I am trying to prove this category has an initial object. However, I cannot seem to think of the initial object. I think once I know the object I can make the proof.

Let $N$ be a fixed rng (ring without identity) and $C$ be the category whose objects are rng homomorphisms $f: N \rightarrow R$ where $R$ is a ring (has identity). The morphisms of $C$ are the commutative diagrams

$\space \space N \space \space \space \space \space \space \space \space \space \space \space \space \space \space N$

$f$ $\downarrow$ $\space \space$ $\longrightarrow$ $\space \space \downarrow g$

$\space \space R$ $\space \space \space \space \space \space \space \space \space \space \space \space \space S$

I apologize for my diagram. Basically, denoting the horizontal arrow by $\alpha$, we have that $\alpha : R \rightarrow S$ is a ring homomorphism that makes the diamgram commute. So, $\alpha(f(n)) = g(n)$.

Could you please help me think of the initial object in this category.


You're looking for a universal ring $\tilde{N}$ living over the rng $N$. This would probably be something like the image of $N$ under an adjoint to the forgetful functor $\mathsf{Ring}\to\mathsf{Rng}$. Consider the ring $\tilde{N}$ which is $N\oplus\Bbb Z$ as an abelian group, with multiplication given by $$ (r,n)(s,m) := (rs + ns + mr, nm). $$

You can check that this ring does indeed have an identity ($1_{\tilde{N}} = (0,1)$), and there's a natural map $N\to \tilde{N}$ given by $n\mapsto (n,0)$, so this is a good candidate for initial object in this category.

You essentially need to check that a map $f : N\to R$ induces a map $\tilde{f} : \tilde{N}\to R$ making the relevant diagram commute (and that it is unique). Try the map $\tilde{f} : (r,n)\mapsto f(r) + n\cdot 1_R$ (it should be checked that this is actually a homomorphism).

  • $\begingroup$ I thought maybe that since $Z$ is initial in the category of rings that I should try something $f: N \rightarrow Z$. Is this perhaps the right track? $\endgroup$
    – user392191
    Nov 23 '16 at 1:58
  • $\begingroup$ There's not necessarily a map $N\to\Bbb Z$. However, you might be able to find a universal ring over the rng $N$. I'll put a construction in the edit $\endgroup$
    – Stahl
    Nov 23 '16 at 1:59
  • $\begingroup$ So, then the unique $\alpha: N-tilda \rightarrow S$ is just the projection map from the first component? Which is forced to be unique? $\endgroup$
    – user392191
    Nov 23 '16 at 2:07
  • $\begingroup$ Careful, you want $(0,1)\in\tilde{N}$ to map to $1_R\in R$. So while $\tilde{f}(s,0) = f(s)$, you also need $\tilde{f}(0,1) = 1_R$, which forces $\tilde{f}(0,n) = n\cdot 1_R$, which forces $\tilde{f}(s,n) = f(s) + n$ if you want $\tilde{f}$ to be a homomorphism. $\endgroup$
    – Stahl
    Nov 23 '16 at 2:15

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