Rng

A rng rʊŋ is a generalized ring which may lack a multiplicative identity. That is, a rng $(𝑅, +, \cdot)$ consists of an Abelian group $(𝑅, +)$ called addition and a Semigroup $(𝑅, \cdot)$ called multiplication, with the extra conditions #m/def/ring

left-distributivity $𝑎 \cdot (𝑏 + 𝑐) = (𝑎 \cdot 𝑏) + 𝑎 \cdot 𝑐)$
right-distributivity $(𝑏 + 𝑐) \cdot 𝑎 = (𝑏 \cdot 𝑎) + (𝑐 \cdot 𝑎)$

These are precisely the semigroup objects in $𝖠 𝖻$ .

Examples

An example of a rng that is not a ring is the even integers

2 ℤ = {2 𝑘 ∣ 𝑘 \in ℤ}

with the ordinary operations of integer addition and multiplication.

Properties

Let $𝑎, 𝑏 \in 𝑅$ and $𝑛, 𝑚 \in ℕ$

$𝑎 0 = 0 𝑎 = 0$
$𝑎 (- 𝑏) = (- 𝑎) 𝑏 = - (𝑎 𝑏)$
$(- 𝑎) (- 𝑏) = 𝑎 𝑏$
$𝑎 (𝑏 - 𝑐) = 𝑎 𝑏 - 𝑎 𝑐$ and $(𝑏 - 𝑐) 𝑎 = 𝑏 𝑎 - 𝑐 𝑎$
$(𝑛 𝑎) (𝑚 𝑏) = (𝑛 𝑛) (𝑎 𝑏)$

Proof of 1–5

Clearly $0 + 𝑎 0 = 𝑎 0 = 𝑎 (0 + 0) = 𝑎 0 + 𝑎 0$ so $𝑎 0 = 0$ , and likewise for $0 𝑎$ , proving absorption. Similarly $𝑎 (- 𝑏) = 𝑎 (- 𝑏) + 𝑎 𝑏 - 𝑎 𝑏 = 𝑎 (𝑏 - 𝑏) - 𝑎 𝑏 = 𝑎 0 - 𝑎 𝑏 = - 𝑎 𝑏$ and likewise for $(- 𝑎) 𝑏$ , proving ^P2. It follows that $(- 𝑎) (- 𝑏) = - (𝑎 (- 𝑏)) = - (- (𝑎 𝑏)) = 𝑎 𝑏$ , proving ^P3. Note $𝑎 (𝑏 - 𝑐) = 𝑎 (𝑏 + (- 𝑐)) = 𝑎 𝑏 + 𝑎 (- 𝑐) = 𝑎 𝑏 - 𝑎 𝑐$ , and likewise for right-distributivity, proving ^P4.

Now

(𝑛 \sum 𝑎) (𝑚 \sum 𝑏) = 𝑛 \sum (𝑎 𝑚 \sum 𝑏) = 𝑛 \sum 𝑚 \sum (𝑎 𝑏) = 𝑛 𝑚 \sum (𝑎 𝑏)

proving ^P5.

#state/tidy | #lang/en | #SemBr