p}(-, \mathbb{Z}_p): {\rm mod}\ R^{op} \to {\rm mod}\ R, $$ where $R^{op}$ is the opposite algebra, is a good way to answer your question. More precisely, projective $R^{op}$-modules will be sent to injective $R$-modules by this functor $D$.
The rest of the post is simply a detailed computation of the images of the projective $R^{op}$-modules by $D$.
It is not hard to see that $R^{op}$ is isomorphic to the ring $\pmatrix{\mathbb{Z}_p & \mathbb{Z}_p \\ 0 & \mathbb{Z}_p}$, which decomposes as the direct sum of the two projectives $Q_1=\pmatrix{\mathbb{Z}_p & 0 \\ 0 & 0}$ and $Q_2=\pmatrix{0 & \mathbb{Z}_p \\ 0 & \mathbb{Z}_p}$. We need to compute the image of $Q_1$ and $Q_2$ by the duality functor $D$.
Clearly, $DQ_1$ is simple, since it is one-dimensional. To know whether it is isomorphic to $S_1$ or $S_2$, multiply by idempotent elements of $R$: if $f\in DQ_1$ and $\pmatrix{x & 0\\ 0 & 0}\in Q_1$, then $$ (\pmatrix{1 & 0 \\ 0 & 0}\cdot f)(\pmatrix{x & 0\\ 0 & 0}) = f(\pmatrix{x & 0\\ 0 & 0}\pmatrix{1 & 0\\ 0 & 0})=f(\pmatrix{x & 0\\ 0 & 0}) $$ and $$ (\pmatrix{0 & 0 \\ 0 & 1}\cdot f)(\pmatrix{x & 0\\ 0 & 0}) = f(\pmatrix{x & 0\\ 0 & 0}\pmatrix{0 & 0\\ 0 & 1})=f(\pmatrix{0 & 0\\ 0 & 0}) = 0, $$ so $\pmatrix{1 & 0 \\ 0 & 0}\cdot f = f$ and $\pmatrix{0 & 0 \\ 0 & 1}\cdot f = 0$. This means that the action of $R$ on $DQ_1$ is the same as that on the module called $S_2$ in your post. Thus $DQ_1$ is isomorphic to $S_2$.
We now have to compute $DQ_2$. Note that, since $Q_2$ is the projective cover of the simple module $S=R^{op}/rad(R^{op})$, we get that $DQ_2$ is the injective hull of $DS$, which is isomorphic to $S_1$. All that remains is to show that $DQ_2$ is isomorphic to $\pmatrix{\mathbb{Z}_p & 0 \\ \mathbb{Z}_p & 0}$. An explicit isomorphism is not hard to find.