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March 14th, 2014, 05:40 AM   #1
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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 .
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March 14th, 2014, 11:50 AM   #2
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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).
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March 15th, 2014, 04:56 AM   #3
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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?
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March 15th, 2014, 08:00 AM   #4
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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}.
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March 15th, 2014, 10:00 AM   #5
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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?
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March 16th, 2014, 01:00 PM   #6
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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.
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March 17th, 2014, 10:14 AM   #7
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Re: Units of this ring: Z?[x]/(x + 1)?Is my solution correc

ok thanks for helping
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