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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?

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    $\begingroup$ $0$ 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. $\endgroup$ – Derek Elkins Apr 2 at 7:42
  • $\begingroup$ oho..this was the point I was really missing (mis-reading). $\endgroup$ – Beginner Apr 2 at 7:43
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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 $R$.

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  • $\begingroup$ Shouldn't the homomorphism be $f(n) = n\cdot 1_R$? $\endgroup$ – Shervin Sorouri Apr 2 at 8:01
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    $\begingroup$ @ShervinSorouri Yes, of course. Thank you. $\endgroup$ – Arthur Apr 2 at 8:07
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The zero ring is a ring with only one element, an element which serves as both the multiplicative and additive identity for the ring. Ring homomorphisms must preserve the respective identities; that is, the additive identity of the domain maps to that of the codomain, and the multiplicative identity of the domain maps to that of the codomain.

Since the zero ring has only one element, and we can't send a single element to two elements, there is no morphism from from the zero ring to any object other than itself.

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