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Assignment 1 – Parts 1, 2, & 3 – Math 413/612 (All-1) Cube Roots of Unity

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Assignment 1 – Parts 1, 2, & 3 – Math 413/612 (All-1) Cube Roots of Unity Assignment 1 { Parts 1, 2, & 3 { Math 413/612 (All-1) Cube roots of unity. We saw in class that the geometric argument we used to show p p p that Z[ −1] and Z[ −2] are Euclideans domains doesn't work for Z[ −3] (and we p will see soon enough that Z[ −3] is not a Euclidean domain). However, there's a ring p related to Z[ −3] for which the argument works. Let p 1 −3 ! := e2πi=3 = − + : 2 2 This is something we call a primitive cube root of unity: cube root because !3 = 1 (and primitive because !n 6= 1 for n = 1; 2). Its complex conjugate is p 1 −3 ! = e−2πi=3 = − − : 2 2 Let R = Z[!] = fa + b! : a; b 2 Zg and for a; b 2 R, let f(a + b!) = ja + b!j2, i.e. f(a + b!) = (a + b!)(a + b!): (a) Show that R is a subring of C. (b) Show that ! = !2 = !−1. (c) For a; b 2 R, show that f(a + b!) = a2 − ab + b2. (d) Plot some of lattice points Z[!] in C. p (e) In analogy with Z[i] and Z[ −2], give a geometric proof that f is a Euclidean function for R. p (All-2) The Golden Ratio. Similarly, we can reuse the argument we used for Z[ 2] in some cases. Let p 1 + 5 ' := ; 2 p 1− 5 the Golden Ratio! Its \conjugate" is ' = 2 . Let R = Z['] = fa + b' : a; b 2 Zg and for a; b 2 R, let f(a + b') = j(a + b')(a + b')j: (a) Show that R is a subring of R. (b) For a; b 2 R, show that f(a + b') = ja2 + ab − b2j. (c) Show that f is a Euclidean function for R. 1 (All-3) Discrete valuations. (a) Let p be a prime number. For a 2 Z define vp(a) to be the power of p in the prime factorization of a. For a=b 2 Q, define vp(a=b) := vp(a) − vp(b): Show that vp is a valuation on Q. (b) Let K be a field and let F := K(x) = fP (x)=Q(x): P; Q 2 K[x]g be the field of rational functions over K. Define v1(P=Q) := deg(Q) − deg(P ): Show that v1 is a valuation on F . X n (All-4) Let K be a field and let f = anx 2 K((x)) be a non-zero formal Laurent series. n2Z Show that f is invertible in K((x)). (All-5) Irreducible elements in Euclidean domains. Suppose R is a Euclidean domain with Euclidean function f. In class, we showed that if d is a proper divisor of a, then f(d) < f(a). (a) Show that if R is not a field, then f must take on at least two different values. (b) Show that if u is a unit and a 2 R n 0, then f(ua) = f(a). (c) In Question (612-1) below, it is shown that if u 2 Rn0 is such that f(u) is minimal (i.e. f(u) ≤ f(a) for all a 2 R), then u is a unit. Show that the set of units is exactly those nonzero elements of minimal f value. (d) Show that if p 2 R has second lowest f value, then f is irreducible (i.e. show that if p 2 R n 0 is such that f(p) ≤ f(a) for all nonzero a 2 R n R×, then p is irreducible). (e) Conclude that if R is a Euclidean domain that is not a field, then there is at least one irreducible in R. (This completes the argument given in class that Euclidean domains are Factorization domains.) 2 413 only: (413-1) Fractions and integral domains. The rational numbers consist of fractions whose nu- merator and denominator are integers. Two different fractions may be equal, like 1=2 = 2=4. So, the rational numbers are really certain equivalence classes of pairs of integers. This constructions can be generalized to arbitrary integral domain (and latter we'll generalize it to arbitrary commutative rings) and that is what this exercise is about. Let R be an integral domain. A multiplicative subset of R is S ⊆ R such that (i) a; b 2 S implies ab 2 S, (ii)0 62 S. Let F := f(r; s): r 2 R; s 2 Sg and define a relation ∼ on F by (r; s) ∼ (r0; s0) if rs0 = r0s: (a) (WI) Show that ∼ is an equivalence relation. r (b) (WI) For r 2 R and s 2 S, let s be the equivalence class of (r; s) 2 F. Denote the set of these equivalence classes by S−1R. Define r r r s + r s 1 + 2 := 1 2 2 1 s1 s2 s1s2 and r r r r 1 · 2 := 1 2 : s1 s2 s1s2 Show that these two operations are well-defined. (c) One can show that + and · turn S−1R into a commutative ring (but that's not s a very interesting proof so you can skip it). Show that, for any s 2 S, s is the identity in S−1R. −1 rs (d) Define ι : R ! S R by '(r) := s (where s is any element of S). Show that ι is a well-defined (independent of s) injective ring homomorphism. In this way, we can think of R as a subring of S−1R. 3 (e) Show that for all s 2 S, ι(s) is a unit in S−1R. Thus, this construction is a way of making the elements of S units. Accordingly, S−1R is called the ring of fractions of R with respect to S. (f) Show that R n f0g is a multiplicative subset of R. When S = R n f0g, S−1R is called the field of fractions of R and denoted Frac(R). Thus, any integral domain R can be viewed as a subring of the field Frac(R), which is the analogue of the passage from Z to Q. 612 only: (612-1) More on Euclidean functions and domains. In class, we said a Euclidean function on an integral domain R is a function f : R n 0 ! Z≥0 having the following two properties for α; β 2 R: (i) if β 6= 0, then there exists q; r 2 R such that α = βq +r and r = 0 or f(r) < f(β), (ii) if α; β 6= 0, then f(α) ≤ f(αβ). (a) Suppose R is an integral domain with a function fe : R n 0 ! Z≥0 satisfying (i). Define f : R n 0 ! Z≥0 by f(α) := min(ff(αβ): β 2 R n 0g) = min(ff(γ): γ 2 (α) n 0g): Show that f satisfies both (i) and (ii), i.e. that f is a Euclidean function for R. In other words, for defining the notion of a Euclidean domain, property (ii) is unnecessary. Let's call a function satisfying (i) a weakly Euclidean function on R. (b) Suppose f is a weakly Euclidean function for R. Let u 2 R n 0 be such that f(u) ≤ f(α) for all α 2 R n 0. Show that u is a unit in R. (c) Suppose that fi : R n 0 ! Z≥0 is a family of weakly Euclidean functions. For α 2 R; α 6= 0, define f(α) := inf fi(α): i Show that f is a weakly Euclidean function. (Remark: this shows that on a Euclidean domain there is a minimal weakly Euclidean function. For Z, its value on an non-zero integer n is one less than number of bits in the binary expansion of n.) (612-2) Let R be an integral domain and let a; b 2 R. In class, we mentioned that if g is a gcd of a and b, then another element g0 is a gcd of a and b if and only if g and g0 are 4 associates. We will write g = gcd(a; b) to mean that g and its associates are the gcds of a and b. Suppose R is a GCD domain. (a) Let a; b; c 2 R. Show that gcd(ab; ac) = a gcd(b; c). (b) If gcd(a; b) = 1 and gcd(a; c) = 1, then gcd(a; bc) = 1. (c) Show that every irreducible is prime in R. 5.
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