Ring Laws

This is a really basic derivation of elementary facts about rings, just for the record. I follow Milne's definition of a ring (R, +, 0, -, ⋅ 1) as a commutative group (R, +, 0, -) equipped with an additional operation ⋅ and a distinguished 1 ∈ R such that (R, ⋅, 1) is a monoid, and ⋅ distributes over + both on the left and on the right: for all a, b, c ∈ R

(a + b)⋅c = a⋅c + b⋅c
a⋅(b + c) = a⋅b + a⋅c

Lemma for any a, b ∈ R, a⋅0 = 0 = 0⋅b

   a⋅0 = 0
⇐ { Cancellation on the left }
   a⋅b + a⋅0 = a⋅b + 0
≡ { Distributivity on the right }
   a⋅(b + 0) = a⋅b + 0
≡ { Group neutral, twice }
   a⋅b = a⋅b
≡
   true

The other direction is analogous:

   0⋅b = 0
⇐ { Cancellation on the right }
   0⋅b + a⋅b = 0 + a⋅b
≡ { Distributivity on the right }
   (0 + a)⋅b = 0 + a⋅b
≡ { Group neutral, twice }
   a⋅b = a⋅b
≡
   true

Theorem for any a, b ∈ R, (-a)⋅b = -(a⋅b) = a⋅(-b)

   (-a)⋅b = -(a⋅b)
⇐ { Cancellation on the right }
   (-a)⋅b + a⋅b = -(a⋅b) + a⋅b
≡ { Distributivity on the right, group inverse }
   (-a + a)⋅b = 0
≡ { Group inverse }
   0⋅b = 0
≡ { Lemma }
   true

The other direction is analogous:

   a⋅(-b) = -(a⋅b)
⇐ { Cancellation on the left }
   a⋅b + a⋅(-b) = a⋅b + -(a⋅b)
≡ { Distributivity on the left, group inverse }
   a⋅(b + -b) = 0
≡ { Group inverse }
   a⋅0 = 0
≡ { Lemma }
   true

Also, the following is immediate:

Theorem If 1 = 0, then R = {0}

Proof: note that, for arbitrary a, b ∈ R:

   a
= { Definition }
   a⋅1
= { Hypothesis }
   a⋅0
= { Lemma }
   0
= { Lemma }
   0⋅b
= { Hypothesis }
   1⋅b
= { Definition }
   b

which shows that there is at most one element in R.