 My Math Forum Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correct?
 User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

 March 14th, 2014, 05:40 AM #1 Member   Joined: Sep 2013 Posts: 91 Thanks: 2 Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correct? Z? [x] / (x˛+1) = {f* : f?Z?[x] } f* is the congruence class of f modulo( x˛+1) (at least i hope i know it right) We are searching those f?Z?[x] polynomials for which exists g?Z?[x] that fg?1 mod( x˛+1). This means fg=1+a*(x˛+1) where a is either 0 or 1. If a=0, fg=1 so f,g=1.If a=1, fg=x˛+2=(?2-ix)*(?2+ix). So our units are: u?=1 , u?=?2-ix , u?=?2+ix . March 14th, 2014, 11:50 AM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correc fg = 1 (mod x^2 + 1) means: fg = 1 + k(x)(x^2+1), where k(x) can be any polynomial in Z2[x]. It is useful to note that Z2[x]/(x^2 + 1) only has 4 elements: 0 + (x^2+1) 1 + (x^2+1) x + (x^2+1) x+1 + (x^2+1). Show x + (x^2+1) is a unit, and that x+1 + (x^2+1) is a zero-divisor. Using symbols like "?2" and "i" is counter-productive, Z2 is NOT a subfield of the complex numbers, and "2" has little meaning (except if you use the convention that 2 = 0 (mod 2)). Unlike in the normal real field, in Z2, -a = a, so the square root of -1 is the same as the square root of 1. There is only one square root of 1, 1 itself, since in Z2[x]: x^2 - 1 = 0 is the same as: (x - 1)(x - 1) = (x - 1)^2 = 0 because (x - 1)^2 = x^2 + x + x + 1 = x^2 + 1 (since x + x = 1x + 1x = (1 + 1)x = 0x = 0, because 1 + 1 = 0 in Z2). March 15th, 2014, 04:56 AM #3 Member   Joined: Sep 2013 Posts: 91 Thanks: 2 Re: Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correc why only 4 elements?isn't it 8?there is also 0,1,x,x+1 another question:how can we work with x^2 in Z2,when we have only 0 and 1? March 15th, 2014, 08:00 AM #4 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correc When you form a quotient ring, the elements are NOT the elements of the "original ring" but rather EQUIVALENCE CLASSES of these elements modulo the ideal. Here, the ideal is I = (x^2 + 1) ={ k(x) in Z2[x] such that k(x) = h(x)(x^2 + 1), for some polynomial h(x) in Z2[x]}. We say that f(x) = g(x) (mod I) if f(x) - g(x) is in I, that is, in this case, if f(x) - g(x) = h(x)(x^2 + 1) (that is the difference of f(x) and g(x) has x^2 + 1 as a factor). Z2 is a field (albiet a small one), so the division algorithm holds for Z2[x]. Thus given ANY polynomial f(x) in Z2[x], we can write: f(x) = q(x)(x^2 + 1) + r(x), with deg(r) < 2, or r = 0 (the 0-polynomial). Thus we have: [f(x)] = f(x) + I = [r(x)] = r(x) + I, since f(x) - r(x) = q(x)(x^2 + 1) is in I. So it suffices to find polynomials of degree less than 2 (that is, of degree 0 (constant polynomials), and degree 1 (linear polynomials)) and the 0-polynomial as representatives of these cosets. Since we have just TWO possible choices for coefficients, we have a total of 2^2 = 4 representative cosets in all. "0" , "1", "x" and "x+1" are NOT elements of Z2[x]/(x^2 + 1}, they are elements of Z2[x]. However, their equivalence classes (cosets) are in Z2[x]/(x^2 + 1). x is not an element of Z2, it is an INDETERMINATE (this is just a fancy word for "formal symbol" which has no necessary intrinsic meaning). x^2 is the square of the (linear) polynomial x. Since x^2 = x^2 + 0 = x^2 + (1 + 1) = (x^2 + 1) + 1, it follows that: x^2 + I = 1 + I. Often, one writes u = x + I, for the coset of x in Z2[x], and identifies a in Z2 with its corresponding coset a + I. Then Z2[x]/(x^2+1) is isomorphic to the ring: {a + bu: a,b in Z2} where addition is given by: (a + bu) + (a' + b'u) = (a+a') + (b+b')u and multiplication is given by: (a + bu)(a' + b'u) = (aa' + bb') + (ab' + a'b)u This ring is small enough to give these sums and products explicitly: 0+0 = 0 0+1 = 1 0+u = u 0+(1+u) = 1+u 1+1 = 0 1+u = 1+u 1+(1+u) = u u+u = 0 u+(1+u) = 1 (1+u)+(1+u) = 0 and the other sums by commutativity of addition 0*0 = 0 0*1 = 0 0*u = 0 0*(1+u) = 0 1*1 = 1 1*u = u 1*(1+u) = 1+u u*u = 1 u*(1+u) = 1+u (1+u)*(1+u) = 0 and the other products likewise by commutativity of multiplication. This is NOT a field, we have the 0-divisor 1+u, which therefore has no multiplicative inverse. The units of this ring are {1,u}. March 15th, 2014, 10:00 AM #5 Member   Joined: Sep 2013 Posts: 91 Thanks: 2 Re: Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correc so the elements of Z?[x]/(x˛ + 1) is x^2+1 , x^2 , x^2+x+1 , x^2+x ? and how can the unit be 1 when it's not the element of the ring? March 16th, 2014, 01:00 PM #6 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correc When we write: f(x) + I = f(x) + (x^2 + 1) we're not just taking the polynomial f(x) + x^2 + 1, we are taking f(x) + any MULTIPLE of x^2 + 1. This is a VERY large set. The multiplicative identity of this quotient ring is NOT "1", it's 1 + (x^2 +1), which is the COSET of 1, and includes: 1 = 1+0 = 1+0(x^2 + 1) x^2 = x^2 + 1 + 1 = 1 + x^2 + 1 x^3 + x + 1 = 1 + x(x^2 + 1) x^3 + x^2 + x = 1 + (x + 1)(x^2 + 1) and infinitely many others. Although each COSET is infinite, we only have finitely many different cosets: the coset 0 is in (which is the ideal generated by x^2 + 1 itself), the coset that 1 is in, the coset that x is in, and the coset that x+1 is in. Every other element of Z2[x] falls into one of these 4 cosets. March 17th, 2014, 10:14 AM #7 Member   Joined: Sep 2013 Posts: 91 Thanks: 2 Re: Units of this ring: Z?[x]/(x˛ + 1)?Is my solution correc ok thanks for helping  Tags 1is, correct, ring, solution, units, zx or x˛ Search tags for this page

,

### how find i (xÂ˛ 1) generated ideal in z [x]

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rain Calculus 1 November 19th, 2013 08:22 AM rain Calculus 4 October 12th, 2013 01:08 PM weekStudent Linear Algebra 1 January 10th, 2013 12:10 PM SarahT Calculus 1 March 6th, 2012 11:29 AM davedave Calculus 1 January 31st, 2012 02:48 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      