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

[kobulingam]

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

Homework Statement

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.

Homework Equations

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

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...

 

[Hurkyl]

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.

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>.​