By Peter J. Cameron
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Extra info for Notes on Algebraic Structures [Lecture notes]
Example text
We use a − b as shorthand for a + (−b). In the next proof we need to know that −(−b) = b. This holds for the following reason. We have, by (A3), b + (−b) = (−b) + b = 0, so that b is an additive inverse of −b. Also, of course, −(−b) is an additive inverse of −b. By the uniqueness of additive inverse, −(−b) = b, as required. In particular, a − (−b) = a + (−(−b)) = a + b. 10 (Second Subring Test) Let R be a ring, and let S be a non-empty subset of R. Then S is a subring of R if the following condition holds: for all a, b ∈ S, we have a − b, ab ∈ S.
When are two fractions equal? • How do we add and multiply fractions? Thus, we start with the set X consisting of all ordered pairs (a, b), with a, b ∈ R and b = 0. ) The ordered pair (a, b) will “represent” the fraction a/b. So at this point we have to answer the first question above: when does a/b = c/d? Multiplying up by bd, we see that this holds if and only if ad = bc. Thus, we define a relation ∼ on X by the rule (a, b) ∼ (c, d) if and only if ad = bc. We have to show that this is an equivalence relation.
Suppose that r = 0. Now b = ac for some element c; so a = acq + r. Thus, r = a(1 − cq), and since r = 0 we have d(r) ≥ d(a), contrary to assumption. So r = 0. Then b | a; since we are given that a | b, it follows that a and b are associates. 29 (a) A Euclidean domain is a principal ideal domain. (b) A Euclidean domain is a unique factorisation domain. Proof (a) Let R be a Euclidean domain, and let I be an ideal in R. If I = {0}, then certainly I = 0 and I is principal. So suppose that I is not {0}.
