Equivalent Definition of Projective Cover

Question

according to Rotman's text "Homological Algebra", he defines a projective cover as follows,

Now I want to show that if an epimorphism $\epsilon$ from $P$ to $M$ where $P$ is projective, is a projective cover of $M$, then $\text{ker}(\epsilon)$ is superfluous.
I know to show this, assume that $\text{ker}(\epsilon) + N = P$ where $N \subseteq P$ a submodule, then I need to show that $N = P$ but I don't know how to proceed next using the def of projective cover.
I know I need to choose some projective module $Q$ to fit in the diagram and use it. Initially, I want to choose $Q = N$ but then I need $N$ to be projective and this is ture if I know $N$ is a summand of $P$. Thank you!!

Answer 1

The result is simply not true.

For instance consider $\mathbb{Z\to Z/2}$ the canonical projection. Let $P$ be a projective $\mathbb Z$-module. Then it is free (projective $\mathbb Z$-modules are always free), so it's of the form $\mathbb Z^{(I)}$ for some $I$, and so any epic $\psi: P\to \mathbb Z/2$ is of the form $\bigoplus_I f_i$ where $f_i : \mathbb Z \to \mathbb Z/2$, where at least one of the $f_i$ is nonzero (and if it's nonzero, then it's the usual projection).

Therefore there is an epic lift $P\to \mathbb Z$.

However, the kernel is clearly not superfluous, as $2\mathbb Z + 3\mathbb Z = \mathbb Z$.

The kernel will be superfluous if and only if every lift is epic, which need not be the case, as $\mathbb Z\overset{3}\to \mathbb Z\to \mathbb Z/2$ shows.

For the claim where you assume all lifts are epic, an easy solution consists in picking a projective module $Q$ together with an epimorphism $f:Q\to N$. It then suffices to check that $\psi := \epsilon\circ f$ is an epimorphism (check this ! this is where you use $N+\ker(\epsilon) = P$), so that $\varphi = f$ is a lift, and by assumption it must be an epimorphism : $P=N$ (since the image of $f$ is contained in $N$ by definition)

Answer 2

Consider the alternative definition of a cover, which automatically implies the definition given by Rotman:

Definition. Let $\mathcal{F}$ be a class of $R$-modules. If $M$ is an $R$-module, a morphism $\phi:F\to M$ is an $\mathcal{F}$-cover if the following two conditions hold:

1. For any map $\psi:F'\to M$ with $F'\in\mathcal{F}$ there is a $\alpha\in\text{Hom}_{R}(F',F)$ such that $\psi = \phi\circ \alpha$.
2. If one sets $F'=F$ as in $(1)$, then $\alpha$ can only be chosen to be in $\text{Aut}(F)$.

Now, if you set $\mathcal{P}$ to be the class of projective covers, it can quickly be deduced that any projective cover will be projective and you therefore this is equivalent to Rotman's definition.

The other, more common definition is the following:

Definition An epimorphism $\phi:P\to M$ is a projective cover if $P$ is projective and $\text{Ker}(\phi)$ is superfluous.

Therefore your question is wanting to show that a morphism $\phi:P\to M$ is a projective cover (as in the second definition) if and only if it is a $\mathcal{P}$-cover, where $\mathcal{P}$ is class of projective modules. This result is true:

Theorem. Let $f:P\to M$ be a morphism with $P$ projective. The following are equivalent:

1. $\phi$ is a $\mathcal{P}$-cover;
2. $\phi$ is a projective cover.

Proof. The implication $(2\implies 1)$ is not very challenging. For the other direction, suppose that $P=Q+\text{Ker}(\phi)$. Then the restriction $\phi_{Q}:Q\to M$ is surjective so there is a map $\alpha:P\to Q$ such that $\phi = \phi_{Q}\circ\alpha$ since $P$ is projective, moreover one can see that $\phi = \phi\alpha$. But since $\phi$ is a $\mathcal{P}$-cover, it follows that $\alpha$ is an automorphism of $P$, but this forces $L=P$, and therefore $\text{ker}(\phi)$ is superfluous. Therefore $(1\implies 2)$ holds.

So what is a problem with the above answer? Well projective covers dont usually exist. In particular the above answer shows that $\mathbb{Z}_{2}$ does not have a projective $\mathbb{Z}$-cover. This is, in fact, the exact example Rotman gives in showing that projective covers don't usually exist.

Rings over which finitely generated modules have a projective cover are called semi-perfect rings, while rings over which every module has a projective cover are called perfect rings. Note that being perfect is equivalent to every flat module being projective, and every module over any ring has a flat cover.