# Initial object in category of rings (with unity)

An object $$I$$ in a category $$\mathcal{A}$$ is initial if for each $$\mathcal{A}$$-object $$X$$, there is a unique morphism $$I\rightarrow X$$.

An object $$J$$ in a category $$\mathcal{A}$$ is final if for each $$\mathcal{A}$$-object $$X$$, there is a unique morphism $$X\rightarrow J$$.

These are definitions from Cohn's Basic Algebra.

Let $$\mathcal{R}$$ be the category of rings. In this category $$0$$ ring is final object, and $$\mathbb{Z}$$ is initial object.

Q. In defining rings, the author mentioned that it should contain $$1$$, multiplicative identity; it is not necessarily distinguished from $$0$$. The author call $$0$$ to be trivial ring (see p. 79, Section 4.1). In defining ring homomorphism, author says that the multiplicative identity should go to multiplicative identity. I am not getting then why the zero ring can not be initial object? The map $$0\mapsto 0\in R$$ is unique ring homomorphism from ring $$\{0\}$$ to any ring $$R$$, am I right?

• $0$ needs to be sent to $0$ and $1$ needs to be sent to $1$. If $0=1$ as they do in the trivial ring, then there is no way to map $0$ to both $0$ and $1$ of any non-trivial ring. – Derek Elkins Apr 2 at 7:42
• oho..this was the point I was really missing (mis-reading). – Beginner Apr 2 at 7:43

The function $$f:\{0\}\to R$$ with $$f(0) = 0$$ doesn't map the multiplicative identity of $$\{0\}$$ to the multiplicative identity of $$R$$ (unless $$R$$ is also a zero ring).
$$\Bbb Z$$ is the initial object of rings with multiplicative identity, as $$f:\Bbb Z\to R$$ with $$f(n) = n\cdot 1_R$$ is the unique unit-preserving ring homomorphism from $$\Bbb Z$$ to $$R$$.
• Shouldn't the homomorphism be $f(n) = n\cdot 1_R$? – Shervin Sorouri Apr 2 at 8:01