Torsion module and its socle

Question

A torsion abelian group has nonzero socle and is an essential extension of it. Let $R$ be a commutative ring. If $M$ is a unital, torsion $R$-module with nonzero socle, is $M$ an essential extension of its socle?

Answer 1

Here is an example.

Example: Let $R=k[x,y]$ be a polynomial ring in two variables, and let $$M=R/yR \oplus R/(x,y)R = M_1 \oplus M_2.$$ Then y acts as 0 on M, but y is a regular element on R, so M is not only torsion but "bounded". M has socle $M_2$, but $M_1 \cap M_2 = 0$, so the socle $M_2$ is not essential.

Verifying the example is easier if you imagine M as an $S=k[x]$ module, then $M_1 \cong S$ obviously has zero socle as an S-module, and thus as an R-module. $M_2 \cong k$ is simple, so its own socle.

Here is my thought process (removing dead-ends on singular modules). Basically reduce to injectives, then reduce to R/P, and check the condition on R/P. Dim 2 rings are mostly fine, especially ones with lots of primes.

Lemma 1: An essential extension of a torsion-module is torsion.

Lemma 2: Every module has a maximal essential extension, namely its injective envelope.

Lemma 3: Over a commutative noetherian ring, every injective module is a direct sum of injective envelopes of $R/P$ where P is a prime ideal of R.

Lemma 4: A cyclic module $R/I$ is torsion if and only if I contains a regular element.

Lemma 5: A cyclic module $R/I$ has a non-essential socle if I is the intersection of infinitely many prime ideals.

Lemma 6: What we really need is a nonzero non-maximal prime ideal in a noetherian domain.