求证环的下述性质等价：(1)任意$R$-模投射；(2)任意$R$-模内射

Hey there! Let's work through this classic ring theory problem from Dummit & Foote (page 403, Problem 6) step by step. We need to show that these two conditions on a ring $R$ are equivalent:

Condition (1): Every $R$-module is projective #

Condition (2): Every $R$-module is injective #

First, let's refresh key characterizations that make this proof straightforward—these are the "workhorse" definitions you'll want to lean on:

• An $R$-module $P$ is projective if and only if every short exact sequence of $R$-modules $0 \to A \to B \to P \to 0$ splits (i.e., $B \cong A \oplus P$).
• An $R$-module $I$ is injective if and only if every short exact sequence of $R$-modules $0 \to I \to B \to C \to 0$ splits.

We'll prove each implication directly using these splitting properties.

Proving $(1) \implies (2)$ #

Suppose every $R$-module is projective. Let $I$ be any arbitrary $R$-module—we need to show $I$ is injective.

Take any short exact sequence of the form:
$$0 \to I \xrightarrow{f} B \xrightarrow{g} C \to 0$$
By condition (1), $C$ is a projective $R$-module. From the projective module characterization above, any short exact sequence ending in a projective module splits. That means there exists a homomorphism $h: C \to B$ such that $g \circ h = \text{id}_C$, which implies $B = \text{im}(f) \oplus \text{im}(h) \cong I \oplus C$.

This is exactly the condition we need to conclude $I$ is injective! Since $I$ was arbitrary, every $R$-module is injective—so condition (2) holds.

Proving $(2) \implies (1)$ #

Now suppose every $R$-module is injective. Let $P$ be any arbitrary $R$-module—we need to show $P$ is projective.

Take any short exact sequence of the form:
$$0 \to A \xrightarrow{f} B \xrightarrow{g} P \to 0$$
By condition (2), $A$ is an injective $R$-module. From the injective module characterization, any short exact sequence starting with an injective module splits. That means there exists a homomorphism $h: B \to A$ such that $h \circ f = \text{id}_A$, which implies $B = \text{im}(f) \oplus \ker(h) \cong A \oplus P$.

This tells us the sequence splits, which by definition means $P$ is projective. Since $P$ was arbitrary, every $R$-module is projective—so condition (1) holds.

A Quick Side Note: Semisimple Rings #

If you've learned about semisimple rings, you can also frame this equivalence through that lens:

• A ring $R$ is semisimple if and only if every short exact sequence of $R$-modules splits.
• If every module is projective, every short exact sequence splits (so $R$ is semisimple), which in turn implies every module is injective.
• If every module is injective, every short exact sequence splits (so $R$ is semisimple), which in turn implies every module is projective.

This ties the two conditions together neatly, but the direct proof above works perfectly even without invoking semisimple rings explicitly.

Let me know if you want to unpack any of these characterizations or steps further!

内容的提问来源于stack exchange，提问作者RipCheck