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1.3. ALGEBRAIC DEFINITIONS 23
which is also given by Maple’s simplify applied to (1.3) (simplify applied to
(1.4) leaves it unchanged). Mathematica’s FullSimplify gives
2 (sin(3x) − sin(x)) (cos(x)) 5 . (1.6)
However, an even simpler form is found by Eureqa 10 :
(cos(x)) 4 sin(4x). (1.7)
We note that (1.4) and (1.5) are the results of algorithmic procedures, pushing
identities in one direction or the other, applied to (1.3), while (1.6) and (1.7)
are half-way positions, which happen to be shorter than the algorithmic results.
1.3 Algebraic Definitions
In this section we give some classic definitions from algebra, which we will return
to throughout this book. Other concepts are defined as they occur, but these
ones are assumed.
Definition 6 A set R is said to be a ring if it has two binary operations + and
∗, a unary operation − and a distinguished element 0, such that, for all a, b
and c in R:
1. a + (b + c) = (a + b) + c (associativity of +);
2. a ∗ (b ∗ c) = (a ∗ b) ∗ c (associativity of ∗);
3. a + b = b + a (commutativity of +);
4. a + (−a) = 0;
5. a + 0 = a;
6. a ∗ (b + c) = (a ∗ b) + (a ∗ c) (distributivity of ∗ over +);
7. a ∗ b = b ∗ a (commutativity of ∗).
Not every text includes the last clause, and they would call a ‘commutative ring’
what we have called simply a ‘ring’. In the absence of the last clause, we will
refer to a ‘non-commutative ring’.
Definition 7 If R is a (possibly non-commutative) ring and ∅ ̸= I ⊆ R, then
we say that I is a (left-)ideal of R, written I ⊳ R, if the following two conditions
are satisfied 11 :
10 http://creativemachines.cornell.edu/eureqa. However, we should note that Maple
(resp. Mathematica) has proved that (1.5) (resp. (1.6)) is equivalent to (1.2), whereas Eureqa
merely claims that (1.7) fits the same data points as [a finite sample of] (1.2). Nevertheless,
Eureqa’s capability to find a short equivalent in essentially Carette’s sense [Car04] is impressive,
and in any case, knowing what to prove, any algebra system can prove equivalence.
11 We write f − g ∈ I since then 0 = f − f ∈ I, and then f + g = f − (0 − g) ∈ I.