What is the injective hull of the ring $R=\mathbb{R}[x]/(x^2)$?

Question

Let $R$ be a ring with unity and $M$ a right $R$-module. Recall that the injective hull $E(M_R)$ of $M_R$ is the maximal essential minimal injective extension of $M_R$.

Let $R=\mathbb{R}[x]/(x^2)$. What is $E(R_R)$ ?

I appreciate if the details are included in the answer.

Thanks in advance.

Answer 1

Actually, $R$ is already injective as a module over itself (i.e. it is a self-injective ring.)

Since there's only three ideals, you could just use the Baer criterion. If you have a homomorphism $I\to R$ where $I$ is an ideal, prove that it extends to a homomorphism $R\to R$.

If $I$ is $\{0\}$ or $R$ you are already done. The last ideal is $I=(x+(x^2))$. Suppose $f:I\to R$. The homomorphism is determined entirely by what it does with $x+(x^2)$. Considering annihilators, it must be mapped into $(x+(x^2))$. So say $f(x+(x^2))=\alpha x+(x^2)$. Let $g:R\to R$ be defined by $g(1+(x^2))=\alpha + (x^2)$. Then it's easy to check that $g$ extends $f$.

There's also a famous theorem about commutative Noetherian rings whose proper quotients are self-injective here:

Levy, Lawrence. "Commutative rings whose homomorphic images are self-injective." Pacific Journal of Mathematics 18.1 (1966): 149-153.

Being a Dedekind domain qualifies, and since $\mathbb R[x]$ is Dedekind, that quotient is self-injective.

Answer 2

The ring has Krull dimension 0 and the socle of $R$ is just $x$, so the rank of its socle is 1. Thus $R$ is an injective $R$ module, so $R$ is its own injective hull.

Here is a sketch of the proof of why krull dimension 0 and socle rank 1 means injective. (See theorem A.29 in 24 hours of local cohomology).

Since the ring is Noetherian and Krull dimension 0, it is Artinian. Thus the inclusion map $\text{soc}(R) \hookrightarrow R$ is an essential extension (prove this!). Since $\text{rank}_{\mathbb{R}}\text{soc}(R) = 1$, $\text{soc}(R) \cong \mathbb{R}$. Thus there is a map $\text{soc}(R) \to E_{R}(\mathbb{R})$. Since $E_R(\mathbb{R})$ is injective and $\text{soc}(R) \hookrightarrow R$, there is a map from $R$ into $E_R(\mathbb{R})$. Because $\text{soc}(R) \to E_R(\mathbb{R})$ is an essential extension and the composition of the map $ \text{soc}(R) \to R \to E_R(\mathbb{R})$ is injective, the map $R \to E_R(\mathbb{R})$ is injective. Thus $R$ sits inside $E_R(\mathbb{R})$. One then has the length of $E_R(\mathbb{R})$ is equal to the length of $R$ (see Corollary A.28 in 24 hours of local cohomology), which implies the two are in fact equal.