About modules over finite direct product of fields ( finite direct product of local rings is local ? )

Question

I'm thinking about a certain issue now and I'm confused so I'm asking a question.

Let $A$ be a commutative ring which is a finite direct prodcut of fields. Let $M$ be a $A$-modue. Then, by Every $A$-module is projective if and only if $A$ is a finite direct product of fields., $M$ is projective ( True? ). We can view $A$ as a finite direct product of local rings. If finite direct product of local rings is local ( True? C.f. Is direct limit of local rings a local ring? ), then since projective modules over local rings are free, $M$ is free $A$-module. But it seems that there is a counter example : A projective module over a finite direct product of fields. Where did mistake occur?

Answer 1

I think the breakdown in this line of reasoning is that the second question does not state a result that includes direct products. It concerns "the direct limit of local rings" which is a limit over a directed set, but that is not the case for a direct product.

Direct products are limits over discrete categories, meaning that they cannot satisfy the directedness condition (because "discrete" means there are no morphisms between distinct objects. When you have two distinct objects in a directed system, there must be morphisms to link them to a common thing above both.)

So the middle post is not actually aiming to say that the direct product of local rings is local, only that a direct system of local ring is local. That direct products of local rings are not local is easily shown to be false, as you find in Jyrki's comment here, and elsewhere in other solutions (although I could not find a question dedicated to the issue.)

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To add some context, "all right $R$ modules are projective" is a well-known equivalent definition of semisimple rings. Finite products of fields are just exactly the commutative semisimple rings. Incidentally, semisimple and local rings do not share a lot in common: the intersection of their classes is the class of division rings.

Answer 2

You are making things too complicated for yourself. Let $A = K_1 \times \dots \times K_n$ be a finite direct product of fields. You can verify quite directly that every $A$-module has the form

$$V_1 \oplus \dots \oplus V_n$$

where each $V_i$ is a $K_i$-vector space. These modules are always projective, but they are free iff each $V_i$ has the same dimension.

Edit: Here are some more details as a series of exercises.

1. Let $R, S$ be two rings and let $M$ be an $R \times S$-module. Let $e_R = (1, 0), e_S = (0, 1) \in R \times S$. Show that $e_R M$ is an $R$-module, $e_S M$ is an $S$-module, and $M = e_R M \oplus e_S M$ as $R \times S$-modules, where $R \times S$ acts via the projection onto $R$ on the first factor and via the projection onto $S$ on the second factor.

2. Generalize the previous exercise to a finite product $R_1 \times \dots \times R_n$.

3. Now consider a finite product $A = K_1 \times \dots \times K_n$ of fields. In terms of the above decomposition, what does the free $A$-module on a set $X$ look like?

The error, as rschwieb says, is that finite direct products are not an example of a direct limit, so the linked post about direct limits of local rings is irrelevant. $K_1 \times \dots \times K_n$ is never local for $n \ge 2$, and in fact has exactly $n$ maximal ideals, one for each factor. Geometrically $\text{Spec } K_1 \times \dots \times K_n$ is the disjoint union of $n$ points $\text{Spec } K_1 \sqcup \dots \sqcup \text{Spec } K_n$, and modules break up into modules corresponding to each of these points, which is a simple way of thinking about the above decomposition.

This is also an excellent advertisement for deprecating the terminology "direct limit" because, contrary to the name, direct limits are a colimit (while products are a genuine limit), in modern terminology. The modern term should be "directed colimit."