Re: Algebra with root and Z_2[x].

On Fri, 23 Nov 2007, mina_world wrote:
"William Elliot" <marsh@xxxxxxxxxxxxxxxxxx> wrote in message
On Fri, 23 Nov 2007, mina_world wrote:

Prove that every polynomial of degree 1, 2, or 4 in Z_2[x]
has a root in Z_2[x] / <x^4 + x + 1>.

If P(x) has degree 1, then P(x) = x + a, which has root.
If P(x) has degree 2, then P(x) = x^2 + ax + b.
Case x^2 + ax + 1
subcase x^2 + 1 = (x + 1)^2, has root.
subcase x^2 + x + 1, has root x^2 + x + 1

x^4 + x^2 + 1 + x^2 + x + 1 + 1

If P(x) has degree 4, then P(x) = x^4 + ax^3 + bx^2 + cx + d.
Case x^4 + ax^3 + bx^2 + cx + 1
subcase a = c = 0. Use previous results and
(x^2 + bx + 1)^2 = x^4 + bx^2 + 1

????

(x^2 + rx + 1)(x^2 + sx + 1)
= x^4 + rx^3 + x^2 + sx^3 + rsx^2 + sx + x^2 + sx + 1
= x^4 + (r + s)x^3 + rs.x^2 + (r + s)x + 1

Thus when a = c, use the results for degree 2.
Cases left:
x^4 + x^3 + bx^2 + 1
x^4 + bx^2 + x + 1
which upon examination, leaves
x^4 + x^3 + 1
x^4 + x + 1

a = x^2 + 1; a^2 = x^4 + 1 = x; a^3 = x^3 + x; a^4 = x^2
a^4 + a + 1 = 0

If f(x) = x^4 + x^2 + 1.
a = x^2 + x + 1 + <x^4 + x + 1> is a root of f(x).
Because, f(a) = x^8 + x^2 + 1 + <x^4 + x + 1>.

Since (x^4 + x + 1)^2 = x^8 + x^2 + 1,
f(a) = 0 + <x^4 + x + 1>.

so, "a" is a root of f(x) in Z_2[x] / <x^4 + x + 1>.


If f(x) = x^4 + x^3 + 1...
If f(x) = x^4 + x^3 + x^2 + x + 1...
complex...



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