Re: Isomorphism between Z_2[x]/<x^3+x+1> and Z_2[x]/<x^3+x^2+1>

On Sat, 10 Nov 2007 21:59:18 EST, Meg Weiss
<megweiss@xxxxxxxxx>
wrote:

On Sun, 11 Nov 2007 00:21:32 GMT,
snipthis.aigret@xxxxxxxxxxxxx (Leon
Aigret) wrote:

On Sat, 10 Nov 2007 05:44:25 EST, Meg Weiss
<megweiss@xxxxxxxxx>
wrote:

Hello,

I have a question about an isomorphism between
Z_2[x]/<x^3+x+1> and Z_2[x]/<x^3+x^2+1>.

Define f(x) = x^3 + x + 1, and g(x) = x^3 + x^2
+
1.
If we let f(a) = 0, then we have an extension
field

Z_2(a) which consists of 8 elements

{0, 1, a, a^2, a^3, a^4, a^5, a^6}

with identities such that
1 = 1
a = a
a^2 = a^2
a^3 = a + 1
a^4 = a^2 + a
a^5 = a^2 + a + 1
a^6 = a^2 + 1

and
f(1) = 1
f(a) = 0
f(a^2) = 0
f(a^3) = a^4
f(a^4) = 0
f(a^5) = a^2
f(a^6) = a

Similarly, if we let g(b) = 0, we have an
extension

field Z_2(b) which consists of 8 elements

{0, 1, b, b^2, b^3, b^4, b^5, b^6}

with identities such that
1 = 1
b = b
b^2 = b^2
b^3 = b^2 + 1
b^4 = b^2 + b + 1
b^5 = b + 1
b^6 = b^2 + b

and
g(1) = 1
g(b) = 0
g(b^2) = 0
g(b^3) = b^5
g(b^4) = 0
g(b^5) = b^6
g(b^6) = b^3

Do next homomorphisms s_i:Z_2(a) -> Z_2(b) give
isomorphism?
s_1(a) = b,
s_2(a) = b^2,
s_3(a) = b^4.

(s_i maps a root to a root.)

Should s_i not map roots of f to roots of f
instead
of roots of g?

Indeed.

And it provided a natural hint:

To find an isomorphism, you need to map a to an
element of Z_2(b)
which "acts like a". In particular, since f(a) =
0,
any homomorphism
from Z_2(a) to Z_2(b) must map a to an element a'
of
Z_2(b) such that
f(a')=0.

quasi

Could you show me such isomorphism by my example?


Is next argument right?
By identities
1 = 1
a = a
a^2 = a^2
a^3 = a + 1
a^4 = a^2 + a
a^5 = a^2 + a + 1
a^6 = a^2 + 1

and
f(1) = 1
f(a) = 0
f(a^2) = 0
f(a^3) = a^4
f(a^4) = 0
f(a^5) = a^2
f(a^6) = a

compared with
1 = 1
b = b
b^2 = b^2
b^3 = b^2 + 1
b^4 = b^2 + b + 1
b^5 = b + 1
b^6 = b^2 + b

can we say

s_1(a) = b (since f(a) = 0)
s_2(a) = b^2 (since f(a^2) = 0)
s_3(a) = b^6
(since f(a^4) = f(a^2 + a) =0 and b^6 = b^2 + b )

is an isomorphism we are looking for?

Are these 3 different maps?

An isomorphism is a single function.

There is no homomorphism from Z_2(a) to Z_2(b) which
maps a to b.

Since f(a) = 0 but f(b) is not zero (check this --
calculate f(b))

An isomorphism from Z_2(a) to Z_2(b) must map a to
some element of
Z_2(b) such that f of that element is 0.

You can always try the elements of Z_2(b) one at a
time.

quasi

I figued it out, thanks for the help!
.

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