I've been working through the exercises of Auslander, Reiten, and Smalø's Representation Theory of Artin Algebras, and have gotten stuck on their Exercise 4.12, which asks:
Let $\Lambda$ be an Artin algebra such that $\Lambda/\mathfrak{r}\cong \operatorname{soc}(\Lambda)$. Prove that $\Lambda$ is self-injective.
where $\mathfrak{r}$ is the Jacobson radical of $\Lambda$.
Previously in the book, there is reasoning that resolves the problem in the case that $\Lambda$ is commutative. A result from earlier says that if $A$ is any semisimple $\Lambda$-module, and $P\to A$ is its projective cover, then it has an injective envelope $A\to D(P^*)$, where $A^* := \mbox{Hom}_\Lambda(A, \Lambda)$ as a right module and $D$ is the standard duality between $\Lambda$-mod and $\Lambda^{\operatorname{op}}$-mod.
So in particular, since $\Lambda\to \Lambda/\mathfrak{r}$ is a projective cover, we likewise have an injective envelope $\Lambda/\mathfrak{r}\to D(\Lambda^*)$. By assumption, we have an injective envelope $\mbox{soc}(\Lambda)\to D(\Lambda^*)$. On the other hand, for any module $A$, the injective envelope of $\mbox{soc}(A)$ factors through that of $A$, so there is an injective envelope $\Lambda\to D(\Lambda^*)$.
Now, $\Lambda^* := \mbox{Hom}_\Lambda(\Lambda, \Lambda)$ is just isomorphic to $\Lambda$ as a right $\Lambda$-module. If $\Lambda$ is commutative, we have that its length as a left module and its length as a right module are the same; since duality preserves length, we have $\ell(\Lambda) = \ell(D(\Lambda^*))$, which means the injective envelope $\Lambda\to D(\Lambda^*)$ is an isomorphism, and $\Lambda$ is injective.
However, Exercise 4.12 makes no assumption of commutativity. Will this reasoning still work in this case?
It would be nice to show that the lengths of any Artin algebra as a left and right module over itself are the same, but since that's not true of general Artin rings I'm not sure if I can hope for that -- is this true? Alternatively, it would suffice to show under this assumption that the isomorphism of $\Lambda/\mathfrak{r}$ and $\mbox{soc}(\Lambda)$ also holds for $\Lambda^{\operatorname{op}}$, since the reasoning above shows regardless of commutativity that the left length of $\Lambda$ is less than or equal to its right length, but I don't know how I could relate the left and right socles of a ring.
It's not true for a general Artin algebra that its length as a right module is equal to its length as a left module. For example, the Artin $\mathbb{R}$-algebra $$\pmatrix{\mathbb{R}&\mathbb{C}\\ 0&\mathbb{C}}$$ has length $3$ as a right module, but length $4$ as a left module.
But this doesn't matter if you just consider lengths over the centre of the Artin algebra instead of over the algebra itself. Suppose $\Lambda$ is an Artin $R$-algebra. Then $$\ell_R(\Lambda)=\ell_R(D(\Lambda^*)),$$ and so an injective $\Lambda$-module homomorphism (or even just $R$-module homomorphism) $\Lambda\to D(\Lambda^*)$ must be an isomorphism.