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Evelyn Haley 10/15/07 Question: Find the GCD of 1155 and 468 Various people used different methods with some using the Euclidean Algorithm. Formally the Euclidean Algorithm is Let m,n be integers and m>0. Then there exist integers q,r with 0≤r<m such that n=qm + r The integers q,r are uniquely determined by these conditions.1 How does the Euclidean algorithm relate to our discussion of making analogies between the set of integers and polynomials? We will see this equation later when we discuss finding irreducible integers and polynomials. It is important to further discuss rings and fields. Let us look more closely at the similarities between the ring of integers and the ring of polynomials over ℚ Definition of Ring (as per class discussion) A ring has 2 operations which are denoted + (operation that is commutative) * (operation that may or may not be commutative) under + (abelian group) closure associativity unique identity each element has unique inverse commutative under * closure associativity distributive laws over addition a(b+c) = a*b +a*c and (b+c)a = b*a + b*c **Formal Definition** A ring R is a set whose objects can be added and multiplied, (i.e. we are given associations (x,y) ↦ x+y and (x,y)↦xy from pairs of elements of ℝ , into ℝ satisfying the following conditions: RI 1. Under addition ℝ is an additive abelian group RI 2. For all x, y, z є ℝ, we have x(y+z) = xy +xz and (y+z)x = yx + yz RI3. For all x, y, z є ℝ, we have associativity (xy)z=x(yz) RI4. There exists an element eєℝ such that ex=xe=x for all xєℝ 1 It is this last RI4 that leads to our discussion of why these two rings are special. Not all mathematicians include this statement in their definition. Why? Because this is special to the ring of integers and the ring of polynomials. The ring of integers has a multiplicative inverse = 1 and the ring of polynomials has the polynomial f(x)=1 as the multiplicative inverse. To supplement this, while neither of these Rings are a field, they both have invertible elements called Units Unit Integers = ±1 Polynomial Unit = any non zero constant is invertible We now will look at the definition of a field. A field is a set with all the properties stated above for rings plus commutativity but in a field we also want to have a multiplicative inverse and multiplicative identity for non zero elements. More formally stated: **A commutative ring such that the subset of nonzero elements forms a group under multiplication is called a field. In a field we have 1≠0 and a field has no divisors of zero.** 1 Let us note here that it is sometimes written that in a field there exists a multiplicative identity, 1≠ 0 that is not the same as the additive identity. (but it is more common not to be a part of the definition.) What would our rings look like if we allowed 1=0. We would not have a unique identity, unique additive inverse or multiplicative inverse. It could also be shown that our field would have one element. So in summation if all non-zero elements of a ring are units (i.e. they are all invertible) and multiplication is also commutative, then have a field. Now we will look at examples of Rings we have seen - The ℚ form ring that is also a field - The ℝ and ℂ are rings that are also fields - The polynomials with coefficients that lie in a ring or field, form a ring. - Matrices Let’s further discuss matrices. To start we said that we need addition to be commutative 1 2 C 3 1 K 1 2 K 1 2 0 4 = K 1 3 C K 1 3 1 2 3 1 2 4 = 0 4 2 4 This demonstrates the commutative property of matrix addition. 0 0 Next, we need an additive identity; The 0 matrix is the additive identity . 0 0 1 0 The matrix is the multiplicative identity 0 1 However, not all matrices are invertible. What would we need to have matrices be invertible? The defining condition for matrices to be invertible would be that the units in n x n matrices would have non-zero determinants. So to summarize, we have associativity in addition, commutativity in addition, an additive identity, and a multiplicative identity. What we do not have is multiplicative commutativity and there is no multiplicative inverse for each element of the ring. Since we need this for a field, the matrices are a ring but not a field. Let's look at more examples of rings ℤ/mℤ, specifically lets look at ℤ/3ℤ or ℤ/4ℤ These are rings; each has an additive identity, multiplicative identity, and each is commutative. Do they have invertible elements?? No because not all non zero elements are invertible Let’s look at the integers Mod 3 0 1 2 0 0 0 0 1 