# The category of (completable) rings has enough projectives in it

I am working on functors and projective resolutions and of course the issue of "Enough projectives" comes up. I know $R$-modules have enough but I am curious about the category of rings in general? Does it have enough? I could also add the question of rings that can undergo completion, does it have enough? Is there a name for this category on it's own right? My first feel is "completable rings"

The issue is I have tried to look for the answers but so far failed any resource for it and don't even know how one would go about proving it, so any resource that I can be geared toward along with the answer would be greatly appriciated.

• What do you mean by "rings that can undergo completion"? Filtered rings? – Najib Idrissi Jan 23 '16 at 13:47
• I am not familiar with what "Filtered rings" mean, I mean rings where we have a norm and we complete the ring such that any cauchy sequence converges to another element within the ring, like how we get reals/p-adic numbers from rational numbers – Zelos Malum Jan 23 '16 at 13:48
• Cauchy completion (of metric spaces) is not functorial with respect to continuous maps though, so you'll also have to say what the morphisms in your category are... – Zhen Lin Jan 23 '16 at 17:13
• The category of rings primarely at first, modules second if workable. – Zelos Malum Jan 23 '16 at 17:14

The category of rings (or more generally the category of algebras over a ring, a ring being a $\mathbb{Z}$-algebra) has enough projectives. Let $A$ be a ring. Since the category of abelian group is known to have enough projectives, there is a projective abelian group $P$ and a surjective group morphism $q : P \to A$. Now consider the tensor algebra on $P$: $$T(P) = \bigoplus_{k = 0}^\infty A^{\otimes k}$$ (the tensor product is over $\mathbb{Z}$). This is a ring: the product is concatenation of tensors $$(x_1 \otimes \dots \otimes x_n) \cdot (y_1 \otimes \dots \otimes y_m) := x_1 \otimes \dots \otimes x_n \otimes y_1 \otimes \dots \otimes y_m,$$ and the unit is the unit of $A^{\otimes 0} = \mathbb{Z}$. It maps into $A$ by sending $x_1 \otimes \dots \otimes x_n$ to $q(x_1) \dots q(x_n)$, the map being obviously surjective.

Moreover $T(P)$ is projective. Let $f : B \to C$ be a surjective ring morphism, and $g : T(P) \to C$ be a ring morphism. Compose $g$ with the injection $i : P = P^{\otimes 1} \to T(P)$ to get a group morphism $g \circ i : P \to C$. Since $P$ is projective as an abelian group, $g \circ i$ admits a lift $h : P \to B$ such that $f \circ h = g \circ i$. Now let

\begin{align} \bar{h} : T(P) & \to B \\ x_1 \otimes \dots \otimes x_n & \mapsto h(x_1) \dots h(x_n) \end{align}

Then this is obviously a ring morphism, and $f \circ \bar{h} \circ i = f \circ h = g \circ i$. Since a ring morphism $m : T(P) \to Z$ is uniquely determined by the abelian group morphism $m \circ i$ (this is easy to verify), it follows that $f \circ \bar{h} = g$, and so $T(P)$ is projective as claimed.

• hmmm got any documents elsehwere where I can read upo n this too? – Zelos Malum Jan 23 '16 at 14:04
• Weibel's book on homological algebra I guess. Is my answer not detailed enough? – Najib Idrissi Jan 23 '16 at 14:11
• Well it depends a bit on how one defined "detailed", the primary reason is I want a source and do you mean Weibel "Introduction to homological algebra"? The second reason that even related to detailness is that I like seeing things and check up on theorems and more so I can follow the line of reasoning and a book is important to me there so I can go back and forth, I am working through your answer – Zelos Malum Jan 23 '16 at 14:14
• Your $P$ is the resolution chain I take it and when you use $q$ as a morphism, what morphism is it exactly? – Zelos Malum Jan 23 '16 at 14:18
• $P$ is a projective abelian group and $q$ is a surjective morphism. It exists because the category of abelian groups has enough projectives. (Yes, I mean "Introduction to homological algebra"). – Najib Idrissi Jan 23 '16 at 14:29