2009-01-05

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, cR

(a + b)⋅c = ac + bc
a⋅(b + c) = ab + ac

Lemma for any a, bR, a⋅0 = 0 = 0⋅b

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

The other direction is analogous:

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

Theorem for any a, bR, (-a)⋅b = -(ab) = a⋅(-b)

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

The other direction is analogous:

   a⋅(-b) = -(ab)
⇐ { Cancellation on the left }
   ab + a⋅(-b) = ab + -(ab)
≡ { 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, bR:

   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:

Frank Atanassow said...

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

Matías Giovannini said...

@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.