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My question pertains to the proof of Proposition 4.7 on page 203 of Jacobson's Basic Algebra II (the Dover edition). The proposition says the following: if $R$ is semi-simple (which for Jacobson means "is a subdirect product of simple rings") and (two-sided) Artinian, then $R$ is a "direct sum $\bigoplus_{i=1}^n R_i$" of simple rings $R_i$. I object to his use of the term "direct sum," when he certainly must mean direct product, and intends a ring isomorphism. In any case, this is at the heart of what I don't understand about his proof. Before I go into that, let me state the lemma he uses, with which I have no problem.

Lemma: Let $M$ be an Artinian module over some ring $R$ that is a subdirect product of simple modules. Then $M$ is a direct sum of finitely many simple submodules (i.e. semi-simple with only finitely many simple factors).

Again, the lemma is fine. Now for his proof of the proposition. He defines $M(R)$ to be the subring of $\mathrm{End}((R,+))$ (additive group of $R$) generated by the left and right multiplication maps $x\mapsto rx$ and $x\mapsto xr$ for $r\in R$. Fine so far. He remarks that the additive group of $R$ is a left $M(R)$-module for which the $M(R)$-submodules are precisely the (two-sided) ideals of $R$. I believe this. Now he says "[the] hypothesis that $R$ is semi-simple is equivalent to the following: $R$ as $M(R)$-module is a subdirect product of irreducible $M(R)$-modules." If $R$ is semi-simple, then there are maximal left ideals $I_j$ such that $R\hookrightarrow\prod_j R/I_j$, and each $R/I_j$ is certainly going to be a simple $M(R)$-module. But I don't see why the converse is true: why should a statement about $R$ as an $M(R)$-module translate to a statement about $R$ as a ring?

Granting the equivalence for now, we conclude that $R$ is a subdirect product of simple $M(R)$-modules. Since it is an Artinian $M(R)$-module (being an Artinian ring for which the $M(R)$-submodules are exactly the ideals), the lemma implies that $R$ is a direct sum of a finite number of simple $M(R)$-modules. Again, what has this to do with the ring structure of $R$? Is this $M(R)$ cooked up so that its modules all have ring structures? Am I missing the point entirely?

This proposition is crucial in Jacobson's proof of the the Artin-Wedderburn theorem (where he incidentally continues to use the term "direct sum" applied to rings when I think he intends "direct product," which is confusing, but this is beside the point), so I'd really like to understand it (or know whether or not it's flawed, and if it is, whether or not it can be salvaged).

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The key here is that $M(R)$ is generated by the left and right multiplication maps of $R$.

Let $L_r$ and $R_r$ denote left and right multiplication by $r \in R$.

$R$ is a left $R$-module under the natural action $r \cdot x = rx$ (action = left multiplication). Likewise, $R$ is a right $R$-module under the natural action $x \cdot r = xr$ (action = right multiplication). So $R$ can be viewed as a bimodule ($R$ acting on both the left and right). Being a submodule mean being closed under subtraction as well as the left and right action (left and right multiplication). Thus submodules = (two-sided) ideals.

The ring $M(R)$ encodes both of these actions of $R$ in one package. $R$ is naturally a left $M(R)$-module. Let $L_r \cdot x = L_r(x) = rx$. Likewise, $R_r \cdot x = R_r(x) =xr$. A submodule would be closed under subtraction and the $M(R)$-action. This is exactly the same as being closed under left and right multiplication. Again submodules = ideals.

Thus $R$ as a $R$-bimodule is the same as $R$ as a left $M(R)$-module.

Again, the key is that $M(R)$ generated by the left and right multiplication maps of $R$, so $R$'s ring structure is bound up in the structure of $M(R)$.

Now to address your other objection... You seem to have made a mental distinction between direct product and direct sum. For both of these there are two flavors: internal and external.

Most authors seemlessly move between internal and external products/sums since the isomorphisms are natural (in a technical "categorical" sense).

When Jacobson says that $R$ is a "direct sum". He means that $R$ is both an internal and external direct sum of subrings. Let me explain in the case of two subrings:

Suppose $R = A \oplus B$ (internal). Then the mapping $(a,b) \mapsto a+b$ gives an isomorphism from the external direct sum of $A$ and $B$ to the internal direct sum of $A$ and $B$. So an internal sum automatically gives an external one.

Conversely, if $R$ is isomorphic to an external direct sum of two rings, say $\varphi :R \to X \oplus Y$. Then defining $\varphi^{-1}(X\oplus \{0\})=A$ and $\varphi^{-1}(\{0\} \oplus Y=B$, we get (after a routine check) that $A \oplus B = R$. Thus $R$ is an internal direct product of $A$ and $B$.

In the end internal vs. external is more or less, for the most part, just a switch in notation.

Now sums and products are different monsters in general. However, direct sums and direct products are identical if there are only finitely many terms involved. Thus internal sums and internal products of finitely many subrings are exactly the same thing. The same goes for external sums and products of finitely many rings.

Moreover, for finite or infinite, external and internal sums are canonically isomorphic. The same goes for internal or external products.

Thus in the finite case everything is either exactly the same or canonically isomorphic, so there's no need to distinguish between the different constructions.

I hope this helps! :)

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  • $\begingroup$ Dear @Bill, Thanks very much for your thorough response. Unfortunately I'm still puzzled by the issue that, ostensibly, what Jacobson gets is an $M(R)$-module isomorphism between $R$ and a direct sum of simple $M(R)$-modules, but what he's claiming to have is an isomorphism of rings between $R$ and a direct product of simple rings. The lemma he invokes is manifestly about module isomorphisms. Also, just to clarify, the reason I object to using the term "direct sum" when one is talking about ring structures is that, while the direct sum of $R$-modules is the same as the direct product in the $\endgroup$ – Keenan Kidwell Jul 19 '15 at 23:41
  • $\begingroup$ case of finitely many factors, the direct product of finitely many rings is generally not the coproduct (i.e. categorical direct sum) of rings (I require rings to have $1$ and for ring maps to send $1$ to $1$, as does Jacobson, I think...I don't have the first volume of his book :)). $\endgroup$ – Keenan Kidwell Jul 19 '15 at 23:42
  • $\begingroup$ It's not an isomorphism of rings really. We have that $R$ is decomposed as a $M(R)$-module internally. So $R = I_1 \oplus I_2 \oplus \cdots$. This says that every element of $R$ can be written uniquely as a sum $r_1+r_2+\dots$ where $r_j \in I_j$. [This is what the lemma gives us.] Next, we have (from previous arguments or my answer above) that each $I_j$ is in fact an ideal of $R$. Therefore, $R = I_1 \oplus I_2 \oplus \cdots$ is a direct sum of ideals. Finally, irreducible = simple in this context. Done. $\endgroup$ – Bill Cook Jul 20 '15 at 16:47
  • $\begingroup$ I see what you mean about sums/products now... The direct sum / product, for rings $R_\alpha$, $\alpha \in I$ is defined as follows: The direct product is the cartesian product of $R_\alpha$'s given the natural coordinatewise operations. This is the product ring. If you take the subring where all but finitely many coordinates are 0 (this is a subring in the "rings without unity" sense it is not a subring in Jacobson's sense) you get an external direct sum of $R_\alpha$'s. So this isn't in the category of rings with 1 (unless we are only dealing with a finite collection of $R_\alpha$'s). $\endgroup$ – Bill Cook Jul 20 '15 at 16:53
  • $\begingroup$ On the module side there's no issue with identities. All that aside, in the end none of this matters in your current context since all products/sums are finite. :) $\endgroup$ – Bill Cook Jul 20 '15 at 16:53

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