Divisibility in Integral Domains Recall: A commutative ring with unity D is called an integral domain if D contains no zero divisors. That is, if a, b D and ab =0,wehaveeithera =0or b =0. 2 Recall: An element x of a ring R is called a unit if there is an element y in R such that xy = yx =1,where1isunityinR. Some examples of integral domains: 1. All fields, such as Q, R, C,andZp where p is prime 2. The ring of integers Z 3. The ring of polynomials D[x] if D is an integral domain 4. The Gaussian integers, a + bi : a, b Z { 2 } Definition Let D be an integral domain. If a, b D,wesaythata is a divisor of b,orthatb is a multiple of a,orthata b,ifthere is an element x in D such that ax =2b. | Example In Z, 3 is a divisor of 18, but 4 is not, since there is no integer x such that 4x = 18. NOTE THE DIFFERENCE FROM OUR USUAL DEFINITION: Example In R, 3 and 4 are both divisors of 18, since there are real numbers x and y such that 3x = 18 and 4y = 18. Definition Let D be an integral domain. If a, b D and a = ub,whereu is a unit in D,thena and b are called associates. Equivalently, a and b are associates if both2 a b and b a. | | Example In Z,1and-1areassociates. Definition A nonzero element a in a ring R is called an irreducible if a is not a unit and if whenever a = bc,theneitherb or c is a unit in R.Equivalently,ifa is an irreducible if whenever a = bc,theneithera b or a c. | | Definition A nonzero element a in a ring R is called prime if a is not a unit and if a bc implies either a b or a c. | | | Theorem 10 Every prime element x of an integral domain D is an irreducible. Proof Suppose x is prime in D, and suppose x = bc for some b, c D.Sincex = bc,thenclearlyx bc, so by the definition of prime, either x b or x c.Hencex is irreducible. 2 | | | Note that we still need the distinct definitions, because primes are not necessarily irreducible if we’re not in an integral domain. Example 3isirreducibleinZ, since it is not a unit and if 3=bc with b, c Z,theneitherb = 1 and c = 3 or b = 3 and c = 1;thatis,either3 b or 3 c.It’salsoprime:if3 bc,theneither32b or 3 c. ± ⌥ ± ⌥ | | | | | Math 280 Fall 2015 Part 3: Rings There are lots of interesting examples to consider here, but we’ll leave some for next semester. Just one last thing for Math 280: Definition Suppose D is an integral domain. If 1. we can write every nonzero, non-unit element x D as a product of irreducibles, and 2. this factorization into irreducibles is unique, up to2 associates and order then we say D is a unique factorization domain, often denoted UFD. Example Our most convenient example here is the integers, and it’s the one that will be the most useful to any future teachers. We know we can factor numbers as a product of primes: 24 = 2 2 2 3 · · · and that there’s only one way to do so, except for rearranging the terms (we wouldn’t consider 2 2 3 2 to be a new factorization). The “up to associates” part means that we also don’t consider 2 2 2 3 to be a· new· factorization.· ·− ·− · Example Any field F would also be a UFD, as would F [x]. This means the polynomials in x with real coefficients form a UFD. So the entire semester just brought us up to Algebra I: polynomials can be factored. This may seem trivial (and annoying) but there are some important points here, particularly if you’ll be teaching math someday: 1. You’ve probably encountered polynomials that “can not be factored,” like x2 +3x +1. If you wanted to find the zeros 3 p13 of such a polynomial, you’d have used the quadratic formula, and found that the zeros are − ±2 . But the polynomial can be factored, just not with integers: 3+p13 3 p13 x2 +3x +1= x + x + − 2 ! 2 ! So what we mean when we say it can’t be factored is that it can’t be factored in Z[x]. 2. Suppose you’re teaching your algebra class to factor, and you ask for the factorization of 2x2 +6x +4.Onestudent suggests (2x + 4)(x + 1), while another suggests (x + 2)(2x + 2), and a third suggest 2(x + 2)(x + 1). You’ve got a good class: they are all correct, though only the third has factored completely. Just know that there is only one way to factor a polynomial, so if you found two, they are equivalent (up to associates and order). 3. You’ll also have students suggest that x(x + 3) + 2 is a perfectly valid representation of x2 +3x +2, and they’re right. It is equivalent. However, it is not factored, which really does mean writing a number or a polynomial (or any other ring element) as a product of irreducibles in the ring. The student’s representation is not factored, as it is not a product of irreducibles. Math 280 Fall 2015 Part 3: Rings.