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What are the projective objects in the category of rings with identity ?


  1. The only projectives I could find so far are $\{ 0\}$ and $\mathbb{Z}$.

  2. If $R$ is projective and $\mathbb{Z}\langle R\rangle$ is the free ring generated by the set $R$ then there is a surjective ring hom. $\kappa: \mathbb{Z}\langle R\rangle \to R$ and, by projectivity, a ring hom. $i: R \to \mathbb{Z}\langle R\rangle$ with $\kappa \circ i = id_R$. In particular $R$ has characteristic $0$ and is free of zero-divisors.

  3. Surprisingly, polynomial rings aren't projective. This is because the embeddings $D[x_1,...,x_n] \hookrightarrow Quot(D[x_1,...,x_n])$ ($D$ a comm. domain) are epimorphisms.

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$\{ 0 \}$ isn't projective, unless your ring homomorphisms aren't unital. I guess you are asking about projective objects with respect to the class of all epimorphisms; otherwise, if you are happy to talk about projective objects with respect to all regular epimorphisms, then the projective rings are precisely the retracts of freely-generated rings. – Zhen Lin Feb 15 '13 at 8:14
Ringhomomorphisms are supposed to preserve the identity and, indeed, I'm looking for projectives with respect to all epimorphisms. – Ralph Feb 15 '13 at 8:36

OK, think I got it: $\mathbb{Z}$ is the only projective.

For let $R$ be a projective. There's an embedding $i:R \hookrightarrow \mathbb{Z}\langle R\rangle$ by the comment above and if $R\neq 0$ (what I assume from now on), $\mathbb{Z} \le R$. Note that the inclusion $$\rho: \mathbb{Z}\langle R\rangle \hookrightarrow \mathbb{Q}\langle R\rangle$$ is an epimorphism.

Let $w \in\mathbb{Z}\langle R\rangle$ be in the image of $i$. Similar to polynomial rings, $w$ can be uniquely written $$w=\sum_{n\ge 0} \sum_{(r_1,...,r_n) \in R^n}a_{r_1,...,r_n}r_1 \cdots r_n$$ where only finitely many $a_{r_1,...,r_n} \in \mathbb{Z}$ are non-zero. Choose $b \in \mathbb{Z}$ with $b > \max |a_{r_1,...,r_n}|$ for all $(r_1,...,r_n) \in R^n$ and define a ring homomorphism $$f: \mathbb{Z}\langle R\rangle \to \mathbb{Q}\langle R\rangle,\;r \mapsto r/b\quad (r \in R).$$

By projectivity of $R$ there's a hom. $g: R \to \mathbb{Z}\langle R\rangle$ such that $\rho \circ g=f\circ i$. In particular, $$\text{im}(f \circ i) \le \mathbb{Z}\langle R\rangle.\tag{$\ast$}$$ Since $f(w)=\sum_{n\ge 0}\sum_{(r_1,...,r_n) \in R^n}\frac{a_{r_1,...,r_n}}{b^n}r_1 \cdots r_n$ is in $\mathbb{Z}\langle R\rangle$ iff $a_{r_1,...,r_n}=0$ for all $n>0$, $(\ast)$ implies $w=a_0\in \mathbb{Z}$. Hence $\mathbb{Z} \le i(R) \le \mathbb{Z}$, i.e. $R \cong i(R)=\mathbb{Z}$.

Conversely, it's trivial to check that $0,\mathbb{Z}$ are in fact projective. q.e.d.

Addendum (by Lord_Farin): As remarked in the comments, $0$ is not projective, because any homomorphism $0 \to B$ forces $B = 0$, while no homomorphism $A \to B$ with $A$ nonzero can admit $0 \to A$, as $A$ is nonzero.

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How can $0$ be projective? Any ring homomorphism to $0$ is surjective, but the only split epimorphisms to $0$ are isomorphisms. – Zhen Lin Feb 15 '13 at 12:58
I know what a projective object is. So let me spell it out for you: If we have a ring homomorphism $0 \to B$, then $B$ must be the zero ring. Then $f : A \to B$ is automatically surjective (hence epic), but there is no chance of getting a ring homomorphism $0 \to A$ unless $A = 0$ as well. But there are plenty of non-zero rings. So $0$ is not projective. – Zhen Lin Feb 17 '13 at 23:24
Why must $B$ be the zero ring ? – Ralph Feb 17 '13 at 23:46
@Ralph: Because a (unit-preserving) homomorphism $0 \to B$ sends $0$ to $1$. – Hurkyl Feb 17 '13 at 23:49
@Zhen Lin: Yes, of course you are right. Thanks for your explanation. – Ralph Feb 28 '13 at 16:25

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