# Z_2 /<u^4+u+1> isomorphism Z_2 /<u^4+u^3+u^2+u+1>

1. Nov 5, 2008

### kobulingam

Z_2/<u^4+u+1> isomorphism Z_2/<u^4+u^3+u^2+u+1>

1. The problem statement, all variables and given/known data

How to figure an isomorphism from
Z_2/<u^4 + u +1> to Z_2/<u^4 + u^3 + u^2 + u + 1>

What I can now show (after a page and a half of work) is that the two polynomials generating the ideals are irreducible over Z_2.

2. Relevant equations

I've been able to prove that the elements creating the ideas are both irreducible polynomials.

3. The attempt at a solution

I can show proof that the ideals are irreducible, but I don't think we need to reuse that part in remaining solution. Essentially u^4 + u +1 has no linear factors by factor theorem (neither 0 nor 1 root), so only possibility is that it could be factored into 2 irreducible quadratics, and there is only once such quadratic in Z_2. Squared this quadratic and didn't get u^4 + u +1. Thus u^4 + u +1 is irreducible.

Similarly, u^4 + u^3 + u^2 + u + 1 has no linear factors (neither 0 nor 1 is a root), so only possibility is that it's the product of an irreducible quadratic and irreducible cubic. There are only 2 possible such cubics. Multiplying each of these cubics with the irreducible quadratic does not give u^4 + u^3 + u^2 + u + 1. Thus u^4 + u^3 + u^2 + u + 1 irreducible over Z_2. I am guessing this is the easy part of the answer, yet this itself stretched me fully...

2. Nov 5, 2008

### Hurkyl

Staff Emeritus
Re: Z_2/<u^4+u+1> isomorphism Z_2/<u^4+u^3+u^2+u+1>

Er, you mean two irreducible quadratics, don't you?

Anyways, don't you know any theorems (or can comptue one) that tell you something about all possible homomorphisms $R[x] / \langle f(x) \rangle \to S$, where R and S are rings?

(If you need a hint, first consider $R[x] \to S$)

Incidentally, a useful syntactic tip is to use different indeterminate variables in your two different rings. I.E. write them as
Z_2/<u^4 + u +1>

and
Z_2[v]/<v^4 + v^3 + v^2 + v + 1>.​