There is no projective covers for $\mathrm{Frac}(D)=K$ over integral domain $D$ which is not a field.

Question

Let $D$ be an integral domain which is not a field and $K=\operatorname{Frac}(D)$. I want to prove that $K$ does not have a projective cover as $D$-module. My idea is to use the fact that if a module $M_{R}$ is projective and injective, then $M_{R}=\lbrace 0 \rbrace$. This fact is proved here

A problem about an $R$-module that is both injective and projective.

Also, it can be proved that $K$ is an injective $D$-module. So I study the particular case of $\mathbb{Q}_{\mathbb{Z}}$ which is not projective. My idea is to prove that if there is a projective cover $\rho:P \to K$, then we must get that $P \cong K$ so that $K$ must be projective and injective, then $K= \lbrace 0 \rbrace$. Any help in order finish up this problem?

Answer 1

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I think you're missing the following lemma:

Let $M$ be a flat module. $M$ has a projective cover iff $M$ is projective.

Remark: this is a special case of the fact that a projective cover of a module, when it exists, is equivalent to the flat cover of that module (which always exists).

The following can be done in far more generality, cf Wisbauer's Foundations of Module and Ring Theory, chapter $7$. I just wanted to give a short mostly self-contained proof here.

First let me make sure that we're on the same page about

Definitions An exact sequence $\mathcal{C} = 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ of $R$-modules is called pure exact if it satisfies either of the following equivalent conditions: (1) $N \otimes_R \mathcal{C}$ is exact for all $R$-modules $N$ (2) $\operatorname{Hom}_R(M, \mathcal{C})$ is exact for all finitely presented modules $M$. (See Lam's Lectures on Modules and Rings, 4.89 for details). An $R$-module $M$ is called pure projective if for every pure exact sequence $\mathcal{C}$, $\operatorname{Hom}_R(M, \mathcal{C})$ is exact. (See Stacks Definition 46.8.1 and Wisbauer 33.6 for equivalent conditions, e.g. that pure exact sequences ending in $M$ split). A submodule $N \subseteq M$ is superfluous if for any other submodule $H$ of $M$, $N + H = M \implies H = M$. A projective cover of an $R$-module $M$ is a projective module $P$ together with a surjection $P \rightarrow M$, the kernel of which is a superfluous submodule of $P$.

A couple notes about these definitions. If $N$ is a superfluous direct summand of $M$, then $N$ is clearly $0$. A projective module is clearly pure projective, and using characterization (2) of pure exact sequences, it becomes apparent that a finitely presented module is pure projective too. It's easy to see that pure projective modules are closed under direct sums.

Lemma: Let $K$ be a finitely generated submodule of a projective module $P$. Then $P/K$ is pure projective.

Proof: Suffice it to assume that $P$ is free, as the proof generalizes easily. Note that $M$ is contained in a finite free submodule $F'$ of $F$, and that $F/F' \cong (F/K)/(F'/K)$ is a free module and that $F'/K$ is finitely presented. Thus $0 \rightarrow F'/K \rightarrow F/K \rightarrow (F/K)/(F'/K) \rightarrow 0$ splits, and hence $F/K$ is a direct sum of a free module and a finitely presented module, so is pure projective. $\square$

Actually "projective" could have been replaced with "pure projective" in the above, but a little more work would need to be done. You can refer to Stacks Lemma 46.8.2 or Wisbauer 33.6 for the fact that pure projectives are direct summands of finite sums of finitely presented modules, and then proceed similarly.

Lemma: Let $0 \rightarrow K \rightarrow P \rightarrow N \rightarrow 0$ a pure exact sequence where $K$ is a superfluous submodule of $P$ and $P$ is projective. Then $K = 0$.

Proof Let $K'$ be any finite submodule of $K$.
We get a commutative diagram with exact rows as follows: \begin{CD} 0 @>>> K' @>>> P @>>h'> P/K' @>>> 0 \\ & @VVaV @VV=V @VVcV \\ 0 @>>> K @>>> P @>>h> P/K @>>> 0 \end{CD}

where $c$ is surjective, $a$ is injective. The previous lemma shows that $P/K'$ is pure projective, so by definition the map $c$ lifts to a map $d$ such that $hd = c$. Deduce from the superfluousness of $K \subseteq P$ and the surjectivity of $c$ that $d$ is surjective. Note, using commutativity of the diagram, that the kernel of $dh'$ is a submodule of $K$, and hence is a superfluous submodule of $P$. The map $dh': P \rightarrow P$ is a surjection onto a projective module, so it splits, i.e. the kernel of $dh'$ is a (superfluous!) direct summand of $P$, so it must be $0$. Since $K'$ is contained in the kernel of $dh'$, in fact $K'$ is $0$. But $K'$ was an arbitrary finite submodule of $K$, so we conclude $K = 0$. $\square$.

Main point If $M$ is flat and has a projective cover $P$, then $M$ is projective.

Proof: Suppose that $P$ is a projective cover of a module $M$. By definition, that means there is an exact sequence $0 \rightarrow K \rightarrow P \rightarrow M \rightarrow 0$, where $K$ is a superfluous submodule of $P$. If $M$ is flat, then this is a pure exact sequence (Stacks Lemma 10.38.12). By the preceding lemma, $K$ is in fact $0$, and $P \cong M$. $\square$

From here you can solve your problem in a few different ways. You want to show that the field of fractions of a non-field domain doesn't have a projective cover. Equivalently, according to the above, you want to show that the field of fractions is never projective.

Method (1) You know that modules can't be both injective and projective, and you know that the field of fractions is injective.

Method (2) Observe that if an extension of rings $R \subseteq T$ makes $T$ a projective $R$-module, then $T$ is even faithfully flat. The key is that $T_\mathfrak{p} \not= 0$ for any prime $\mathfrak{p}$ of $R$, and then you can argue as in the comments to this post. Now another feature of the field of fractions is that, being a localization, it is an epimorphism in the category of commutative rings. A faithfully flat epimorphism is always an isomorphism (Stacks Lemma 10.106.7).

As a corollary to Method (2) we can state:

Let $R \subsetneq T$ be a flat epimorphic extension of rings. Then $T$ does not have a projective cover. In particular this holds for any $T = S^{-1}R$ where $S$ is a multiplicative set generated by non-zerodivisors of $R$.