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March 14th, 2014, 05:40 AM  #1 
Member Joined: Sep 2013 Posts: 91 Thanks: 2  Units of this ring: Z?[x]/(x² + 1)?Is my solution correct?
Z? [x] / (x²+1) = {f* : f?Z?[x] } f* is the congruence class of f modulo( x²+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( x²+1). This means fg=1+a*(x²+1) where a is either 0 or 1. If a=0, fg=1 so f,g=1.If a=1, fg=x²+2=(?2ix)*(?2+ix). So our units are: u?=1 , u?=?2ix , 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]/(x² + 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 zerodivisor. Using symbols like "?2" and "i" is counterproductive, 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]/(x² + 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]/(x² + 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 0polynomial). 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 0polynomial 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 0divisor 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]/(x² + 1)?Is my solution correc
so the elements of Z?[x]/(x² + 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]/(x² + 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]/(x² + 1)?Is my solution correc
ok thanks for helping 

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