Categorical Perspective on Kernels, Rings, and Rngs: Exploring Zero Objects and Adjunctions

Kernel in Category with Zero ObjectWhy the Ideal of a Ring is Not a SubringRngThe Free-Forget Adjoint Between Rng and Ring

Kernel in Category with Zero Object

Let $\mathcal{C}$$\mathcal{C}$ be a category with a zero object, denoted as $0$$0$. For convenience, we assume that equalizers exist in $\mathcal{C}$$\mathcal{C}$.

Definition. For any $X,Y\in \mathrm{Ob}(\mathcal{C})$$X,Y \in \mathrm{Ob}(\mathcal{C})$, there exists a unique $0$$0$ morphism $0:X\to Y$$0: X \to Y$. It is defined by the composition:

$$\begin{array}{}\text{(1)}& X⟶0⟶Y\end{array}$$

It is unique since $0$$0$ is both an initial and terminal object.

Definition. Let $f:X\to Y$$f: X \to Y$ be a morphism in $\mathcal{C}$$\mathcal{C}$, we define $\ker (f):=\mathrm{Eq}(f,0)$$\mathrm{ker}(f) := \mathrm{Eq}(f,0)$.

Example. Let ${\mathrm{Set}}_{\ast }$$\mathrm{Set}_*$ be the category of sets with base point, i.e., the coslice category $\ast /\mathrm{Set}$$*/\mathrm{Set}$. Then it is easy to see that $\{\ast \}$$\{*\}$ is the $0$$0$ object. For $f:X_{\ast }\to Y_{\ast }$$f: X_* \to Y_*$, the kernel of $f$$f$ is equal to $f^{-1}(\ast )$$f^{-1}(*)$.

Lemma. Let $0:X\to Y$$0: X \to Y$ be the zero morphism. Then for any $f:X^{\prime }\to X$$f: X' \to X$, $g:Y\to Y^{\prime }$$g: Y \to Y'$, $0\circ f=0$$0 \circ f = 0$, $g\circ 0=0$$g \circ 0 = 0$.

Proof.

$$\begin{array}{}\text{(2)}& X^{\prime }⟶X⟶0⟶Y⟶Y^{\prime }\end{array}$$

Proposition. $f$$f$ is a monomorphism implies $\ker (f)=0$$\mathrm{ker}(f) = 0$.

Proof. If $f$$f$ is mono, and $f\circ g=0\circ g=0=f\circ 0⟹g=0$$f \circ g = 0 \circ g = 0 = f \circ 0 \implies g = 0$, i.e., the unique arrow equalizing $f$$f$ and $0$$0$ is $0$$0$. $◻$$\square$

Why the Ideal of a Ring is Not a Subring

Notation. We will use $\ker (f)$$\mathrm{ker}(f)$ to refer to both the object and the morphism at the same time.

The reason is that $\mathrm{Ring}$$\mathrm{Ring}$ has no zero object. When we define ring homomorphisms, we require that $f:R\to S$$f: R \to S$, $f(1_R)=1_S$$f(1_R) = 1_S$. Since $B(f)$$B(f)$ should be an additive functor from $B(R)\to B(S)$$B(R) \to B(S)$. You should view $B$$B$ as a functor from $\mathrm{Ring}$$\mathrm{Ring}$ to $\mathrm{Cat}$$\mathrm{Cat}$. The fact that $f(1_R)=1_S$$f(1_R) = 1_S$ implies that $\mathbb{Z}$$\mathbb{Z}$ is the initial object in $\mathrm{Ring}$$\mathrm{Ring}$. You could use this fact to prove Fermat's Little Theorem.

Indeed, for a ring homomorphism $f$$f$, the $\ker (f)$$\mathrm{ker}(f)$ is the equalizer in ${\mathrm{Set}}_{\ast }$$\mathrm{Set}_*$. Suppose $\ker (f)$$\mathrm{ker}(f)$ is a subring of $R$$R$, then $f\circ \ker (f)=0$$f \circ \mathrm{ker}(f) = 0$. Hence $f(1_R)=0$$f(1_R) = 0$, therefore $\ker (f)$$\mathrm{ker}(f)$ could not be a ring homomorphism, hence it is not a subring.

Rng

Definition. A Rng is just a ring (not necessarily containing $1$$1$). The rng homomorphism $f$$f$ should preserve $+$$+$ and $\times$$\times$.

Proposition. In the category of Rng, the $0$$0$ ring is the zero object, hence $\ker (f)$$\mathrm{ker}(f)$ is a rng as well.

Proof. Obviously. $◻$$\square$

The Free-Forget Adjoint Between Rng and Ring

Let $R$$R$ be a Rng, and consider $R\oplus \mathbb{Z}$$R \oplus \mathbb{Z}$ with the multiplication defined as $(r+n)(s+m)=rs+rm+ns+nm$$(r+n)(s+m) = rs + rm + ns + nm$. Then it becomes a unital ring with identity $1\in \mathbb{Z}$$1 \in \mathbb{Z}$.

For a Rng homomorphism $f:R\to S$$f: R \to S$, we could extend $f$$f$ to $f\oplus {\mathrm{id}}_{\mathbb{Z}}:R\oplus \mathbb{Z}\to S\oplus \mathbb{Z}$$f \oplus \mathrm{id}_{\mathbb{Z}}: R \oplus \mathbb{Z} \to S \oplus \mathbb{Z}$.

$$\begin{array}{}\text{(3)}& f(r+n)=f(r)+n\end{array}$$

This is the left adjoint of the forgetful functor $F:\mathrm{Ring}\to \mathrm{Rng}$$F: \mathrm{Ring} \to \mathrm{Rng}$.

$$\begin{array}{}\text{(4)}& {\mathrm{Hom}}_{\mathrm{Ring}}(R\oplus \mathbb{Z},S)\cong {\mathrm{Hom}}_{\mathrm{Rng}}(R,F(S))\end{array}$$