0 1 2 2 0 2 1 So 0,1 have multiplicative identity. The integers mod 3 also form a field because all the non-zero elements are invertible Let’s look at the integers Mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 1 and 3 are units, but 2 is not a unit. The integers mod 4 are not a field because not all non zero elements are invertible Then why are the integers mod 4 unique? One idea was that it is unique because we can find elements that when multiply together the product is zero when neither of the elements are zero. Actually this is a very special point, because when we are teaching math in high school, we say that when we multiply two elements together and get zero, then one of them must be zero. It is what enables us to solve for example, x2 - 4 = 0 (x-2)(x+2) = 0 The fact that ℝ, (or ℤ or ℚ) has no zero divisors, tells us that either x=2 or x=-2. An important definition to bring up is the following: If a ring has no 'zero divisors' then a ring is called an Integral Domain So in the integers mod 4, 2 is a zero divisor in the integers mod 6, 2 and 3 are zero divisors We bring this up because we just stated that ℤ/4ℤ has a zero divisor, and that is special, so when a ring does not have zero divisors we need to have a different name for it.(Integral Domain) How are we relating all this discussion on rings and polynomials to the Euclidean Algorithm we did in the beginning of class. To start to connect, we will first look at primes in integers and polynomials. Definitiona prime is a non unit such that whenever p=ab, either a or b is a unit 7=1*7, 7=-1*-7 Professor explained that since 1 is not prime in the integers, then if we are using the above definition, we need to also exclude units in P(R) from being considered irreducible. In other words, the polynomial 5 is not considered irreducible because ∃ number 1/5 such that 5*1/5 =1.Also, -8 is not considered irreducible because again, ∃ number 1/8 such that -8*1/8= -1 Therefore we need to define what exactly we mean when we use the term "irreducible polynomial" An irreducible polynomial f(x) is a non-constant polynomial (non-zero, non-unit) such that whenever you can factor f(x) = g(x) h(x) then either g(x) or h(x) is a unit If you can pull out only a constant then f(x) is not reducible. We also said it was not reducible because the degree did not decrease. Now we can look at the GCD. Greatest Common Divisor Composite integers factor uniquely into products of prime numbers and polynomials factor uniquely into products of irreducible polynomials Example : x 2- 4 = (x-2)(x+2) which means x-2 divides x 2 -4 35 = 7*5 Given a,b є ℤ, with b non-zero, we write a= bq + r (this is the Euclidean Algorithm) but again we want to get something smaller than we started with. So we need r =0 or r<b or more correctly |r| < |b| 5= 3*1 + 2 or 3= 0*5 + 3 or 3= 5 * 1 = -2 3=5(-1) + 8 so looking at these which one "feels " like we reduced it or broke it down in some way? Joe said "we want to have r<a and thus throw out 3= 0*5 + 3, because if you just take out the bq, the remainder should be less than a" Tom wanted to know, "Why do we have to throw any out?" Professor states that we want to have one canonical choice of q and r Tom also suggested that in the Euclidean Algorithm we need |a| ≥ bq Now, to extend this to our analogy of polynomials and integers: We discussed primes so, the notion of a unit came up and the notion of size came up. We can use the Euclidean Algorithm with integers using this idea of primes, quotients and remainders and discuss the size of these integers. Now to look at polynomials, we discussed that we want to divide 2 polynomials and somehow make sense of size. In other words how the sizes of the quotient and remainders relate to the size of the original polynomials. We are using the Euclidean Algorithm to reduce the integers and polynomials to something that is irreducible. More Formally A ring with a Euclidean valuation (measure of size) is a domain (meaning a ring with no zero divisors) with a function, such that v: R→ℤ, ℤ≥0 s.t for all a,b ≠ 0 v(a)≤ v(ab) and s.t. for all a,b≠ 0 there exists q,r there exists a = qb + r, r=0 or v(r) < v(b) So why might we use this method as a tool for finding the GCD 40= 24(10) + 16 24= 16(1) + 8 16= 8(2) + 0 when you get to zero the remainder in the previous equation is the GCD Tom had a definition of the GCD of 2 natural numbers a,b is divisible by n, if: n divides a and n divides b and if n is the largest number that divides both. We left with thinking about why this works.... 1. Lang,S Undergraduate Algebra 3rd Edition 2000.