Monoid ring

Let $𝑅$ be a ring and $𝑀$ be a monoid¹. The monoid ring $𝑅 [𝑀]$ is the extension ring of $𝑅$ by adjoining $𝑀$ in the most general way maintaining the monoid product as ring multiplication, #m/def/ring as formalized by the Universal property. Thus it is an $K$ -monoid constructed from the free module $𝑅 (𝑀)$ .

Universal property

Let $𝑅$ be a ring and $𝑀$ be a monoid. The associated monoid ring is a triple consisting of a ring $𝑅 [𝑀]$ , a ring homomorphism $𝜄 : 𝑅 \to 𝑅 [𝑀]$ , and a monoid homomorphism $𝜇 : 𝑀 \to 𝑅 [𝑀]$ ; such that given any ring $𝑇$ , ring homomorphism $𝑖 : 𝑅 \to 𝑇$ , and monoid homomorphism $𝑚 : 𝑀 \to 𝑇$ there exists a unique ring homomorphism $𝑓 : 𝑅 [𝑀] \to 𝑇$ such that the following commutes in $𝖲 𝖾 𝗍$

This admits a unique extension to a bifunctor $(-) [-] : 𝖱 𝗂 𝗇 𝗀 \times 𝖬 𝗈 𝗇 \to 𝖱 𝗂 𝗇 𝗀$ such that

𝜄 : Π 1 \Rightarrow (-) [-] : 𝖱 𝗂 𝗇 𝗀 \times 𝖬 𝗈 𝗇 \to 𝖱 𝗂 𝗇 𝗀 𝜇 : Π 2 \Rightarrow (-) [-] : 𝖱 𝗂 𝗇 𝗀 \times 𝖬 𝗈 𝗇 \to 𝖬 𝗈 𝗇

become natural transformations.

Construction as maps

As with the free module, $𝑅 [𝑀]$ may be constructed as the set of maps of finite support $𝑀 \to 𝑅$ , where we identify $𝑚 \in 𝑀$ with $𝜇 (𝑚) = 𝛿 𝑚 : 𝑡 \mapsto [𝑚 = 𝑡]$ invoking an Iverson bracket, and elements of $𝑅$ with constant functions. For $𝑎, 𝑏 \in 𝑅 [𝑀]$ , the product is given by

[𝑎 * 𝑏] (𝑥) = \sum 𝑚 𝑛 = 𝑥 𝑎 (𝑚) 𝑏 (𝑛)

Proof of universal property

Clearly $𝑅 [𝑀]$ is an abelian group under pointwise addition. The convolution operation is associative, since

[𝑎 * (𝑏 * 𝑐)] (𝑥) = \sum 𝑚 𝑛 = 𝑥 𝑎 (𝑚) [𝑏 * 𝑐] (𝑛) = \sum 𝑚 𝑛 = 𝑥 \sum 𝑘 ℓ = 𝑛 𝑎 (𝑚) 𝑏 (𝑘) 𝑐 (ℓ) = \sum 𝛼 𝛽 𝛾 = 𝑥 𝑎 (𝛼) 𝑏 (𝛽) 𝑐 (𝛾) = \sum 𝑚 𝑛 = 𝑥 \sum 𝑘 ℓ = 𝑚 𝑎 (𝑘) 𝑏 (ℓ) 𝑐 (𝑛) = \sum 𝑚 𝑛 = 𝑥 [𝑎 * 𝑏] (𝑚) 𝑐 (𝑛) = [(𝑎 * 𝑏) * 𝑐] (𝑥)

and a multiplicative identity is given by $𝜄 (1) = 1$ . Clearly $𝜄$ is a Ring monomorphism, and $𝜇$ is a monoid monomorphism since

[𝛿 𝑚 * 𝛿 𝑛] (𝑥) = \sum 𝑘 ℓ = 𝑥 𝛿 𝑚 (𝑘) 𝛿 𝑛 (ℓ) = 𝛿 𝑚 𝑛 (𝑥)

Now suppose $𝑇, 𝑖, 𝑚$ is another such triple. For the diagram to commute, we require that $𝑓 (𝑟) = 𝑖 (𝑟)$ for all $𝑟 \in 𝑅$ and that $𝑓 (𝛿 𝑛) = 𝑚 (𝑛)$ for all $𝑚 \in 𝑀$ . For $𝑓$ to be a ring homomorphism, it follows

𝑓 (𝑟 𝛿 𝑛) = 𝑖 (𝑟) 𝑚 (𝑛)

and thus for $𝑎 \in 𝑅 [𝑀]$

𝑓 (𝑎) = \sum 𝑛 \in 𝑀 𝑖 (𝑎 (𝑛)) 𝑚 (𝑛)

which is unique, as required.

Monoid ring

Universal property

Construction as maps

See also

Footnotes