I want to show that in $\mathbb{Z}_2[x]$, the ideal $\langle x^2+x+1\rangle$ is maximal ideal.

I have listed out the table and I can see that the quotient ring is isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2$, but I can't find a ring homomorphism $f:\mathbb{Z}_2[x]\rightarrow \mathbb{Z}_2\oplus \mathbb{Z}_2$ such that the kernel of this homomorphism is exactly the ideal $\langle x^2+x+1\rangle$.

Also, can anyone give me any general suggestion of finding such homomorphisms?

  • $\begingroup$ Joey: is this modified question exactly the question in your mind? $\endgroup$ – Groups Sep 30 '15 at 5:44
  • $\begingroup$ yes this is exactly the question im asking $\endgroup$ – Mark Sep 30 '15 at 5:45
  • 3
    $\begingroup$ something is misunderstanding here: since the ideal is maximal, quotient would be a field of order $4$; but $\mathbb{Z}_2\oplus \mathbb{Z}_2$ is not a field with usual point-wise multiplication. $\endgroup$ – Groups Sep 30 '15 at 5:48
  • $\begingroup$ Something is wrong here. Since $(1)^2+1+1=1$ and $0^2+0+1=1$, this polynomial is irreduicible. Thus the ideal is prime, and so the quotient is a domain, but since finite domains are fields, so is the quotient, but $\mathbb{Z}/2^2$ is not, so something about your isomorphism is wrong. $\endgroup$ – Pax Kivimae Sep 30 '15 at 6:07
  • $\begingroup$ So can we get a homomorphism from ℤ2[x] to some ring such that the kernel is exactly ⟨x2+x+1⟩, I try to think about such a homomorphism, but I can't $\endgroup$ – Mark Sep 30 '15 at 14:42

This ideal is maximal since any element $ax+b+ < x^2+x+1>$ has an inverse. To find the inverse see that $(ax+b)(cx+d)=(acx^2+(ad+bc)x+bd=(ad+bc+1)x+(bd+1)$ modulo $<x^2+x+1>$. So if $a,b$ both are not zero we can solve for $c,d$ from the equations: $ad+bc+1=0, bd+1=1$.

You got the quotient ring wrong: for example $x^2= x+1$ modulo $x^2+x+1$ but all the elements in $\mathbb{Z}_2\times \mathbb{Z}_2$ are idempotents.


It is a maximal ideal because $x^2+x+1$, a quadratic polynomial with no root in $\mathbf Z_2$, is irreducible in $\mathbf Z_2[x]$ and in PIDs irreducible elements generate maximal ideals.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.