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⋅ca⋅(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**.

## 2 comments:

I am curious to know how you typeset these proofs. Did you format them by hand?

@Frank: yes, I write the HTML by hand on a text editor. If you inspect the stylesheet you'll see a section marked "/* Modifications */" with the classes I use to format equations, code, etc. The crucial bit is to have the displayed block be preformatted but use the proportional font for the rest of the page.

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