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A ring $R$ is left-perfect if every left $R$-module $M$ admits a projective cover, i.e., an epimorphism $\varphi:P\to M$ with $P$ projective, such that $\ker \varphi\subset P$ is superfluous; i.e. for every submodule $N\subset P$, we have $\ker \varphi + N=P \Rightarrow N=P$.

There are some equivalences with this condition which are very interesting. However, I would like to stick with the formulation above.

Why is $\mathbb{Z}$ not a perfect ring?

Here's my attempt. Let's consider $\mathbb{Z}_2=\mathbb{Z}/2\mathbb{Z}$. Over $\mathbb{Z}$, projectives are free; let's then consider an epimorphism $\varphi:F\to \mathbb{Z}_2$ with $F$ a free abelian group.

Surjectivity guarantees the existence of $x\in F$ such that $\varphi(x)=1$. Then, $\varphi(3x)=1$ too, and in fact $F=\ker \varphi + \langle 3x\rangle$. Indeed, if $u\in F$, then:

  • If $\varphi(u)=0$, then of course $u\in \ker \varphi$ and it's done.
  • If $\varphi(u)=1$, then $u=(u-3x)+3x$ with $3x\in \langle 3x\rangle$ and $u-3x\in \ker \varphi$ wince $\varphi(u-3x)=\varphi(u)-3\varphi(x)=1-1=0$.

If $\ker \varphi$ were superfluous in $F$, it would mean $F=\langle 3x\rangle$.


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Dear Bruno: I'd state the argument as follows: Let $\varphi$ be an epimorphism from a free $\mathbb Z$-module $F$ onto $\mathbb F_2$. Assume that the kernel $S$ is superfluous, and let $f$ be an element of $F$ mapping to $1$. Then $F=S+\mathbb Zf$, and thus $F=\mathbb Zf$. So we can assume $F=\mathbb Z$, and thus $S=(2)$. But $\mathbb Z=(2)+(3)$, a contradiction. – Pierre-Yves Gaillard Feb 19 '12 at 16:04
Dear @Pierre: how is it that "...and thus $S=(2)$"? – Bruno Stonek Feb 19 '12 at 16:24
Dear Bruno: There is a unique $\mathbb Z$-linear epimorphism from $\mathbb Z$ onto $\mathbb F_2$, and its kernel is $(2)$. – Pierre-Yves Gaillard Feb 19 '12 at 16:38
@Pierre: Ah, I get it now :) Thank you! It seems the 3 was a red herring, completely unnecessary. Would you care to elaborate it into an answer, so I can upvote it? Or I could do it myself, whichever suits you best. – Bruno Stonek Feb 19 '12 at 16:46
Dear Bruno: Thanks for your kind invitation. I posted an answer. – Pierre-Yves Gaillard Feb 19 '12 at 17:10
up vote 2 down vote accepted

To prove that $\mathbb Z$ is not perfect, let $\varphi$ be an epimorphism from a free $\mathbb Z$-module $F$ onto $\mathbb F_2$, and assume by contradiction that the kernel $S$ is superfluous.

Let $f$ be an element of $F$ mapping to $1$.

Then $F=S+f\mathbb Z$ because $S$ is a maximal submodule of $F$.

As $S$ is superfluous, we have $F=f\mathbb Z$, and we can assume $F=\mathbb Z$.

Since $\{1\}$ is a basis of $\mathbb Z$ and $\varphi$ is onto, we have $\varphi(1)=1$, and thus $S=2\mathbb Z$.

But we also have $\mathbb Z=2\mathbb Z+3\mathbb Z$, in contradiction with the assumption that $S$ is superfluous.

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It is easy to see that no submodule of a free $\mathbb{Z}$-module can be superfluous, unless it is trivial, so in particular, $\ker\varphi=0$. This, however, is absurd, since that would give you an injection from a free module to $\mathbb{Z}/2\mathbb{Z}$, which certainly cannot happen.

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I added the (non?-)proof of the sum statement to the OP. However, I never really asserted that $\ker \varphi=0$... – Bruno Stonek Feb 19 '12 at 15:50
But I believe your second paragraph finishes the work. What I wrote above proves that $F\simeq \mathbb{Z}$ (it is free and generated by one element), and $\ker \varphi\subset \mathbb{Z}$ being superfluous implies $\ker \varphi=0$ (indeed, superfluous submodules of $\mathbb{Z}$ must be null, since $n\mathbb{Z}\subset \mathbb{Z}$ with $n\neq 0$ can't be superfluous: let $p\neq n$ be prime, and then $n\mathbb{Z}+p\mathbb{Z}=\mathbb{Z}$ with $p\mathbb{Z}\neq \mathbb{Z}$.) This is, as you say, absurd. Is this correct? – Bruno Stonek Feb 19 '12 at 16:00

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