MA 3362 Lecture 23 - Integral Domains

MA 3362 Lecture 23 - Integral Domains Monday, November 17, 2008. Objectives: Define integral domains and order. Integral Domains We’ve talked around divisors of zero as we’ve played with solving quadratic equations. What we’ve seen is that in a commutative ring, the things we do to solve a quadratic equation by factoring work fine. The only problem is that if a · b = 0, when both a and b are not zero, then extra solutions can creep in. This complicates things immensely. It makes sense, therefore, to have a special category of ring that avoids this problem. Definition 1. Let a be an element of a ring R. If there is another element b ∈ R such that b =6 0 and (1) ab =0, then a is said to be a divisor of zero. Now, 0 is always a divisor of zero. The problems we’ve mentioned arise, when something other than zero is also a divisor of zero. Definition 2. If a commutative ring with unity has no non-zero zero divisors, then it is called an integral domain. In an integral domain, therefore, the factoring technique for solving a quadratic equation works fine, and the solutions from the two factors are the only two solutions. We’re going to go in a different direction with integral domains in the little bit of time we have left. Ordering in an integral domain In high school algebra, we have a natural ordering for our numbers. For example, we know what (2) 3 < 5 means. All of the real numbers are comparable this way. Given any two numbers x, y ∈ R, we can say that exactly one of the following is true (3) x < y (4) x = y (5) or x>y. This is called the trichotomy law. We can’t easily say that for R2 or C. Also keep in mind that we’re in algebra. As a ring, R is a set with two binary operations defined on it. A ring doesn’t have a distance defined on it, for example. Is < an algebraic concept? That is, can we define it in terms of the operations? Note that as far as multiplication on Q+ is concerned, the primes are interchangeable, and so order is not necessarily preserved, even if an operation is. Here’s one way of defining order in a ring (or in our case, an integral domain). 1 MA 3362 Lecture 23 - Integral Domains 2 0.1. Positives and negatives. In Z, for example, we can rewrite the trichotomy law as follows. (6) x < y ⇔ x − y < 0 (difference is negative) (7) x = y ⇔ x − y = 0 (difference is zero) (8) or x > y ⇔ x − y > 0 (difference is positive). This doesn’t seem like much, but at least we’ve done stuff that sounds like negative and positive before, the additive inverses. We can now define what an order is in an integral domain. Definition 3. Let D be an integral domain. D will be called an ordered integral domain, if the there is a subset D+ ⊂ D with the following properties. (1) If a,b ∈ D+, then a + b ∈ D+. (2) If a,b ∈ D+, then ab ∈ D+. (3) For each element a ∈ D, exactly one of the following is true: a ∈ D+, a =0, or −a ∈ D+. In other words, D is ordered, if it can be partitioned into positives, negatives, and zero, and the positives are closed. As above, we will say that x > y, if x − y ∈ D+, etc. We will also use notation like x ≥ y to mean x > y or x = y. Quiz 23 1. Suppose R is not an integral domain. Then there are a,b ∈ R, a =6 0, b =6 0, and ab = 0. Could there be an R+ satisfying the properties above? 2. We can reverse the ordering on R, and that’s pretty much the same, right? Are the positives and negatives interchangeable? 3. In an ordered integral domain D, is D+ a subring? 4. Z3 is an integral domain. Can it be an ordered integral domain? Well-ordered integral domains Our normal rings, Z, Q, and R, are nicely ordered (they’re all ordered integral domains). Consider Q+. Is there a smallest element of Q+? That is, is there some element L ∈ Q+ such that L ≤ q for every q ∈ Q+? Definition 4. Let D be an ordered integral domain, and let S ⊂ D. If there is an element L ∈ S such that L ≤ s for every s ∈ S, then L is said to be a least element of S. Definition 5. Let D be an ordered integral domain. If every subset of D+ has a least element, then D is said to be well-ordered. Let’s play with this. Suppose D is a well-ordered integral domain. Therefore, there is a set of positives D+, and D+ has a least element, which we’ll call u. We know that 2u = u + u ∈ D+, since D+ must be closed. Similarly, u + u + u =3u, 4u, etc. are all elements of D+ . Suppose there is a v such that u<v< 2u. This would mean that v − u ∈ D+ and 2u − v ∈ D+. I claim that 2u − v<u, since (9) u − (2u − v) = u − 2u + v = −u + v = v − u ∈ D+. But u is the least element of D+, and so v cannot exist. Similarly, if nu<v< (n + 1)u, then v − nu ∈ D+, (n + 1)u − v ∈ D+, (10) u − ((n + 1)u − v) = u − nu − u + v = v − nu ∈ D+, MA 3362 Lecture 23 - Integral Domains 3 and (n + 1)u − v is an element in D+ smaller than the least element of D+. Can there be anything in D+ other than u, 2u, 3u, etc.? Let E ⊂ D+ be the set of all elements of D+ that don’t take the form u,2u, etc. The set E must have a least element, which we’ll call v. We know that v>u, so v − u ∈ D+. Also v − (v − u) = u ∈ D+, so v − u < v. Therefore, v − u is a positive element less than v, so it must be of the form nu. But then v = (n + 1)u. Contradiction. See what’s going on? What we have here is the following theorem. Theorem 1. All well-ordered integral domains are isomorphic to Z..

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