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.
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.
Post a Comment