A functor from Mon to Cats, and degree of fields extension as a functor.

Since we often need to consider one-object categories associated with monoids, groups, and rings, let us explore this idea further.

Let $M$$M$ be a monoid. We define $B(M)$$B(M)$ to be the one-object category associated with $M$$M$, given by:

$$\mathrm{Ob}(B(M))=\ast ,{\mathrm{Mor}}_{B(M)}(\ast ,\ast )=M$$

with composition given by the monoid operation in $M$$M$.

Indeed, this defines a functor

$$B:\mathbf{Mon}\to \mathbf{Cat}$$

from the category of monoids to the category of (small) categories. For a monoid homomorphism $f:M\to M^{\prime }$$f : M \to M'$, we get a functor

$$B(f):B(M)\to B(M^{\prime })$$

which acts as the identity on the unique object and maps each morphism $m\in M$$m \in M$ to $f(m)\in M^{\prime }$$f(m) \in M'$.

---

Application: The Degree of Field Extensions as a Functor

Let us now consider an interesting application of this construction.

Define a monoid:

$$N:=({\mathbb{N}}^{+}\cup \{\mathrm{\infty }\},\cdot ,1)$$

where multiplication is extended by defining $a\cdot \mathrm{\infty }=\mathrm{\infty }$$a \cdot \infty = \infty$ for all $a\in N$$a \in N$. Intuitively, this models the behavior that an infinite extension dominates any finite degree. (Note: you might view $N$$N$ as isomorphic to $(\mathbb{N},\cdot ,1)$$(\mathbb{N}, \cdot, 1)$, since $\mathrm{\infty }$$\infty$ acts as an absorbing element under multiplication. What is the difference between $\mathrm{\infty }$$\infty$ and $0$$0$ ?)

We now claim that field extension degree defines a functor:

$$\mathrm{deg}:\mathbf{Fields}\to B(N)$$

• On objects: all fields are sent to the unique object $\ast$$*$.

• On morphisms: a field extension $F\hookrightarrow L$$F \hookrightarrow L$ is sent to the element $[L:F]\in N$$[L : F] \in N$.

This assignment is functorial because field degrees satisfy the multiplicativity property:

$$[L:F]=[L:K]\cdot [K:F]$$

for any tower of fields $F\subseteq K\subseteq L$$F \subseteq K \subseteq L$.

Hence, this defines a functor from $\mathbf{Fields}$$\mathbf{Fields}$ to $B(N)$$B(N)$.

Proposition. $[L:F]=1⟺L\cong F$$[L:F]=1\iff L\cong F$.

Proof.

Since $1$$1$ is the only isomorphism in $B(N)$$B(N)$, we have $L\cong F⟹[L:F]=1$$L\cong F\implies [L:F]=1$.

Proposition.
$[L:F]=1⟺L\cong F$$[L : F] = 1 \iff L \cong F$ (as field extensions).

Proof.
Since $1$$1$ is the identity morphism in the monoid $N$$N$, and hence the only isomorphism in $B(N)$$B(N)$, we have:

$$L\cong F⟹[L:F]=1$$

Conversely, suppose $[L:F]=1$$[L : F] = 1$. Then $L$$L$ is a one-dimensional vector space over $F$$F$. That means any element $\ell \in L$$\ell \in L$ can be written uniquely as $\ell =a\cdot 1_L$$\ell = a \cdot 1_L$ for some $a\in F$$a \in F$. In particular, $L=F\cdot 1_L$$L = F \cdot 1_L$, the isomorphism of field is given by $f(a)=a\cdot 1_L$$f(a)=a\cdot 1_L$.$◻$$\quad \square$

---

Definition. Let $\mathcal{C}$$\mathcal C$ be a category. A morphism $f:X\to Y$$f:X\to Y$ is called irreducible if $f=gh$$f=gh$ then $g$$g$ is invertible or $h$$h$ is invertible.

Example. Let $R$$R$ be an integral domain, then $f$$f$ is irreducible iff $B(f)$$B(f)$ is irreducible in $B(R)$$B(R)$.

Proposition. $\mathrm{deg}(f)$$\mathrm{deg}(f)$ is irreducible implies $f$$f$ is irreducible.

Proof. Suppose $f$$f$ is reducible, i.e. $f=gh$$f=gh$, then $\mathrm{deg}(f)=\mathrm{deg}(g)\mathrm{deg}(h)$$\deg(f)=\deg(g)\deg (h)$ and $\mathrm{deg}(g)\ne 1\ne \mathrm{deg}(h)$$\deg(g)\ne 1\ne\deg(h)$. Hence $\mathrm{deg}(f)$$\deg(f)$ is reducible. Contradiction. $◻$$\square$

$B(N)$$B(N)$ has some interesting property.

Easy to see that all the morphism is mono since $ab=ac⟹b=c$$ab=ac\implies b=c$. Also, since it is commutative monoid, all the morphism are epi as well. Hence we have if $a\circ b=b\circ c$$a\circ b=b\circ c$, then $c=a$$c=a$. Let us generalize it.

Let $M$$M$ be a commutative monoid with $ab=ax⟹x=b$$ab=ax\implies x=b$, for example, integral domain. Then if we consider $B(M)$$B(M)$, then we $a\circ b=b\circ c⟹a=c$$a\circ b=b\circ c\implies a=c$.

i.e. Proposition.

The quadrilaterals in a one-object category with $ab=ba$$ab=ba$ and where all morphisms are mono/epi must be parallelograms once you have fixed a pair of opposite sides $(a,a)$$(a,a)$.

hence, for example, if you want to compute

$$[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}(i)]$$

You could apply the deg functor and the consider following diagram:

By above proposition, we have $[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}(i)]=4$$[\mathbb Q(\sqrt[4]{2},i):\mathbb Q(i)]=4$

Here is another application, we could use $B(G),B(R)$$B(G),B(R)$ to define $G$$G$-Set and $R$$R$-Mod.

https://marco-yuze-zheng.blogspot.com/2025/03/category-of-g-set-and-category-of-r.html